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• Jennilyn Nicolas

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PREFACE

TABLE OF CONTENTS This book is designed to give more emphasis in solving problems in Power Plant Engineering subjects due to the increase number of problem solving on recent board examinations. The main purpose is to show the way by presenting subject matter and material that have appeared in many board examinations, together with worked-out solutions that are acceptable to board examiners. By following a rather simple pattern of self-discipline and working out many problems can help you pass the jist time. The contents of each topic are also well designed for familiarization of expected questions that will come out in the actual board examinations. The author can guarantee a good passing grade in Power Plant Engineering subjects if mastery of topics including the principles behind in this reviewer were done.

Description

Thermodynamics Fuels & Combustion Variable Load Problems Steam cycles Boilers Steam engine Steam Turbine Geothermal Plant Diesel Plant Gas Turbine Plant Hydro-electric plant Chimney Machine Foundation Heat Transfer Air-compressor Pumps Fans & Blowers Fluid Mechanics Past Board Examination Elements Refrigeration Air-conditioning Conversion of Units-

Page

1 72

95 97 123

135 142

149 158 188 199 219 226 237

253 276 304

314 331 390

422 453

,,"0_

Thermodynamics

THERl\10DYNAMICS THERMOD YNAI~lICS PROPERTIES

Thermodynamics - 1 (Math-ME Ed Oct. 1998) What is the pressure 8,000 ft (2000 m) below the surface of the ocean? Neglect the compressibility factor, in Sl units. Ao 21.4 Mpa C. 21.0 Mpa B. 20 I Mpa D. 22.3 Mpa SOLUTION:

Sea water

p = '>'1 h U sing typical SG of sea water equal to 1.03 p = (1.03 x 98 J )(2000) p = 20,208 Kpa P c 20.21 Mpa

~~o~=r:2:~m

1

P=wh

Thermodynamics - 2 (Math-ME Ed Oct. 1998) What is the temperature at which water freezes using the Kelvin scale'! C. 278 A. 373 B. 273 D. 406 SOLUTION:

Freezing temperature of water is "K = °C + 273 "K = 0 + 273 "K = 273 Thermodynamics - 3 (Math-ME Ed Oct. 1998) The SI unit of temperature is: A. of B. oK 411.\', B

C. BTU

D. oR

o-c.

::'

~

Thermodynamics

Thermodynamics

.J

St lllJTlON

Thermodynamics - 4 (Math-ME Bd Oct. 1998)

iii ~~

The pressure reading of 35 psi in kpa is: A. 342.72 kpa B. 724.00

h = u + Pv 9500 u + 900(58) II ~c 4280/t-lh/lh -r-

C. 273.40 D. 42730

fi'] i:~'

;

Thermodynamics - 8 (Math-ME Bd Oct. 1997)

SOL UIION'

~

Pg Pg P abs

35 (101325 14.7) 241.25 kpag =

Pabs Pabs

Pg

+

Palm

24l.2" + 10132" 342.57 kpaa

The barometer reads 29.0 inches (737 mm) absolute pressure if a vacuum gage reads 9.5 A. 3202 kpa C B. 3304 k p a . D.

of mercury. What is the psi (66 kpa) in 51? 31.36 kpa 31.86 kpa

SOLUTION: 101.325

Patrn Thermodynamics - 5 (Math-ME Bd Apr. 1999) 1 torr is equivalent to pressure A. 1 atm C 2 mm Hg

Paun

r.,

_ C 14.7 D. 1/760 atm

An5. D

p abs PaLs

(29)(---) 29.92 98.2 kpa Pgage + Pann -66 ~ 98.2 32.2 kpa

Thermodynamics - 9 (Math-ME Bd Oct. 1997)

Thermodynamics - 6 (Math-ME Bd Apr. 1999) What is the standard temperature in the US? A. Fahrenheit B. Rankine

C Celsius D. Kelvin

Ans. A

A fluid with a vapor pressure of 0.2 Pa and a specific gravity of 12 is used in a barometer. If the fluids column height is 1 rn, what is the atmospheric pressure? A. 150.6 kpa C 144.4 Kpa B. 115.5 Kpa D. 117.7 Kpa SOLUTION Pressure = (specific weight)(height) Pressure (12 x 9.81)(1 m) Pressure c= 117.72 Kpa rr;

Thermodynamics - 7 (Math-ME Bd Apr. 1999) Given steam pressure of 900 Ib/fe, temperature of 300°F, specific volume of 5.8 fellb. If the specific enthalpy is 9500 ft-lb/lb, what is the internal energy per Ib of the system? A. 4400 C. 3600 B. 3900 D. 4280

Thermodynamics - 10 (Math-ME Bd Apr. 1997) What is the atmospheric pressure on a planet if the pressure is 100 kpa and the gage pressure is 10 kpa? A. 10 kpa C. so kpa B. 100 kpa D. 90 kpa

Thermodynamics

Thermodynamics

4

';

SOLUTION'

Thermodynamics - 13 Pabs

P atm

+

P gage

If the

100

=

P atm

PanTI

=

90 kpa

temperature inside the corresponding reading in OF? A. 700.60 B. 750.60

+ 10

furnace

is 700 oK, what

is

the

C. 860.60 D. 800.60

SOLUTION:

Thermodynamics - 11 (Math-ME Bd Apr. 1997)

OK

= "C + 32 700= "C + 273 "C ~ 427 of = 9/5 De +- 32 of = 9/5 (427) + 32 °F= 800.6

A column of water 200 cm high will give a pressure equivalent to: A. 9810 dyne/em 2 C. O. JO bar 2 B 0.1 atrn D. 19,620 N/m SOLl:TTON:

h= 200 em h~ 2 m 3 w = 9810 N/m P = wh P = (9810)(2 m) P == 19,620 Nlm 3

r

1 200cm

r:"

Thermodynamics - 14

11' the of scale is twice the °C scale, what will be the corresponding reading in each scale'? A. zz-c and 44°F e. 40 and 80°F B. 160 0e and 320°F D. 1oo-c and 200°F 0e

lher mudynamics - 12 What is the equivalent "R of 400 A n0600R B. 851.15°R

SOLUTION:

0K?

C. 670.2rR D 344.25°R

9/5 -c + 32

z-c

2°e = 9/5 0.668 + 3.326 -i- 0.486 Total volume ~ 448 ft3 Weight of cement per ft° of concrete mixture ~ 94/4.48 3 3 2098 Ibs/ft 3 (27ft /yd )

II

Total volume

289.9 cm '

Specific volume

.~ Total volume

Specific volume

Total mass 289.9

250

566.5 lbs or 567lbs Specific volume

1.1596 cm'rgr

Thermodynamics - 24 A cylindrical tank 2 m diameter, 3 m high is full of oil. If the specific gravity of oil is 0.9, what is the mass of oil in the tank? A 8482 kg C. 1800 kg B. 4500 kg D. 7000 kg

o .,

SOLUTION: Volume Volume Volume Mass .~ Mass = Mass ~ Mass =

n/4 0' h n/4 (2)2 (3) of cylinder = 9.425 m' Density x Volume w x V 3 (09 x rooo kg/m )(9 .425) 8482.3 kg of cylinder

ofc~linder

= =

_----

~_'!1_

Thermodynamics - 26 (Math-ME Bd. Oct. 1997) 100 g of water are mixed with 150 g of alcohol (w = 790 kg/m'). What is the specific gravity of the resulting mixtures, assuming the fluids mixed completely? A. 0.96 C. 0.82 B 0.63 D. 0.86

SOLUTION Total mass Total mass

__

O. J 00 + 0.1 SO 0.250 kg

0.100 o.isu Total volume = - - + ----1000 790 Total volume 2.899 x io' m 3

Thermodynamics - 25 (Math-ME Bd Apr. 1998) Density of mixture

3 100 g of water arc mixed with 150 g of alcohol (w = 790 kg/m ) . What is the specific volume of the resulting mixtures, assuming the fluids mixed completely? 3 C. 0.63 cmvkg A. 0.82 cm /kg D. 1.20 crrr'zkg B. 0.88 cm]/kg

- Density of mixture

0100

Total volume 0.250 2899 x 10-

4

Density Of mixture = 862 kg/rn: Specific Gravity = 86211000 Specific gravity = 0.862

SOLUTION: Total mass = 100 -t- 150 Total mass = 250 grams

Total mass

Thermodynamics - 27 0.150

Total volume

----+---

Total volume

1000 790 2.899 x 10.4 m 3

A spherical tank is full of water that has a total mass of 10.000 kg. If the out side diameter of the tank is 2722 mm, how th ick is the wall of the tank?

I

12

Thermodynamics

nerrnouynumus

1,)

SOLUTION' A. 50 rnrn B. 25 mm

C. 30 mm D. 35 mm

SO}"-,UTI()r~:

v

rn/w V 10,000 kgll,OOO kg/rrr' V = 10 m 3 V = 4/3 it [3 10 = 4/3 it [3 r = 1336 m r = 1336 mm 1 2722/2-1336 ~

1 =

"' t l

O i,

.

i

r,

+-

Q= Q = Q = Q ~

volume flow of water flowing A x vel n/4 D 2 x vel n/4(1)2(10)

Q

7.85jr/sec

=

~

:r; I

..-- ; -- t+

1ft

i

J

'lS

EI%1§

I

I IV

Thermodynamics - 30 A certain fluid is flowing in a 0.5 m x 0.3 m channel at the rate of 3 m/sec and has a specific volume of 0.0012 m 3/kg. Dctermine the mass of water flowing in kg/sec. A 380 kg/sec C. 375 kg/sec i3 390 kg/sec D.370kg/sec

24.49 mm

Thermodynamics - 28

a.3m

A cylindrical t~WK is filled with water at the rate of 5.000 gal/min. The height of water ill the tank after minutes is 20.42 ft. What is the diameter of the tank'? A 30 ft C. 20 ft B. 25 ft D.9m

Q

=

Q

Q Q

SOLUTION:

m

20.42

~-_Q --+

Thermodynamics - 29 Water is flowing through a 1 foot diameter pipe at the rate of W It/sec, What is the volume flow of water flowing? A. 7.50 ft3/sec C. 7.85 m 3/sec D. 0.22 mzsec B. 7.95 ft3/sec

A x vel

(05 X 0.3)(3) OA5 m ' /sec =

OA5 After J5 minutes, V c, 5000(15) ~, 75,000 gal V = 75,000x(l fe/7ASI gal) V = 10,025.39 ft3 V c. n'4 DC h 2 10,025.39 = n:/4 D (20.42) D = 25ft.

P-O

3m/s r = c _ ~

SOLUTION:

=

fiX v

rn (0.0012) 375 kg/sec

=

O.5m

J.

............ ..

IICI ".uuy"u,"'U.. . ...,

_'_~,.a

......

v

UWS OF THERMODYNAMICS Thermodynamics - 33

Thermodynamics - 31 (Math-ME Bd. Apr. 1998) One useful equation used is the change of enthalpy of compressible liquid with constant specific heat is: hs ub2 - h,u b l = c(T, ub1 - T.u b l ) + V(P.ub2 - P.ub l ) where: T. ubu = temperature at state n p. u bn = pressure at state n v = specific volume of liquid Water with enthalpy with C,ubp = 4.187 KJIkg_oK and v = 1.00 x 10 to rd the _3 power cu.mIkg has the following states: State I: T su b l = 19°C P. ub l = 1.013 x 10 to the 5th power Pa State II: T, ub2 = 30°C P.u b2 = 0.113 Mpa What is the change in enthalpy from state I to state II? A. 46.0 Kpa/kg C. 46.0 KJlkg B. 56.0 KJlkg D. 46.0 KNlkg SOLUTION:

h 2 - h. h z - hi h z - h,

The flow energy of 124 IiImin of a fluid passing a boundary to a system is 2 KJ/sec. What is the pressure at this point. A. 100 Kpa C. 1,000 Kpa B. 140.39 psi D. 871 Kpa SOLUTION: W = Pressure x Volume

2 KJ/sec x 60sec/min = P(0.124 m 3/min) P = 967.74 Kpa x 14.7/101.325

P = 140.39 psi

Thermodynamics - 34 (Math-ME Bd Apr. 1996) Cp(Tz-T 1) + v(P 2-P t ) 4.187(30 - 19) + 0.001(113 - 101.3)

46KJ/kg

Thermodynamics - 32

Steam at 1000 Ib/fr pressure and 300 0 R has a specific volume of 6.5 fe/lb and a specific enthalpy of 9800 ft-Ib/lb. Find the internal energy per pound mass of steam. A. 5400 C. 6400 D. 2500 B. 3300 SOLUTION:

What is the potential energy of a 500 kg body if it is dropped to a height of 100 m? A. 490.50 KJ C. 560.50 KJ B. 765.50 KJ D. 645.48 KJ

h=u+Pv

9800 = u + 1000(6.5) u = 3300 ft-lbfllb m

SOLUTION: Potential Potential Potential Potential

Energy Energy Energy Energy

= = = =

mxz 500 x 100 50,000 kg.m x 0.00981 KNlkg 490.50 KJ

Thermodynamics - 35 Air and fuel enter a furnace used for home heating. The air has an enthalpy of 302 KJ/kg and the fuel has an enthalpy of 43,207 KJ/kg. The gases leaving the furnace have an enthalpy of 616 KJ/kg. There

Thermodynamics

Thermodynamics

re 17 kg air/ kg fuel. The house requires 17.6 KW of heat, What is the lei consumption per day? A 85 kg C. 45 kg B 41 kg D. 68 kg SOLUTION: By mass balance: rn, + m, = m g m/mr = 17 m, = 17mr i 7mr + m, = mg

mg

18mr

C.C

gas

~ --.

air ....... rna

EY-RJ*.A~e

---

heat ...... 17.kw

fuel mf

Thermodynamics - 3 7 (ME Rd. Apr. 1995) The enthalpy of air is increased by 139.586 KJ/kg in a compressor, The ~ate of air flow is 16.42 kg/min. The power input is 48.2 KW. Which of the following values most nearly equals the heat loss from the compressor in KW? C. -9.95 A. -10.0 B. +10.2 D. +9.95 SOLUTION U.42k

W

By heat balance: rn, h, mrhr= m g h g (l7rnrJ(302) -t mt ~,

e, ,"

ec

0.318/0.5010

c..

63.5%

SOLUTION: T 2/T 1 = (v/vd- ' T 2/(300+273) = (16/1)14 T 2 = 1737.01 10 K

I __

,I

I'hermodynamics - 99 (ME Bd. Apr. 1995)

0

1

P

rocess 2 to 3 is constant pressure: V)V 2 = T)/T 2 r, = V)N 2 = 2031/1737.011 r, cc 1.169 1 r k -I e = I---{ c } r k k-l k(r -I) c

2

3

4

T,=2031°K T,=3000K P,=100okPa

1

Therrnodvna nucs - 101 (ME Bd. Apr. 1991) Determine the air-standard efficiency of an engine operating on the' diesel cvcle with clearance of 8'Yo when the suction pressure is 99.9: Kpa and the fuel is injected for 6'Yo of the stroke. Assume K = 1.4.

SOLUTION

v r

1 (1.169) J ( - 1 e=I---{ } (l6)IH 1.4(1.169-1)

I', k- I

\I, - V;

e = 65.98%

5I

"

=

k -

kr r '

i

1) (

(J06V,)

v.

008V" \I, - 008V il v 014 V)

()

Il() \"

r

Thermodynamics

52

rc

=

0.06V

v.v.

P·2· (0).

VJ

C;;

Thermodynamics

3

PURE SUBSTANCE

=

0.14V O

4

0.08V o 1.175 i + 0.08

=

,1 '

IA~

0.08 13.5 1

=

(13.5)14-1 60.02%

{

Find the enthalpy at 100 psi and 97% quality, h r = 298.55 Btu/lb; h rg = 889.119 Btu/lb. A. 1,170 Btu/lb C. 1,734 Btu/lb B. 1,161 Btu/lb D. 1,803 Btu/lb

LV

vp

$.08 Vp

e = 1e

P,=99.97

Thermodynamics - 103 (Math-ME Bd. Oct. 1999)

rk = - -

r,

53

V2

rc = rc

r

1

p

(1.75)~4 - 1 } 1.4(1.75-1)

SOLUTION: h

=

he + x h eg

Thermodynamics - 102 (Math-ME Bd Apr. 1996)

h

=

298.55 + 0.97(889.119)

A heat engine (Carnot cycle) has its intake and exhaust temperature of 157°C and 100°C, respectively. What is its efficiency? A. 12.65% C. 15.35% B. 14.75% D. r3.7'>%

h

=

1161 Btu/lb

Thermodynamics - 104 (Power-ME Bd. Oct. 1999) SOLUTION: T H = 157 + 273 T H = 430 0K T L = 100 + 273 T L = 373°K Efficiency

=

T'l~l -c

TH -TL

TH 430- 373

Efficiency

,v

=

430

Efficiency = 13.25%

'1-

6J. •

j

180 grams of saturated water of temperature 95°C undergoes evaporation process until all vapor completely vaporized. Determine the changed in volume. At 95°C, Vr = 0.00 I 0397 mvkg, V g = 1.9819m3/kg 3 C. 0.2565 rrr' A. 0.1656 m B. 0.4235 m' D. 0.3656 m 3

2

SOLUTION:

T

s Volume = Volume = Volume = Volume =

Specific Volume x mass (v g - vr) x m (1.9819 - 0.0010397)(0.18 kg) 0.3565 m 3

Thermodynamics - 105

s

Five kilograms of saturated liquid at 120 Kpa is heated until its moisture content is 5%. Find the work done for this process.

54

Thermodynamics

Thermodynamics A. 813.59 KJ/kg B. 643.23 KJ/kg SOLUTION:

C. 542.34 KJ(kg D. 753.12 KJ/kg

Thermodynamics - 107

For constant pressure process,

W

=

P(V2 -

55

VI)

From steam table: At 120 kpa VI = vr at 120 Kpa (sat. liquid) v g = 1.4284 roJ/kg VI = 0.0010473 mJ/kg V2 = vr + x Vrg x = 1- Y x = I - 0.05 x ~ 0.95 V2 0.00J0473 + 0.95(1.4284 - 0.0010473) V, = 1.357 Vi = 120(1.357 - 0.0010473) W = 162.73 KJ/kg (5 kg) W = 813.59 KJ

5

Steam at 2 Mpa and 250°C in a rigid cylinder is cooled until the quality is 50%. Find the heat rejected from the cylinder. At 2 Mpa and 250°C 3 V = 0.11144 m /kg u = 2679.6 KJ/kg At 2 Mpa, (saturated) vr = 0.0011767 rrr'zkg v g = 0.09963 mJ/kg Ur = 906.44 Urg = 1693.8 A. -432.23 KJ/kgC. -834.55 KJ/kg B -926.26 KJ/kgD. 1082.34 KJ/kg SOLUTION

Q = (U 2

Thermodynamics - 106 Twenty kilograms of water at 40°C is confined in a rigid vessel. The heat is supplied until all the water is completely vaporized. Find the heat added in KJ. C. 45,252 KJ A. 45,422 KJ D. 65,233 KJ B. 43,122 KJ

-

UI)

U 1 = 2679.6 U. = U, + x Urg U 2 = 906.44 + 0.5(1693.8) U2 1753.34 KJ/kg

Q Q

= CC

(175334 - 2679.6) -926.26 KJ/kg

Thermodynamics - 108 SOLUTION: V=c

For rigid vessel,

Q UI U2

=

v,)

(VI =

m (U 2

Ud = U, (saturated liquid) = U g (saturated vapor)

Q == m (U rg ) Q = 20 (2262.6) =

t = 40°C

-

Q = m (U, - U r)

Q

me = 20kg

45,252 KJ

1

Q

Find the entropy in KJ/kg-K at 90% moisture of a IMpa steam-water mixture? At 1 Mpa: Sg = 6.5865 Srg = 4.4478 A. 4.87 C. 2.583 B. 6.34 D. 4.36 SOLUTION:

x x

I - 0.9 0.10

Thermodynamics

')6

Thermodynamics

S ~' Sf + X Sfg Sfg = Sg - Sf 44478 = 6.5865 - Sf Sf = 2.1387

SOLUTION: For isothermal process, t)=

Thermodynamics - 109

SOLUTION: S = Sf + X Sfg 4 = 2.2525 + x (6.4953 - 2.25 IS) x = 0.412

h

hf X hfg h = 814.93 + 0.412(1972.7) h

Thermodynamics - I I I

A tank contains exactly one kilogram of water consisting of liquid and vapor in equilibrium at I Mpa. If the liquid and vapor each occupy one-half the volume of the tank, what is the enthalpy of the contents of the tank? />. 644.40 KJ/kg C. 8331i0 KJ/kg B. 774.40 KJ/kg D. 435.2lJ KJ/kg

At I Mpa

Vf= 00011273

Vfg

hIe 762.81

hfg " 20 I 53

~c

=

Let V

.--(Sat. Vapor)

Sf' 1.8607 Sfg = 4.9606 At 300 Kpa and 151.86°C S = 70888 KJ/kg A 652.34 KJ/kg B. 535.16 KJ/kg

~.

total volume of tan-, in! == Vl/Vl. 1/2 V mi.

Mixture with 80% quality at 500 Kpa is heated isothermally until its pressure is 300 Kpa. Find the heat added during the process. At 500 Kpa

mL

00011273 443.54 V

m,

v«. 1" :2 V

m.

01944 .' 572 V m

m, x

-r-;

I'; \

C. 983.44 KJ/kg

D. 765.34 KJ/kg

0.19444

=

SOLLJTlON:

1627.71 KJ/kg

Thermodynamics - 110

t2

Q = T (S2 - S) ) SI = Sf + X Sfg SI = 18607 + 0.3(49606) SI = 5.829 S2 = 7.0888 Q = (151.86 + 273)(7.0888 - 5.829) Q = 535.16 KJ/kg

= 2.1387 + 0.10(44478) S = 2.583 KJ/kg-"K

At 1.3 Mpa, mixture steam and water has an entropy of 4 KJ/kg_°K. Find the enthalpy of the mixture. At 13 Mpa: sf~22515 hf = 8 14 .93 Sg = 6.4953 hfg = 1972.7 A. 1627.71 KJikg C. 1234.45 KJ/kg B. 1533.33 KJ/kg D 1734.45 KJ/kg

5'7

+- ,n L

L::::

, rn, = 1 kg (Sat. Liquid]

r -3 -

--

m.

me

'2_

iV'I.,v,' r .1.. v=;v 2 ' -

.4-_

58

Thermodynamics 2.572V

x

c. ].672 D. 3.230

A. 156 B. 2.12

2.572V + 443.542V

SOLUTlON:

0.005765 = h, + xh rg h = 762.81 + 0.005765(2015.3) h = 774.43 KJ/kg x h

59

Thermodynamics

=

From steam tables: At 70 bar(7 Mpa) and 65°C VI = 0.001017 mJ/kg At 50 bart S'Mpa) and 700 V2 = 0.06081 rrr'zkg

0K(427°C)

Thermodynamics - III (ME Bd. Oct. 1996) ml

A vessel with a volume of 1 m contains liquid water and water vapor in equilibrium at 600 Kpa. The liquid water has a mass of 1 kg. Using stea m tables, calculate the mass of water vapor. A.3.16kg C. 1.57kg B. 0.99 kg D. 1.89 kg

m2

=

01 I

J

v) = Q2 I V2

AxV\

A(]OO)

0.00 I0 17

0.06081

---

VI

=

1.672 m/sec

SOLUTlON: From steam tables, at 600 Kpa:

~ ~i~~~ -'m

v. vg

>

=

0.001 ]01 mJ/kg 0.3157 m 'zkg

1L

m,

Volume ofliquid Volume of liquid Volume of liquid

=

= =

mL VL ](0.001101) 0.001101 m'

m

__

Thermodynamics - 114 (ME Bd. Oct. 1991) Vv

t, (Sat. Vapor)

Water substance at 70 bar and 65°C enters a boiler tube of constant inside diameter of 25 mm. The water leaves the boiler tube at 50 bar and 700 at velocity of 100 m/s. Calculate the inlet volume flow(li/sec) A. 0.821 C. 0.344 B.1.561 0.2.133 0K

6S'C

v

7Ob~

600Kpa

Volume of vapor Volume of vapor

= =

1-0.001101 J 0.998899 m

(Sat. Liquid)

Mass of vapor = 0.998899/0.3157 Mass of vapor = 3.164 kg

Thermodynamics - 113 (ME Bd. Oct. 1991) Water substance at 70 bar and 65°C enters a boiler tube of constant ins'ide diameter of 25 mm. The water leaves the boiler tube at 50 bar and 700 0 K at velocity of 100 m/s. Calculate the inlet velocity(m/sec)

SOLUTION:

V,

hom steam tables: At 70 bar(7 Mpa) and 65°C 3 VI = 0.001017 m /k g

Ef '

-

:

At 50 bar(5 Mpa) and 700 0 K ( 4 2 7 ° C ) 3 V2 = 0.06081 m /k g

mt QI

=

I

VI

m2 =

Q2

I

V2

700 0K

~ba'

V,=100m/s

Thermodynamics

Thermodynamics

60

AxV j A(lOO) --0.001017 0.06081 V I = 1.672 mJsec

Q\ = A x vet, QI = rc/4 (0.025/(1.672) Q\ = 0.8207li/sec

61

Thermodynamics - 120 (ME Bd. Apr. 1989) Steam at the rate of 600 kg/hr is produced by a steady flow system boiler from feedwater entering at 40°C. Find the rate at which heat is transformed in Kcal per hour if enthalpy of steam is 660 Kcal/kg and of the feedwater at 40 Kcal/kg. C. 345,200 A 372,000 B. 387,000 D. 312.444 SOLUTION

Thermodynamics - 115 (ME Bd. June 1990) One Ib (0.455 kg) of a mixture of steam and water at 160 psia(l 103.2 Kpa) is in rigid vessel. Heat is added to the vessel until the contents are at 560 psia (3861.2 Kpa) and 600°F (315.55°C). Determine the quantity of heat in KJ added to the water and steam in the tank. A. 1423.70 C. 1562.34 B. 1392.34 D. 1294.45

Rate at which heat is transformed

~

ms(h, - h-)

Rate at which heat is transformed

=

600(660 - 40)

372,000 Kcal/hr

Rate at which heat is transformed

Thermodynamics - 12] (ME Rd. Oct. 1988) SOLUTION: For a rigid vessel, the volume is constant: From steam tables: At 1103.2 Kpa 3/kg vr ,~ 0.0011332 m J/kg v g = 0.17704m U. = 780.65 KJ/kg U rg = 1805.8 KJ/kg At 3861.2 Kpa and 315 .55°C, 3/kg V2 = 0.06378 m U 2 = 2761.3 KJ/kg Solving for the quality of mixture: Vj = V2 vr+ XVrg = V2 0.0011332 + x(0.17704-0.00 11332) = 0.06378 x = 0.3561 = 35.61% Solving for U\: U = U r + xUfg U, = 780.65 + 0.3561(1805.8) U\ = 1423.70 KJ/kg Heat added = m(U 2 - U I) Heatadded = 0.455(2761.3 - 1423.7) Heat added = 608.6 KJ

Steam leaves an industrial boiler at 827.4 Kpa and 171.6°C.·A portion of the steam is passed through a throttling calorimeter and is exhausted to the atmosphere when the calorimeter pressure is 10].4 Kpa. How much moisture does the steam leaving the boiler contain if the temperature of the steam at the calorimeter is] ]5.6°C? A 3.78% C. 456% B. 308% D. 2.34% SOLUTION:

I

~

E At 827.4 Kpa (171.6°C):Jl

I

r

Calorimeter

~ 101.4kPa 115.6°C

h; = 727.25 KJ/kg C I 17r>C

PIV/RT 1 (I 37.80)(0.029)/(0.287)(300) 00464 kg P 2V 2/RT 2 (413.4)(0.029)/(0.287)(450) 0.0928 kg

Heat loss = Heat gain rn, C v2 (t 2 - tf) = rn, cv l (t f - t,) 0.OlJ2S(0.7I 6)(1 77 - tf) = 0.0464(0.7I6)(tf - 27) tf = 127°C T, = 127 + 273 T r = 400 L\.s = m c, In(T tiT1) L\.SI = 0.0464(0.716) In(400/300) L\.SI = 0.00956 L\. S 2 = 0.0928(0.716) 1n(400/450) L\.SI = -0.00783 L\.s = 0.00956 - 0.00783 L\.s = 0.00173 KJ/"C 0K

h n = 2779.25 - 2010.7 1.03,MPa :f. C.10MPa, 125°C h n ~ 768.55 KJ/kg For throttling process: hi = h2 h, + xh., = h2 76855 -+- x(20 10.7) ~ 2726.6 x 09738 x 97.38% y .~ 100 - 97.38 Y = 2.62%

Thermodynamics - 125 Thermodynamics - 123 (ME Rd. Apr. 1996) A vessel of 0.058 m 3 capacity is well insulated and is divided equally by a rigid conducting diaphragm. Initially both halves contain air at pressure of 137.8 Kpa and 413.4 Kpa and temperature of 27°C and 177°C respectively. What is the increase of entropy of the system in

Using steam table, find the enthalpy of steam at 250 kpa if its specific volume is 0.3598 m 3/kg A. 1625.86 KJlkg C. 1543.45 KJlkg B. 1785.34 KJlkg D. 1687.55 KJlkg

KJ/OC?

A 1.002 B. 0.5080

C. 0.00173 D. 0.1080

SOLUTION:

At 250 kpa:

Thermodynamics

t\~

h[= 535.37 Kl/kg h,~ = 2181.5 KJ!kg v,' = 0.0010672 m]/kg v g = 0.7187 m]/kg

v = v, -+- XVlp 03598 = 0.0010672 + x(0.7187 - 0.(010672) x r--. 0.49988 Solving for h:

h

=

h,

-+-

xhfg

h = 53537 + 0.49988 (2181.5) n /625.86 KJ/kg z;

Thermodynamics

65

Thermodynamics - 127 Steam enters an isothermal compressor at 400°C and 100 kpa, exit pressure is 10 Mpa, determine the change of enthalpy. A. 198 KJlkg C. 187 KJ/kg B. 178 KJ/kg D. 182 KJ/kg

The

SOLUTION:

At 100 kpa and 400 oe: h = 3278.2 KJ/kg For isothermal process, t 2 = t\ ~~ 400°C

Thermodynamics - 126 A throttling calorimeter is connected to the desuperheated steam line

supplying steam to the auxiliary feed pump on a ship. The line pressure measures 2.5 Mpa. The calorimeter pressure is 110 kpa and 150"C: Determine the entropy of the steam line. A. 6.8 KJ!kg-OK C. 6.6 KJlkg-OK 8. 7.2 KJ/kg-OK D. 7.5KJlkg-OK

At 400"C and 10 Mpa: h = 3096.5 Kl/kg

/"h ". h, - h2 ,)h= 3278.2 - 3096.5 L\h = ] 81. 70 lU/kg

SOLUTION:

At 110 kpa and 150 oe: h2 = 2775.6 Kl/kg At 2.5 Mpa: h. = 962.11 Kl/kg hfg = 1841 KJ/kg Sf = 2.5547 Sfg = 3.7028 For throttling process: (h I = h2 )

hi = h2 h, + xh., 2775.6 + 962.11 + x(1841) x = 0.985 s I - Sf + XSfg s\- 25547 + 0.985(37028) s\ 6.202 KJ/kg-"K Cc

l

Thermodynamics - 128 Stearn enters an adiabatic turbine at 300°C and 400 kpa, It exits as a saturated vapor at 30 kpa, Determine the work done. A. 476.34 KJ/kg C. 436.33 KJlkg B. 441.50 Kl/kg D. 524..34 Kl/kg SOLUTION:

At 300°C and 400 kpa: b, .= 3066.8 Kl/kg At 30 kpa and saturated vapor: h 2 =. h g = 2625.3 KJ/kg

W= hi - h2 W = 3066.8 - 2625.3 W 44/.51U/kg

n~

J _1 ). ._~~,~J.l~_ .~\-"'.__

..

300°0 400Kpa

----------toS

66

Thermodynamin'

Thermodynamics

67

Thermodynamics - 129

Thermodynamics - 131

A 0.5 m ' tank contains saturated steam at 300 kpa. Heat is transferred until pressure reaches 100 kpa. Find the final temperature.

A I kg steam-water mixture at 1.0 Mpa is contained in an inflexible tank. Heat is added until the pressure rises to 3.5 Mpa and 400°C. Determine the heat added. A. 1378.64 KJ C. 1456.78 KJ B. 1532.56KJ D.1635.45KJ

A. 934S"C B ~3::'3uC

C 99.63°C D 103.2'C

SOLl;TJO,\:

SOLUTION: At 300 kpa:

() 6058 m ' kg

\

VI

-

At 100 kpa, ~ 0 00 I 0432 m'/kg Yf J' 694 m3/kg v ~ ,. Since v I is in between Vr and vb at I ()o kpa. then the temperature is equal to the saturation temperature at I no kpa which is equal to 99.631!C.

At 3.5 Mpa and 400°: V2 = 0.08453 KJ/kg-OK U 2 = 2926.4 KJ/kg At 1 Mpa: vr = 0.0011273 m3/kg vg = 0.19444 m3/kg U, = 761.68 m 3 lk g U rg = 1822 KJ/kg for inflexible tank, VJ

=

V2 =

VI

Vr

=

V2

+

XVrg

Thermodynamics - 130

0.08453 = 0.0011273 + x(0.1944 - 0.001127) x = 0.4314

A 500 Ii tank contains a saturated mixture of steam and water at 300"C. Determine the mass of vapor if their volumes are equal. A. 1154 kg C 1345 kg

UJ = + x U rg U I = 761.68 + 0.4314(1822) U I = 1547.76 KJ/kg Q=m(U 2 - U r)

8.]034kg SOLUTION

D.1634

J,g

u,

Q = 1(2926.4 - 1547.76) Q = 1378.64 KJ

At 300°C vb =

mv

-r-

002167 m'/kg V[ ~.

Thermodynamics - 132 (Math-ME Bd. Oct. 1999)

v "

500/2 VI

---

VI 111 v

1000 025 m' 0.25/0.02167

mv

11.54 kg

Atmospheric pressure boils at 212°F. At the vacuum pressure at 24 in Hg, the temperature is 142"F. Find the boiling temperature when the pressure is increased by 45 psia. A. 342.34°F C. 479.13"F B. 526.34°F D. 263.45°F

Thermodynamics

Thermodynamics

68

HEAT CAPACITY

SOLUTION:

'"'

P2 P2 PI PI

= =

14.7 + 45 59.7 psia

=

-24(14.7/29.92) + 14.7

=

2.908

Thermodynamics - 132 (Power-ME Bd Apr. 1998)

By interpolation: t2

59.7

212

14.7

142

2.908

t,-212

59.7-14.7

142

59.7 - 2.908

t2

-

69

T, 212"F

142°F

t2 -212 = 0.7923tr 112.515 t 2 = 478.98°F

14.7+45psi

14.7psi(atmospheric)

-24"Hg

What is the temperature in degree C of 2 liters of water at 30°C after 500 calories of heat have been added to? A. 35.70 C. 38.00 D. 39.75 B. 30.25 SOLUTION:

Q Q Q

30

0

e

HEATER

t,

m c p (t 2 - t.) 500 cal Q=500cal = 0.50 Kcal 0.50 x 4.187 = (2 x I) (4. I 87)(t 2 - 30) t2 = 30.25"C =

=

Thermodynamics - 133 (ME Bd. June 1990) A mass of 0.20 kg of metal having a temperature of lOODC is plunged into 0.04 kg of water at 20 DC. The temperature of the water and metal becomes 48 DC. The latent heat of ice at ODC is 335 KJ/kg-DK, and the specific heat capacity of water is 4.19 KJ/kg-DK. Assuming no heat loss to the surroundings, determine the specific heat capacity of the metal in KJ/kg-DK. A. 0.234 C. 0.754 water Metal B. 0.564 .---"----p. OA51•r - - -• SOLUTION:

0.2kg 0 100 e

0.04 kg 20°C

Mixture

Q Heat loss by metal = heat gam by water (m c p flt)metaJ = (m c p flt)water

70

Thermodynamics 0.20( c m )( 100 - 48) c.,

=

=

Thermodynamics

0.04(4.19)(48 - 20) Heat loss by iron

0.451 KJ/kg- oK

m, cp,(t I

Thermodynamics - 134 (ME Bd. Apr. 1996) What is the total energy required heating in raising the temperature of a given amount of water when the energy applied is 1000 KWH with heat losses of 25%? A. 1000 C. 1333 B. 1500 D. 1250 Q --.

HEATER

=

=

Heat gain by water - t ) 2

rn., c pw (t 3

where: t, = equilibrium temperature after mixing 30(0.4)(220 - t) = 14.33(4.1 87)(t) - 10) 220 - t) = 5(t) - 10) t) = 45°C i\.s = Q,

T]

11000~Wh

Q - 0.25(Q) = 1000 =

t3 )

i\.s ~ m, c p, (t 3 ~tl)

SOLUTION:

Q

-

1

1333 KWH

Loss=O.25Q

Thermodynamics - 135 (Power-ME Bd Oct. 1999) A 30 kg iron was put in a container with 14.33 kg water. The water is at looe and the iron has an initial temperature of 493°K until the iron was in thermal equilibrium with water. Find the change in entropy. (c, for iron = 0.4 KJ/kg°K) A. -12.56 KJf'K C. -25.78 K!f'K B. -43.58 KJf'K D. -6.6 KJf'K Iron

SOLUTION:

t,

Water

30 kg 220°c

1,

'\:

t, t,

-r-

=

493 -273 220°C

Mixture /

D

14.33kg

20 0 e

t3

+ 273

30(0.4)(45- 220)

.6.5

= ------

i\.s

=

45 + 273 -6.6 KJlkg

71

Fuels & Combustion

Fuels & Combustion

72

SOLLTION

FUELS AND COMBUSTION

Fuel

0,

CIj!];O +

Fuels & Combustion - 1

C1JI!;u

A diesel power plant consumes 650 liters of fuel at 26°C in 24 hours with 28°API. Find the fuel rate in kg/hr. A 23.83 C. 22.85 D. 26.85 B. 24.85

~

~

~. A .

~

3.76N z

Product of Combustion ~

CO 2

-i-

7e

~

)(

o'

i

21.5(3.76)N 2

-- ---

Theoretical A/f

I

=

\02.34

")

'X" ""'" . 2.2:-:\_·~~ActuaJ ~

14CO:.; 15H 20

21.5 + 21.5(376)

r t) lv\t1.~7"7 0t'1-~'j.~

H20 +- 3.76 he

71.- :"J') Theoretical A.T

0" )( l''J f/'l I ~"-!1

SOLlJT!ON: 131.5+° API 141.5

Air

2150 2 + 21.5(376)N 7

i:"-Y\-\

141.5

+

.L

f\ M~k~ Nt'; 1'.,

:7

Sg.t 156°C

73

Y))

102.34 (1.15) 117.69 mol air/mol fuel

A/F

Actual AlF

7._--=~"-":i.2..

~~

Fuels & Combustion - 3

Sgat1560C Sg.t156"c =

SG a1 W C SG at 26"C

= =

131.5+28 0.887 0.887[ I - 0.0007(26 - 15.6)] 0.88

Density of fuel = 0.88(1 kg/Ii) Density of fuel = 0.88 kg/Ii

w

m

v

V = 650/24 V = 27.0833 li/hr

0.88 = m/27.0833 m = 23.83 kg/hr

A diesel power plant uses fuel that has a density of892.74 kg/m} at 15.67°C. Find the heating value of fuel.

SOLUTiON: Density of fuel

SG

- - - - - _.. _----

Density of water 892.74 /1000

SG SG

0.89274

"API

o.h,PI Fuels & Combustion - 2 A boiler burns fuel oil with 15% excess air. The fuel oil may be represented by C I4 H 30 • Calculate the molal air fuel ratio. A 14 C. 102.34 B 117.69 D.17.14

C. 43,000 K.J:'kg D. 43562 KJ/kg

A. '(4.911() KJ/kg B. \9,301 Btu/lb

=

1..J 1.5 ------ - 131.5 0.89274 27

Q ~. 41.130 Q =1-],130 Q

cc

r-t-

139PAPJ !39.6(27)

44,899.31 KJ/kg

Fuels & Combustion

74

.e

Flld\

7~

Combustion

Fuels & Combustion - 4

Fuels & Combustion - (,

A certain coal has the following ultimate analysis: C = 69% N2 =5% H 2 == 2.5% S = 7% Moisture = 8% Ash = 5% O 2 = 3.5% Determine the heating value of fuel used. C. 25,002.4 KJ/kg A. 26,961.45 KJ/kg 8. 45,256 KJ/kg D. 26,000 KJ/kg

A diesel engine consumed 945 liters of fuel per day at 30°C. If the fuel was purchased at 15.5°C and 30° API at P5.00/li, determine the cost of fuel to operate the engine per day. C P4888.90 A P5677.50 . O. P5000.0Q B. P4677.50 SOUTlON

SOLUTION:

SG 13 6 ' [

131.5 + 30 SG I 3 6 C = 0.87616 SG,QcC = 0.87616 [I - 00007(30 - 156)J SG w c = 0.8673 v ;O'C SG IS 6'C

o

Qh = 33,820C + 144,2l2(H --) + 9,304S 8 0.035 33,820(0.69) + 144,212(0.025 - - - ) + 9,304(0.07) 8 = 26,961.45 KJlkg

Oh = Oh

1415 =

VI' 6C

945

SG JO'C 0.87616

-------

VIS 6"C

Vt3 6 Cost Cost

Fuels & Combustion - 5 A diesel power plant uses fuel with heating value of 45,038.8 KJ/kg. What is the density of fuel at 30°C? A. 0.900 kg/Ii C. 0.850 kg/Ii B. 0.887 kg/li D. 0.878 kg/Ii SOLUTION:

Oh

= 41,130 + 139.6 °API 45,038.8 = 4],130 + 139.6(oAPI) API = 28

C

=

=

0.8673 = 935.44 li

P500/li (935 44 Ii) P4,677.20

Fuels & Combustion - 7 \. cylindrical tank 3 m long and 2 m diameter is used for oil storage. How many days can the tank supply the engine having 27° API with fuel consumption of 60 kglhr? 3m A.484 C. 7.84, B 5.84 O. 8.84 ~ 2m

SOLUTI.ON

SG I5 6, C =

]41.5 ---

131.5+0 API SG I5 6, C = 0.8872 SG.1 30, C = 0.8872[ 1 - 0.0007(30 - 15.6)] SG ot 30'C = 0.8782 Density of fuel = 0.8782(1 kg/Ii) Density of fuel = 0.8782 kglli

V

~ ;v4 0 2h

V - "1412/(3) V. 4.42478 m ' 1415 '-,( J

t,

f. (

1]1.5~·27

~,"'Okglh

I~~ ·

..

~ '..

~

/0

Fuels & Combustion

SG l 5 6

0

C

2.

77

Fuels & Combustion

0.8927

SOLUTION:

Density of fuel ~. 0.89274(1000 kg/m3) Density of fuel = 892.74 kg/nr' W = m/V 892.74 = 60IV V = 0.0672 m 3/hr

A/F = 11.5C + 34.5(H - 0/8) + 9.3S A/F = 11.5(0.7) + 34.5(0.03 - 0.04/8) + 4.3(0.06) A/F ~~ 9.1705(1.25) A/F = 11.46 kg air/kg fuel

Number of days = 9.42478/0.0672 Number of days = 140.23 hrs Number of days = 5.843 days Fuels & Combustion - 10 (ME Bd. Apr. 1991)

:it A 650 Bhp diesel engine uses fuel oil of 28° API gravity, fuel

Fuels & Combustion - 8 Determine the minimum volume of day tank in nr' of 28° API fuel having a fuel consumption of200 kg/hr. A. 10.43 m' B. 5.41 m'

l

C. 6.87 rn'

(

D. 7.56 m'

consumption is 0.65 lb/Bhp-hr. Cost of fuel is P7.95 per liter. For continuous operation, determine the minimum volume of cubical day tank in em), ambient temperature is 45°C. 3 A. 5,291,880 em C. 5,491,880 em] B. 5,391.880 crn' D. 5,591,880 crn'

SOLUTION: SOLUTION:

1415

SG 15 6°e

=

SG 15 6°e

=

141.5

131.5 + 28 0.887

SG\S6'C

0

0



=

5.4/ m3

Solving for fuel consumption' m. = 0.65(650) m. = 422.5 lb/hr m- = 191.61 kg/hr V r = 191.61/0.869 V f = 220.495 li/hr

Fuels & Combustion - 9 Given the following ultimate analysis: C = 70% O 2 = 4%

N2 = 5% ~ = 6% Moisture H 2 = 3% Ash = 5% Using 25% excess air, determine the actual air fuel ratio A. 11.46 B. 24.85

C. 23.85 D. 26.85

--

131.5-t-28 SG!56'C = 0.887 SG 45 C = SG1WC[l - 0.0007(t - 15.6)] SG 4 5 C = 0.887[1 - 0.0007(45 - 15.6)] SG 4 5, C = 0.86Q Density of fuel =2 0.869(1 kg/li) ~ Density of fuel = 0.869 kg/li ~

Density of fuel = 0.887(1000) Density offue1 = 887 kg/rn ' W = m/V 887 = 200N V = 0.22548 m 3/hr x 24 hrs/day

V

=

=

8%

Volume Volume Volume Volume

of dav of day of day of day

tank tank. tank. tank.

=

= = =

220.495 x 24 hrs 3/1000li 5,291.88 Ii x Im 3 5.291.88 m' x (100)3 e m 3 1m 5,291,880 em'

Fuels & Combustion

Fuels & Combustion

78

S( l[l

79

iT!ON:

Fuels & Combustion - 11 (ME Bd. Apr. 1991) A 650 Bbp diesel engine uses fuel oil of 28°API gravity, fuel consumption is 0.65 lblBbp-br. Cost of fuel is P7.95 per liter. For continuous operation, determine tbe cost of fuel per day at 45°C. A. P42,870.45 C. P42,570.45 B. P42,070.45 D. P42,170.45

SOLUTION: SG 15.6o C =

141j ---

131.5 + 28 SG 1W C = 0.887 SG w c = SG 1W C[1 - 0.0007(t - 15.6)] SG w c = 0.887[1 - 0.0007(45 - 15.6)] SG w c = 0.869 Density offuel = 0.869(1 kg/li) Density of fuel = 0.869 kg/li Solving for fuel consumption: m, = 0.65(650) m, = 422.5 lb/hr IDr= 191.61kglhr v, '" 191.61/\).869 V r = 220.4951i/hr Volume of day tank = 220.495 x 24 hrs Volume of day tank = 5,291.88 li Cost of fuel per day = 5,291.88 li x P7.95/li Cost of fuel per day = P42,070.45

60°F 80°F

= =

SG 15 6"C

15.6°C 26.6°C 1415 = ---

131.5 + 30 0.876 = 0876[1 - 00007(2667 - 1556)] = 0869 250gal/24hrs x 3.7851i/gal 39.431lilhr 39.431(0.869/0.876) 39.1161ilhr 2700/24 112.5 KW 39.l16xP3.00/li Cost per KW-hr = 112.5 Cost per KW-hr = Pl.043/KW-h

SG I5 6"C SG 26 6 "C SG 2 6 6 , C At 26.6°C. m, = At 26.6°C. m. = At 15.6°C, m. = At 15.6°C,mr = Load = Load =

=

Fuels & Combustion - 13 (ME Bd. Oct. 1981) A logging firm in Isabela operates a Diesel Electric Plant to supply its electric energy requirements. During a 24 hour period, the plant consumed 250 gallons of fuel at 80°F and produced 2700 KW-hrs. Industrial fuel used is 30° API and was purchased at P3.00/li at 60°F. Determine the overall thermal efficiency of the plant. A. 26.08% C 43.12% B. 34.23% D. 18.46% SOLUTION:

Fuels & Combustion - 12 (ME Bd, Oct. 1981) A logging firm in Isabela operates a Diesel Electric Plant to supply its electric energy requirements. During a 24 bour period, tbe plant consumed 250 gallons of fuel at 80°F and produced 2700 KW-hrs. Industrial fuel used is 30° API and was purchased at P3.00m at 60°F. Determine cost of fuel to produce one Kwb A. P3.043/KW-h C. P1.043/KW-h B. P4.043/KW-h D. n.043/KW-h

I

Qh Qh Qh

= = =

60°F 80°F

41,130 + 139.6 x °API 41,130 + 139.6(30) 45,3 18 KJ/kg = =

15.6°C 26.6°C

Fuels & Combustion

ISU

SG 1S6 o C

141.5 =

81

Fuels & Combustion Mass ot Iuel

--

-r-

24.933 kg/hr

Qh ~, 41,1:10 + l396CO API) Qh = 41,13 0' 139.6(2 8) Qh = 45,039 KJlkg

131.5+ 30 SG 1W C = 0.876 SG 26 6 C = 0.876[1 - 0.0007( 26.67 - 15.56)] SG 26 6 , C = 0.869 At 26.6°C, IDr = 250gaV 24hrs x 3.7851i/gal At 26.6°C, IDr = 39.431l ilhr Load == 2700/24 Load = 112.5 KW IDr = 39.431l ilhr x 0.869 kg/li x Ihr/360 0sec IDr = 0.00952 kg/sec Power' Output Overall Efficiency = ----- ---=0

Overall Efficien cy

Power Output -----mrQ h

82.5

Overall efficien cy

Overall efficien cy

(24.933 ! 3600)(4 5.039)

26.47%

IDfQ h

Overall efficiency = Overall efficiency

=

112.5 0.00952(45,318) 26.08%

Fuels & Combu stion -14 (ME Bd, Oct. 1981) A diesel electric plant in one of the remote provinces in the South utilizes diesel fuel with an °API of 28 at 15.6°C. The plant consum es 680 liters of diesel fuel at 26.6°C in 24 hrs, while the power genera ted for the same period amoun ts to 1,980 KW-hr s. Determ ine overall therma l efficiency of the plant A. 26.47% B. 12.34%

C. 23.45% D. 34.34%

SOLUT ION: SG 15 6 , C =

141.5 ---

131.5+ 28 SG I5 .6 , C = 0.887 SG 26 6, C = 0.887[ 1 - 0.0007( 26.6 - 15.6)] SG 26 6, C = 0.88 Density = 0.88 x 1 kg/Ii Density = 0.88 kglli Mass of fuel = 680 (0.88)/2 4

-----_._---

Fuels & Combu stion - 15 (ME Rd. Oct. 1991)

A circula r fuel tank 45 feet long and 5.5 feet diamet er is used for oil storage . Calcula te the numbe r of days the supply tank can hold for continu ous operati on at the followi ng conditi ons: Stea m now = 2000 Ibs!h r Steam dry and saturat ed at 200 psia Fecdw ater temper ature = 230°F Boiler Efficien cy = 75% Fuel oil = 34° API C 13.45 A. 1234 D. 2344 B 1758

SOLUT ION: 45ft

From steam tables:

At 200 psi( 1380 Mpa), h, =c 2789.6 KJ1 C:i94CO o 0022H 2 0 ~ o T

Combu stion with excess air: 1460, 0.] 24CO;o + onCO t 0022H 2 + o 584N 2 + (I +x)O (I ix)OI4 6(176) N 2 --> o 394CO , -t 0.022H 20 + o 584N 2 + 0146(3 .76)(1 +-x)N 2 + x(O 14\))0,

~

water: Express ing the percent age of oxygen in the product s excludi ng the

Variab le Load Problem s - 2 0.01 x = 0.11 1 + 0.11 Percent age Theoret ical air Percent age Theoret ical air = III % 0=

KWThe daily energy produc ed in a certain power plant is 480,000 load? e hrs, What is the daily averag A. 10 MW C. 25 MW B. 15 MW D. 20 MW SOLUT ION: Averag e Load

Energy Pr oduced

A verage Load

No. of hours 480,000 /24

A verage Load

20,000 Kw

A verage Load

20Mw

~ 'ariable

96

Load Problems

STEAM CYCLE

Variable Load Problems - 3 The annual energy produced in a 100 MW power plant is 438,000,000 KW-hrs. What is the annual capacity factor of the plant? A. 40% C. 35~'o B. 50% D. 60% SOLUTION: Annual Energy Pr oduced

Annual Capacity Factor

=

Annual Capacity Factor

=

Annual Capacity Factor

~

Plant Capacity x 8760 438,000,000 100,000x 8760 50%

Steam cycle - 1 (ME Bd Oct. 1999) In a Rankine cycle steam enters the turbine at 2.5 Mpa (enthalpies & entropies given) and condenser of 50 Kpa (properties given), what is the thermal efficiency of the cycle? At 2.5 Mpa: h g = 2803.1 KJ/kg Sg = 6.2575 At 50 kpa: Sr = 1.0910 Srg = 6.5029 h, = 340.49 h rg = 2305.4 Vr= 0.0010300 A. 25.5~~~~ C. 34.23% B. 45.23% D. 12.34% SOLUTION:

s = Sr + x Srg 6.2575 = 1.0910 + x(6.5029) x = 0.7945 h z = h, + xh rg h z = 340.49 + 0.7945(2305.4) L-·---------hz = 2172.13 h, = 340.49 KJ/kg h, = h r + vr(P z - PI) h, = 340.49 + 000 I 03(2500 - 50) h, = 342.98

A power plant has a use factor of 50% and capacity factor of 44%. How many hours did it operate during the year? A. 7700 hrs C. 7709 hrs B. 7800 hrs D. 7805 hrs

s""·

SOLUTION: Annual Energy Pr oduced Annual Capacity Factor

Plant Capacity x 8760 Annual Enerzv Pr oduced ~-

=

Plant Capacity x 8760 Energy Produced = 3854.4(Plant Capacity) Use Factor 0.50

T

h, = 2803.1 KJ/kg Solving for hz:

Variable Load Problems - 4

0.44

97

Steam Cycles

Efficiency

Efficiency =

Energy Pr oduced =

=

(hi -h z)-(h 4 -h,) -------

(h j - h 4 ) (2803.1 - 2172.11)- (342.98 - 340.49)

(2803.1 - 342.98)

-----'=-------

Plant Capacity x t Energy Pr oduced

Efficiency = 25.55%

~

Plant Capacity x t Energy Produced = 050(Plant Capacity)t 0.50(Plant Capacity)t = 3854.4(Plant Capacity) t = 7708.8 hrs.

Steam cycle - 2

, In an ideal Rankine cycle, the steam throttle condition is 4.10 Mpa and 440°C. If turbine exhaust is 0.105 Mpa, determine the pump work in KJ per kg.

98

Steam Cycles A. 6.34 B. 5.34

C. 4.17 D. 2.12

SOLUTION: Solving for h.: h, =h fatO.105Mpa h- = 423.24 KJ/kg V3 = 0.0010443 mvkg Solving for h.: Using pump work equation: h, V3(P4 - P3) + h 3. 3/kg V3 = 00010443 m h, = 0.0010443(4100 - lOS) + 423.24 h, ~ 427.412 KJ/kg W p = h4 - h3 W p = 427.412 - 423.24 W p = 4.172 KJ/kg -c;

Steam cycle - 3 A thermal power plant generates 5 MW has also 300 KW power needed for auxiliaries. If the heat generated by fuel is 13,000 KJ/sec, determine the net thermal efficiency. A. 35.78% C. 30.56% B. 36.15% D. 3367% SOLUTION: 5,000 - 300 11 net

Steam Cycles SOLUTION: Solving for h.: At 410 Mpa and 440°C (Table 3) hi = 3305.7 KJ/kg s\ = 6.8911 KK/kg-OK Saving for h 2: At 0.105 Mpa(Table 2) sf=1.3181 h r=423.24 Sfg = 6.0249 h rg = 2254.4 s I = S2 = Sr + XSfg 6.8911 = 1.3 I 81 + x(6.0249) x = 0.925

h2

=

11 net = 36.15%

Steam cycle - 4 In an ideal Rankine cycle, the steam throttle condition is 4.10 Mpa and 440°C. If turbine exhaust is 0.105 Mpa, determine the thermal efficiency of the cycle. A. 20.34% C. 34.44% B. 27.55% D. 43.12%

CD @

G»)

•

1

h, + xh.,

h 2 = 423.24 + 0.925(2254.4) h 2 = 2508.54 KJ/kg Solving for h.; h, = h-at 0.105 Mpa h, = 423.24 KJ/kg Solving for h.: Using pump work equation: h, = Vi(P 4 - P 3 ) + h, 3/kg V3 = 0.0010443 m h4.= 0.0010443(4100 - 105) + 423.24 h, = 427.412KJ/kg

QA = h. - h, QA = 3305.7 - 427.412 QA = 2878.29 KJ/kg

=

13,000

99

WT = Wr = WT = Wp = Wp = w, =

hi - h2 3305.7 - 2508.54 797.16 KJ/kg h, - h, 427.412 - 423.24 4.172 KJ/kg

W ne t = W T - W p W net = 797..16 - 4.172 W n e t = 79299 KJ/kg

11, = W nc,lQA 11, = 792.99/2878.29 11t = 27.55%

s

100

101

Steam Cycles

Steam Cycles

Steam cycle - 6 (ME Bd. Oct. 1989)

1--

Steam cycle - 5 (ME Bd. Oct. 1991) In a Rankine cycle, saturated liquid water at 1 bar is compressed isentropically to 150 bar. Fir..t by heating in a boiler, and then by superheating at constant pressure of 150 bar. the water substance is brought to 750°K. After adiabatic reversible expansion in a turbine to 1 bar, it is then cooled in a condenset to saturated liquid. What is the thermal efficiency of the cycle (%)? A. 23.45% C. 34.24% D. '18.23% B. 16.23%

SOLUTION:

SOLUTION:

CD

At 150 bar(l5 Mpa) and 750 0 K ( 4 7 7 ° C ) h, = 3240.5 KJ/kg SI = 6.2549 KJ/kg-OK At I bar(O.IO Mpa) Sf = 1.3026 Sfg = 6.0568 vr = 0.001043

A steam generating plallt has two 20 MW turbo-generators. Steam is supplied at 1.7 Mpa and 320°C. Exhaust is at 0.006 Mpa. Daily average load factor is 80%. The steam generating units operate at 70% efficiency when using bunker fuel having a heating value of 31,150 'KJlkg and an average steam rate of 5 kg steam/K'W-hr. Calculate the Mtons of fuel oilfbunker fuel required per 24 hours. A. 515 C. 6.17 B. 432 D. 762

Load Factor

Peak Load Av/;;. Load

= S2 = Sf + xSsg 6.2549 = 1.3026 + x(6.0568) x = 0.8176 x = 81.76% hz = 417.46+0.8176(2258) hz = 2263.6 KJ/kg 14 = V3(P 4 - P 3) + h3 14 = 0.0010432(15,000-100)+417.46 14 = 433 KJ/kg Wp = 14 - h, Wp = 433 -417.46 Wp = 15.54 KJ/kg Wr = h, - h z Wr = 3240.5 - 2263.6 Wr = 976.9 KJ/kg SI

Efficiency

976.9 - 15.54 =

=

I

~ 0.8 = - - - 20,000 x 2 Ave. Load = 32,000 KW m,

h f = 417.46 h fg = 2258.0

Efficiency

Ave. Load =

(3240.5 - 433) 34.24%

s

I

I

® @

From Steam Tables: PM .J h, = 3077 KJ/kg Pw h, =, 151.53 KJ/kg 3/kg V3 = '0.0010064 m Solving for 14: 14 = V3(P 4 - P 3) + h) h, = 0.001 0064( 1700 - 6) + 151.53 h, = 153.23 KJ/kg m, = 5(32,000) m, = 160,000 kg/hr m s (h l - h 4 ) llb = mfQ h 160,000(3077 - 153.23) 0.70 = ------'------'m f(31,150) m, =.21,454 kg/hr 21,454(24) Fuel needed for 24 hours operation = 1000 Fuel needed for 24 hours operation = 514.9 Mtons

102

Steam cycle - 7 (ME Bd. Oct. 1994) A back pressure steam turbine of 100,000 KW serves as a prime mover in a cogeneration system. The boiler admits the return water at a temperature of 66°C and produces the steam at 6.5 Mpa and 455°C. Steam then enters a back pressure turbine and expands to the pressure of the process, which is 0.52 Mpa. Assuming a boiler efficiency of 80% and neglecting the effect of pumping and the pressure drops at various location, what is the incremental heat rate for electric? The following enthalpies have been found; turbine entrance = 3306.8 KJ/kg, exit = 2700.8; boiler entrance = 276.23 KJ/kg, exit = 3306.8. A. 21,504 KJIKW-hr C. 23,504 KJIKW-hr B. 22,504 KJIKW-hr D.24,504 KJIKW-hr

WT = turbine work WT = m(h, - h2) WT = m(3306.8 - 2700.8) WT 606 m KW

QA

(m x 3600)(h) - h 4)

(m x 3600)(3306.8 -

...........

L.71.J0.~)

m

•

,

0.8 13,637,565m KJ/hr

Heat rate Heat rate

13,637,565m =

=

SOLUTION:

®

Net Output = 1000 - 0.09( 1000) Net Output = 910 MW Net Output = 910,000 KW Heat generated = m, Qh 9800(907) Heat generated =

Qh m I

(6,388.9 x 4.187)

pw

J

24 x 3600 Heat generated = 2,752,001 KW

Steam Cycle - 9 (ME Bd. Oct. 1995)

llbo

QA

value of 6,388.9 Kcal/kg and the steam generator efficiency is 86%. What is the net station efficiency of the plant in percent? A. 30% C. 33% B. 25% D.38%

Station efficiency = Net output/Heat input Station efficiency = 910,000/2,755.00 I Station efficiency = 33.07%

SOLUTION:

QA

103

Steam Cycles

Steam Cycles

~l

I •

•

606m m 22,504 KJIKW-h,

Steam Cycle - 8 (ME Bd. Oct. 1994) A coal-fired power plant has a turbine-generator rated at 1000 MW gross. The plant required about 9% of this power for its internal operations. It uses 9800 tons of coal per day. The coal has a heating

A superheat steam Rankine cycle has turbine inlet conditions of 17.5 Mpa and 530°C expands in a turbine to 0.007 Mpa. The turbine and pump polytropic efficiencies are 0.9 and 0.7 respectively, pressure losses between pump and turbine inlet are 1.5 Mpa. What should be the pump work in KJ/kg? C. 37.3 A. 17.3 D. 47.3 B. 27.3 SOLUTION:

»;>

V 3 (P4

- P3)

IIp where: Using density of water = 1000 kg/rrr' V3 = 1/1000 3/kg V3 = 0.001 m P4=17.5+1.5 P4 = 19 Mpa P4 = 19,000 Kpa P3 = 0.007 Mpa

104

Steam Cycles P3 IIp

=

7 Kpa

=

0.70 0.001(19,000 - 7)

QA

Steam properties:

At 1.70 Mpaand 370°C: h = 3187.1 KJ/kg S = 7.1081 At 0.17 Mpa: h f = 483.20 Sf = 1.4752 hfg = 2216.0 Sfg = 5.7062 At 65.soC: h, = 274.14 C. 91.24 D. 69

SOLUTION:

h,

=

3187.1 KJ/kg

Solving for h2 : SI

=

S2

=

Sr

h) - h 4 3187.1- 274.14

QA

=' - - - - -

QA

=

0.8 3641.2 KJfkg

Cogeneration efficiency

:.;- Steam Cycle - 10 (ME Bd, Oct. 1995)

A. 78 B. 102.10

=

T]b

W = -----p 0.70 Wp = 27.1 KJ/kg

A steam plant operates with initial pressure of 1.70 Mpa and 370°C temperature and exhaust to a heating system at 0.17 Mpa. The condensate from the heating system is returned to the boiler at 65.5°C and the heating system utilizes from its intended purpose 90% of the energy transferred from the steam it receives. The T]T is 70%. If boiler efficiency is 80%, what is the cogeneration efficiency of the system in percent. Neglect pump work.

+ XSrg

7.1081 = 1.4752+x(5.7062) x .~ 0.9871 h z = h, + xh rg hz = 483.20 + 0.9871(2216) hz = 2670.60 KJ/kg h, = h, = 274.14 KJ/kg WT = (h, - hz)lh W 1 = (3187.1-2670.60)(0.70) WT = 361.55 KJ/kg OR = 0.90(hi - h 3) QR = 0.90(2670.6 - 274.14) QR = 2156.81 KJ/kg

105

Steam Cycles

=

QT+Q R

QA 361.55 + 2156.81

Cogeneration efficiency

=

Cogeneration efficiency

=

3641.2 69.16%

Steam Cycle - 11 (ME Bd. Apr. 1996) In a cogeneration plant, steam enters the turbine at 4 Mpa and 400°C. One fourth of the steam is extracted from the turbine at 600 Kpa pressure for process heating. The remaining steam continues to expand to 10 Kpa, The extracted steam is then condensed and mixed with feed water are constant pressure and the mixture is pumped to the boiler pressure of 4 Mpa. The mass flow rate of steam through the boiler is 30 kg/sec. Disregarding any pressure drops and heat losses in the piping, and assuming the turbine and pump to be isentropic, how much process heat is required in KW? Steam properties: At 4 Mpa and 400°C: h = 3213.6 KJlkg, s = 6.7690 At 600 Kpa: hr = 670.56 Sf = 1.9312 h rg = 2086.3 Sfg = 4.8288 A. 15,646.8 C. 1.9312 B. 2,468.2 D. 1,027.9

SOLUTION:

® 52 = St· + xSrg 6.7690 = 1.9312 +x(4.8288) .-. .. • 1I J. ® x = 1.00(saturated vapv" - ~ 51

hz hz

=

= =

h r+ xh rg 670.56 -I. 1.00(2086.3)

6bj'i,Ij.;1

® p)..

•

y

Steam Cycles

106

Steam Cycles

h2 == 275/,)."7 h, == h-at 600 Kj . h, == 670.56 Q == m p (h, - h 3 ) Q == (30/4)(2756.9 - 670.56) Q == 15,647.5 KW

Steam Cycle - 12 In an ideal Reheat cycle, the steam throttled condition is 8 Mpa and 480°C. The steam is then reheated to 2 Mpa and 460°C. If turbine exhaust is 60°C, determine cycle efficiency. A. 38.3% C. 34.3% B. 24.3% D. 45.2% SOLUTION: At 8 Mpa and 485°C(Table 3) hi == 3348.4 KJ/kg 51 == 6.6586KJ/kg-OK

8ft1Pa

4l10°C3 = Sf + X3 Sig 6.7976 = 0.6493 + x3(7.5009) x3=0.8196 h, = hfJ + xh fg h, = 191.83 + 0.8196(2392.8) h, = 2152.96 KJlkg WT = 30(3410.3 - 2756.89) + (30 - 7.6)(2756.86 ·2152.96) WT = 33,129.66 kw

114

Steam Cycles

Steam Cycles

QR = (rn. - m2l (h, - 14) h, = hc at 10 Kpa 14 = 191.83 KJ/kg QR = (30 - 7.6)(2152.96 - 191.83) QR = 43,929.312 Kw QA = ml(h l-hc4) QA = 30(3410.3 - 191.83) QA = 96,554.1 Kw 33,129.66+43,929.312 Cogeneration efficiency = - - - - - - - 96,554.1 Cogeneration efficiency = 79.81% (No exact answer in the choices)

115

Steam Cycle - 21 Pump work of Rankine cycle is 15 KJ/kg. Density of water entering the pump is 958 kg/rn", If condenser pressure is 100 Kpa, what is the pressure at the entrance of tlie turbine? A. 14.47 Mpa C. 15.67 Mpa B. 20.48 Mpa D. 17.77 Mpa SOLUTION: W p = v(P l - P 4 ) W p = (l/w)(P 1 - P4 ) 15 = (I/958)(P[ - 100) P = 14,470 Kpa P = 14.47 Mpa

Steam Cycle - 20 (ME Bd. Oct. 1997) A heat exchanger was installed purposely to COOl U.50 kg of gas per second. Molecular weight is 28 and k = 1.32. The gas is cooled from 150°C to 80°e. Water is available at the rate of 0.30 kgls and at a temperature of rz-c, Calculate the exit temperature of the water in "C. A. 48 C. 46 B. 42 D. 44

SOLUTION:

R R

= =

r~

8.314/28 0.2969 KJ/kg-K kR

cp

=

cp

=

k-I 1.32(0.2969) 1.32- 1 cp = 1.2247 KJ/kg-K

11t =

180°C

I

•

~

t,,=12 c

t

~

1000-13

2800 11t = 35.25%

HEAT t:XCHANGE

Q

By heat balance:

In a Rankine cycle the turbine work is 1,000 KJ/kg and pump work of 13 KJ/kg. If heat generated by generator is 2800 KJ/kg, what is the efficiency of the cycle? A. 35.25% C. 38.65% D. 30.25% B. 40.75% SOLUTION:

14. ::l

Steam Cycle - 22

1SOoC

Steam Cycle - 23 m w=O.3kgls

Qgain = Qloss

m; cp (t, - to) = fig cpg (t2 - t l ) (0.30)(4. 187)(tb - 12) = (0.5)( 1.2247)(150 - 80) t b = 46.125"C

In a Reheat power plant the difference in enthalpy at the entrance and exit is 550 KJ/kg for first stage and second stage is 750 KJ/kg. If both stages has an efficiency of92% and heat added to boiler is 3,000 KJ/kg. Determine the plant cycle efficiency neglecting the pump work. A. 30% C. 40% B. 35% D. 45%

water required because of high pressure exit velocity if the steam flow rate is 38 kg/s and the cooling water temperature rise is iz'c with an inlet condition of 30°C.

SOLUTION:

llT

=

QA

llT = llT

=

SOLUTION:

3000 39.87%

Steam Cycle - 24 An adiabatic feed pump in a steam cycle delivers water to a steam generator at a temperature of 200°C and a pressure of 10 Mpa. The water enters the pump as a saturated liquid at 180°C. If the power supplied to the pump is 75 kw, determine the mass flow rate. A. 6.23 kg/s B. 8.34 kg/s

C. 7.39 kg/s D. 9.12 kg/s

p...

200"C ~ 180"C

~.

=

=

10 Mpa

P sat h2 h2 h2

= = =

763.22 KJ/kg 3 0.0011274 m /kg = 1.0021 Mpa

If the exit velocity is not considered: At 15 Kpa: h, = 225.94 KJ/kg h fg = 2373.1 KJ/kg hi = h f + X h fg h, = 225.9 + 0.9(2373.1) hi = 2361.73 KJ/kg h2 = h, at 15 kpa h2 = 225.94 KJ/kg Qgain =

38 kg1s 15Kpa (90%quality)

30°C

h,

m.

Qloss

m; (4.187)(12) = 38(2361.73 - 225.94) m; = 1615.38 kg/s

SOLUTION:

vr

C. 13.23 kg/s D. 21.78 kg/s

A. 11.23 kg/s B. 17.23 kg/s

W n + Wn

0.92(550) + 0.92(750)

At 180°C: h,

117

Steam Cycles

Steam Cycles

116

75 Kw

If exit velocity is considered: (1)(240)2 h, = 2361.73 + [(1/ 2) ] 1000 h, = 2390.53 KJ/kg IDw (4.187)(12) = 38(2390.53 - 225.94) m; = 1637.1 kg/s Mass difference Mass difference

vr (P2 - PI) + hi 0.0011274(10,000 - 1002.1) + 763.22 773.36 KJ/kg

W p = m s (h2 - hi) 75 = m s(773.36 - 763.22) rn, = 7.39 kg/s

Steam Cycle - 25 A condenser receives steam from a turbine at 15 kpa, 90% quality, and with a velocity of 240 m/s. Determine the increase in circulating

= =

1637.1 - 1615.38 21.78 kg/s

Steam Cycle - 26 A steam generator has an exit enthalpy of 3195.7 KJfkg at the rate of 10 kg/s. The enthalpy available at the turbine inlet is 3000 KJfkg. Determine the heat lost between boiler outlet and turbine inlet. A.1957kW B. -1957 kw

C.1873kw D. -1873 kw

118

Steam Cycles

Steam Cycles

SOLUTION:

119

h, = 289.23 + 0.0010223 (2000 - 30) h, = 291.24 KJ/kg By mass balance in the heater: Assume supply steam = Ikg m\(h z) -t (I - m.jh, = I h, h, = h-at 2 Mpa h6 = 908.79 KJ/kg ml(2902.5) + (I - m l)(291.25) = 1(908.79) m, = 0.2364 kg extracted steam/kg supply

Q = m (h z - h.) Q = 10(3000 - 3195.7) Q=-1957kw

Steam Cycle - 27 A Rankine cycle has a turbine unit with available enthalpy of 800 KJ/kg. The pump has also 10 KJ/kg energy available. Find the net cycle output of the plant if mass now rate is 5 kg/so A. 2619 kw C. 8745 kw B. 3950 kw D. 4234 kw

SOLUTION:

W net = m, (W T - W p) W net = 5(800 - 10) W net = 3950 kw

Steam Cycle - 29 An adiabatic turbine in a steam generating plant receives steam at a pressure of 7.0 Mpa and 550°C and exhausts at 20 kpa. The turbine inlet is 3 m higher than the turbine exit, the inlet steam velocity is 15 m/s and the exit is 300 m/s. Calculate the turbine work in KJ/kg. A. 1297.45 C. 1093.45 B. 1197.10 D. 1823.45 SOLUTION: At 7.0 Mpa and 550°C: h = 3530.9 KJ/kg . s = 6.9486 KJ/kg-OK At 20 kpa:

Steam Cycle - 28 In a Regenerative cycle, the steam is extracted from the turbine at 2 Mpa and 250°C for feedwater heating and it is mixed with condenser exit at 30 kpa after pumping. Find the fraction of vapor extracted from the turbine. C. 0.5632 A. 0.23464 B. 0.19338 D. 0.3855 SOLUTION:

h, ~~ 251

I C"-550°C

V'==15mi~ ~

,4

w .~ L'-"il • • ~""

hhg - 23:J8.3

1

Sr = 0.8320 Srg = 7.0766 s = Sr + XSrg 6.9486 = 0.8320 + x(7.70766) x = 0.864 h, = 251.4 + (0.864)(2358.3) h2 = 2288.9 KJ/kg

~

\61

20 Kpa V,= 300 m/s

2 Mpa

At 2 Mpa and hz = At 30 kpa, h, = Vr = hs =

250°C: 2902.5 KJ/kg

m, T 250°C

-

30 Kpa

289.23 KJ/kg 0.0010223 m3/kg

14 + V4(PS - P4)

1

HEATER

W T = (hi - h2 ) + 1/2 m (v/ - v/) + (PEl - PEz) 50 Kpa

1-m,

2-300 2

W T = (3530 -2288.9) + WT

=

1197.IOKJ/kg

15

2(1000)

+

3(1 x9.81) 1000

Steam Cycles

120

121

Steam Cycles

Steam Cycle - 30

Steam Cycle - 31

A steam power plant operates on the Rankine cycle. The steam enters the turbine at 7 Mpa and 550°C with a velocity of 30 m/s. It discharges to the condenser at 20 kpa with a velocity of 90 m/s. Calculate the net work in kw for a flow of 37.8 kg/s. C. 34.22 Mw A. 23.23 Mw D. 46.54 Mw B. 53.34 Mw

A Carnot cycle uses steam as the working substance and operates between pressures of. 7 Mpa and 7 kpa. Determine the cycle thermal efficiency. A. 44.17% C. 34.23% B. 54.23% D. 59.44% SOLUTION:

SOLUTION:

At 7.0 Mpa and 550°C: . h = 3530.9 KJ/kg S = 6.9486 KJ/kg-OK At 20 kpa: hf = 251.4 hhg = 2358.3 Sf = 0.8320 Sfg = 7.0766

At 7.0 Mpa: t = 285.88°C At 7 kpa: t = 39°C T H = 285.88 + 273 T H = 558.88°K T L = 39 + 273 TL = 312

w

e = e

6.9486 = 0.8320 + x(7.70766) x = 0.864 h 2 = 251.4+(0.864)(2358.3) = 2288.9 KJ/kg = h f at 20 kpa = 251.4 KJ/kg = 0.001017 m3 /k g = h, + V3 (P4 - P 3 ) = 251.4 + 0.0010 17(7000 - 20) = 258.5 KJ/kg 30 2 _ 90 2 WT = (3530.9 ·2288.9) + - - 2(1000) WT = 1238.4 KJ/kg WOe l = 1238.4 - (258.5 - 251.4) WOe l = 1231.3 KJ/kg (37.8 kg/s) WOel = 46,543.19 kw WOe l = 46.54 Mw

2

285.B80C

1

[;] 3

5,-5,

4

TI-l -TL 5

TI-l

S=Sf+XSfg

h2 h, h] v] h, h, h,

T

558.88 - 312 =

558.88 e = 44.17%

Steam Cycle - 32 A supercritical power plant generates steam at 25 Mpa and 560"C. The condenser pressure is 7.0 kpa.. Determine the exit quality of steam if it expands through a turbine in this power plant. A. 45.66% C. 56.56% rI D. 74.26% '1' B. 68.45%

+

W25MPa

560°C

SOLUTION:

At 25 Mpa and 580°C: h = 3430.5 S = 6.2897

/

/-

7Kpa

r @ 5

Steam Cycles

122 At 7 kpa:

Sc = SCg =

BOILERS

0.5592 7.7167

S = Sc

123

Boilers

+ XSCg

6.2897 = 0.5592 + x(7.7167) x = 74.26%

Boiler - 1 The heating surface area of water tube boiler is 200 m2 , what is the equivalent rated boiler horsepower? A. 217 Hp C. 200 Hp B. 2365.93 Up D. 219.78 Hp

Steam Cycle - 33 Steam enters a turbine at 1.4 Mpa and 320?C. The turbine internal efficiency is 70%, and the total requirement is 800 kw. The exhaust is to the back pressure system, maintained at 175 kpa. Find the steam flow rate. C. 3.23 kg/s A. 2.62 kg/s D. 5.34 kg/s B. 4.23 kg/s

SOLUTION: Rated Boiler horsepower = RS.!0.91 Rated Boiler horsepower = 200/0.91 Rated Boiler horsepower = 219.78 Hp

SOLUTION: Boiler - 2 W=800kw

--.

At 1.4 Mpa and 320°C: h, = 3084.3 SI = 7.0287 At 175 Kpa: Sc = 1.4849 SCg = 5.6868 hc = 489.99 V,= 90 m/s hcg = 2213.6 Solving for the quality: 7.0287 = 1.4849 + x (5.6868) x = 0.9748 h 2 = 489.99 + 0.9748(2213.6) h 2 = 2647.93 KJ/kg WT = IDs (hi - h 2)(llT) 800 = ills (3084.3 - 2647.93)(0.70) ills = 2.62 kg/s

The rated boiler horsepower of a fire tube boiler is 500 Up. What is the heating surface area of the boiler? 2 A. 500 m2 C. 400 m 2 2 B. 300 m D. 550 m SOLUTION: Rated Boiler horsepower = H. S:! 1.1 500 = H.S.! 1.1 U.S, = 550m 2

Boiler - 3 A water tube boiler has a heating surface area of 500 m2 • For a developed boiler hp of 825. Determine the percent rating of the boiler. A. 120.15% B. 160.15%

C. 150.15%

D. 300.15%

Boilers Boilers

124

125

Rated boiler Hp = 281.32 Hp Dev. Boiler Hp xIOO% %R = Rated Boiler Hp Dev. Boiler Hp 2 = ----281.32 Developed Boiler Hp = 562.64 Hp ills (2257 x 1.08) 562.64 = - - - - - " - - - - 35,322 ID, = 8153.02 kg/hr

SOLUTION'

Rated boiler horsepower = 500/0.91 Rated boiler horsepower = 549.45 hp Dev. Boiler Hp xIOO% %R = Rated Boiler Hp %R = 825/549.45 x 100% %R = 150.15%

Boiler - 4 Boiler - 6 The factor of evaporation of a boiler is 1.1 and a steam rate of 0.79 kg/sec. What is the developed boiler horsepower? A. 300 C. 869 B. 200 D. 250 SOLUTION: FE

=

h, - h,

The actual specific evaporation of a certain boiler is 10. Factor of evaporation is 1.05. If the heating value of fuel is 30,000 KJ/kg, find the boiler efficiency. A. 60% C. 70% B.65% D.79% SOLUTION:

h, - h F

--=

2257 2257 x 1.1

Developed Boiler hp Developed Boiler hp Developed Boiler hp

l1b = ills(hs-h F )

ills(hs-h F ) IDrQ h

IDs

35,322 (0.79 x 3600)(2257 x 1.1)

10

ID r

35,322 199.89 Hp

l1b

1O(2257xI.05) =

30,000

l1b = 79% Boiler - 5 The percent rating of water tube boiler is 200%, factor of evaporation 2 of 1.08 and heating surface area is 256 m • Find the rate of evaporation. B. 7,200 kg/hr A. 8,153.02 kg/hr D. 8,500.46 kg/hr B. 5,153.02 kg/hr SOLUTION:

Rated boiler Hp = 256/0.91

Boiler - 7 The AS ME evaporation units of h boiler is 24.827,000 KJ/hr. The boiler auxiliaries consumes 1.5 MW. What is the net boiler efficiency if the heat generated by the fuel is 30,000 KJ/hr? A. 64.75% C. 62.76% B. 68.94% D. 68.54%

127

Boilers

Boilers

lLO

SOLUTiON:

SOLUTiON.

ms(h s - h F )

11ner =

-

Boiler Aux.

mrQ"

(24,827 ,000 I 3600) - (L5xl 000) llne'

(30,000,000 I 3600)

11net ,=

Theo A/F = Theo. NF = Theo. A/F = Actual A/F = Actual A/F =

64.75%

11.5C + 345(H - 0/8) T 4.3S 11.5(0.705) + 34.5(0.045 - 0.06/8) + 4.3(0.03) 9.53 kg air/kg fuel 9.53(1.3) 12.389 kg air/kg fuel

=

h, - h F -2257 (h, - h F ) = FE x 2257 (h, - h f ) = 1.1 x 2257 33,820C + 144,212(H - 0/8) + 9304S, Kl/kg 33,820(0.705) + 144,212(0.045 - 0.06/8) + 9,304(0.03)

=

29,930 KJ/kg

FE~

Boiler - 8 A 100,000 kg of coal supplied two boilers. One has a capacity of 150 kg/hr. How many days to consume the available fuel if the other boiler consumes 200 kg/hr? A. S days C. 15 days D. 12 days B. 7 days

Qh Qh Qh

=

m s (h s

llb = 0.70

-

hF)

mrQ h

175,000( 1.1 x 2257) =

------

m r (29,930)

SOLUTION: m,

m

=

m

=

mrl + mf2 m = ISO + 200

350 kg/hr

No. of days = 100,00/350 No. of days = 285.71 hrs No. of days = 11.9 days

=

rna = rna = PV = 101.325(V) = V

=

21,018.456 kglhr 21,018456(12389) 260,397.651 kg/hr mRT 260,397.651(0.287)(15.6 ~ 273) 3/hr 212,830 m

Boiler - 10 (ME Bd. Apr. 1981) Boiler - 9 (ME Bd. Apr. 1981) The following coal has the following ultimate analysis by weight: C = 70.5% Hz = 4.5% O, = 6.0% N z = 1.0% S = 3.0% Ash = 11 % Moisture = 4% A stocker fired boiler of 175,000 kg/hr steaming capacity uses this 3/hr coal as fuel. Calculate volume of air in m with air at 60°F and 14.7 psia pressure if boiler efficiency is 70% and FE = 1.10. 3/hr A. 212,830 m C. 213,830 rrr'rhr D.. 214.830 m'zhr B.. 2 J 5,830 mY/hr

The following coal has the following ultimate analysis by weight: C = 70.5% Hz = 4.5% Oz = 6.0% N z = 1.0% S = 3.0% Ash = 11 % Moisture = 4% A stocker fired boiler of 175,000 kglhr steaming capacity uses this 3/hr coal as fuel. Volume of air in m with air at 60°F and 14.7 psia pressure. Weight in metric tons of coal needed for 24 hours operation at rated capacity if boiler efficiency is 70% and FE = 1.10. A. 503.443 Mtons C. 502443 Mtons B. 508.443 Mtons D. 504,443 Mtons

128

Boilers (h, - h-) Theo. A/F Theo. A/FTheo AiF ActuaiA/F = Actual A/F·~

129

Boilers

SOLUTION: 1I5e i 34.5(H - 0/8) + 4.3S 115(0.705) j 34.5(0.045 - 0.06/8) + 4.3(0.03) 953 kg air/kg fuel 9.53(13) 12.389 kg air/kg fuel

h, - h F FE = - 2257 (h, - hr ) = FE x 2257: (h, - hr) = 1.1 x 2257 On = 33.820C + 144,212(H - 0/8) + 9304S, KJrKg Oh = 33.820(0.705) -i- 144,212(0.045 - 0.06/8) + 9,304(0.03) Oil = 29.930 KJ/kg ms\h s - h F ) llb

=

0.70 =

mrQ h 175,000(1.1x2257)

rn r (29,930) rn, = 21,018.456 kg/hr

Coal needed in 24 hrs =

21,018.456(24)

1000 . Coal needed in 24 hrs = 504.443 Mtons

Boiler - 11 (ME Bd. Oct. 1982) Two boilers are operating steadily on 91,000 kg of coal contained in a bunker. One boiler is producing 1591 kg of steam per hour at 1.2 factor of evaporation and an efficiency of 65% and another boiler produces 1364 kg of steam per hour at 1.15 factor of evaporation and an efficiency of 60%. How many hours will the coal in the bunker run the boilers if the heating value of coal is 7590 Kcal/kg? C. 230.8 hrs A. 220.8 hrs B. 256.2 hrs D. 453.3 hrs SOLUTION: For Boliler No. I: (h, - h r) = FE x 2257

llb =

=

1.2 x 2257

m s ( h s - h f- ) mrQ"

1591( 1.2 x 2257) 0.65 = - - - - - rn f ( 7590 x 4.l 87) mfI= 208.605 kg/hr For Boiler No.2: (hs-hF ) = FE x 2257 (h, - hr)= 1.15 x 2257 ms(hs-h F ) llb

1591 kg/h

911,)00 kg

.m

Bunker

=

mrQ h 1364(1.15 x 2257) 0.60 = - - - ' - - - - m r (7590 x 4.187) mf2 = 185.673 kg!hr rn- = total fuel consumed mT = rnfl + rnf2 m- = 208.605 + 185.673 m- = 394.278 kg!hr No. of hours = 91,000/394.278 No. of hours = 230.8 hrs

B1

Boiler#1

B2 1364kgfh Boiler#2

Boiler - 12 (ME Bd. Oct. 1984) A steam generating plant consisting of a boiler, an economizer and superheater generates superheated steam at the rate of 50 tons /hr, Feed water enters the boiler at 5 Mpa and 120°C. Steam leaves the superheater at 4.5 Mpa and 320°C. If the coal used has a heating value of 30,000 KJ/kg, calculate the number of tons of coal fired per hour for a gross efficiency of 85%. A.4.89 C. 5.34 B. 6.34 D. 45.5 SOLUTION:

At 4.5 Mpa and 320°C(Table 3), h, = 3000.6 KJ/kg

130

coal that could be used in order to ensure the generation of required

At 5 Mpa and 120°C(Table 4), h F = 507.09 KJ/kg

llb

=

steam.

11kg/c:m' _ h,

SOLUTION: h.

(50)(3000.6 - 507.09)

p

_,I r (30,000)

m, = 4.889 tons/hr

C. 23,556 D. 30,976

A. 28,464 B. 29,977

ms(hs-h F ) mrQ h

0.85 =

131

Boilers

Boilers

a.,..30000kJ/k m,

11 kg/cm 2 x 101.325/1.033 1079.1 Kpa 1.0791 Mpa

=

P

=

p

=

From Table 2,

h,

From Table 1,

Boiler - 13 (ME Bd. Oct. 1986) A water tube boiler has a capacity of 1000 kg/hr of steam. The factor of evaporation is 1.3, boiler rating is 200%, boiler efficiency is 5%, heating surface area is 0.91 m 2/boiler Up, and the heating value of fuel is 18,400 Kcal/kg. The total coal available in the bunker is 50,000 kg. Determine total number of hours to consume the available fuel A. 533.45 C. 634.34 B. 743.12 D. 853.26

llb

=

0.85

o,

hs

=

h,

=

hF hF

m, =50000kglhr

BOILER

h g at 1.0791 Mpa 2781 KJ/kg = hr at 80°C = 334.91 KJ/kg

ms(h s - h F ) mrQ h 50,000(278-1 - 334.91) = -------

4,800Q h =

29,977 KJ/kg

SOLUTION:

(h, - h F) = FE x 2257 (h, - h F)= 1.3 x 2257 ms(hs-h F ) llb = mrQ h 1000(1.3 x 2257) 0.65 = ----'--------'---m r (18,400x4.l87) m, = 58.592 kglhr No. of hours No. of hours

=

50,000/58.592

=

853.36 hrs

Boiler - 14 (ME Bd. Apr. 1984) A boiler operating at 11 kgicm 1 is required to generate a minimum of 50,000 kg/hr of saturated steam. Feed water enters the boiler at 80°C. The furnace is designed to fire coal at an average rate 4,800 kg/hr and boiler efficiency is 85%. Compute the minimum heating value of local

Boiler - 15 (ME Bd. Apr. 1984) A boiler operating at 11 kg/cm 2 is required to generate a minimum of aC. 50,000 kg/hr of saturated steam. Feed water enters the boiler at 80 The furnace is designed to fire coal at an average rate 4,800 kg/hr and boiler efficiency is 85%. Compute the developed boiler horsepower. A. 3462.56 hp C. 2345.67 hp D. 4233.34 hp ~'m. B. 1234.56 hp SOLUTION: P

=

P

=

11 kg/crni x 101.325/1.033 1079.1 Kpa P = 1.0791 Mpa From Table 2, h, = h g at 1.0791 Mpa h, = 2781 KJlkg From Table 1, h F = h, at 80°C

132

Boilers

Boilers At 145°C: h, 610.63 A.65 B. 95

133

z;

hF

334.9i

-

50,000(2781- 334.91)

Developed Boiler Hp Developed Boiler Hp

~.J!h:g

==

C. 88 D. 78

SOLUTION:

35,322 3,462.56 Hp

Tlb

ms(hs-h F ) mfQ h

Boiler - 16 Tlb =

A steam boiler generating 7.1 kg/s of 4.137. Mpa, 426.7°C steam is continuously blown at the rate of 0.31 kg/sec. Feed water enters the economizer at 148.9 0C. The furnace burns 0.75 kg coal/sec of 30,470.6 KJ/kg higher heating value. Calculate the overall thermal efficiency of steam boiler. A. 76.34% C. 82.78% B. 84.23% D. 88.34%

SOLUTION: From Steam Table: h, = 629.87 KJ/kg h, = 3274.1 KJ/kg h, = 109702 KJ/kg 7.1(3274.1) + 0.31(1097.02) - 7.41(629.87) Tlb

0.75(30,470.6)

Tlb == 82.78%

23.5(3195.7 - 610.63) 2.75(25,102)

Tlb = 88%

'*

Boiler - 18 (ME Bd. Apr. 1997) A steam boiler on a test generates 885,000 Ib of steam in a 4-hollr period. The average steam pressure is 400 psia, the average steam temperature is 700°F, and the average temperature of the feedwater supplied to the boiler is 280°F. If the boiler efficiency for the period is 82.5%, and if the coal has a heating value of 13,850 Btu/lb as fired, find the average amount of coal burned in short tons per hour. At 400 psia and 700°F, h, == 1362.7 Btu/lb At 2BO°F, hI = 249.1 Btu/lb A. 9.84 short tons per hour B. 10.75 short tons per hour C. 12.05 short tons per hour D. I 1.45 short tons per hour

SOLUTION: Boiler - 17 23.5 kg of steam per second at 5 Mpa and 400°C is produced by a steam generator. The feedwater enters the economizer at 145°C and leaves at 205°C. The steam leaves the boiler drum with a quality of 98%. The unit consumes 2.75 kg of coal per second as received having a heating value of 25,102 KJ/kg. What would be the overall efficiency of the unit in percent? Steam properties: At 5 Mpa and 400°C: h = 3195.7 KJ/kg At 5 Mpa: h; = 1154.23 hfg = 1640.1 At 205°C: h, = 875.04

rn, = 885,000/4 m, = 221,250 lb/hr Tlb =

0.825

ms(hs-h F ) mfQ h 221,250(1,362.7 - 249.1)

I400 r.-....:..--.

m r (13,850)

mf= 21,563 Ib/hr Q. = 13,850 Btull~ mf= 2 1,563/2000 mf- 10.78 short tons per hr

p.sia

700~F m, = 885,0001b

BOILER

I

280°F

134

Steam Engine

Boilers

135

STEAM ENGINE

Boiler-19 A steam boiler generates 401,430 kg of steam in a 4-hour period. The steam pressure is 2750 kpa and 370°C. The temperature of the water supplied to the steam generator is 138°C. If the steam generator efficiency is 82.5% and the coal has a heating value of 32,200 KJ/kg, find the average amount of coal burned per hour. A. 9771 C. 9563 B. 8734 D. 7354

SOLUTION:

Steam Engine - 1 A steam engine have 10% brake thermal efficiency and delivers 750 kglhr steam. The enthalpy of engine entrance is 2800 KJ/kg and condenser exit is 450 KJ/kg. Determine the brake power of the engine. A. 46 KW C. 49 KW B. 47 KW D. 48 KW SOLUTION:

At 2750 kpa and 370°C: h, = 3166.9 KJ/kg At 138°C: hF = hf at 138°C hF = 580.54 KJ/kg 401,430

Brake Power lltb =

ID s ( h s

0.10

--------

(750 /3600X2800 - 450) Brake Power = 48.96 KW

IDs IDs =

=

- hf2 ) Brake Power

4 x3600 27.877 kg/s IDs(hs-h F )

llb =

IDFQ h

0.825

ID F (32,200) IDF = IDF =

Steam Engine - 2

27.877(3166.9 - 580.54) 2.714 kg/s (3600) 9770.77 kg/hr

The indicated efficiency of a steam engine is 60%. The engine entrance is 2700 KJ/kg and exit is 2000 KJ/kg. if steam consumption is 800 kg/hr and mechanical efficiency is 90%, what is the brake power of the engine? A. 55 KW C. 65 KW B. 84 KW D.70KW

SOLUTION:

Indicated Power llei ffi s

0.60

( h \ - h2 ) Indicated Power

=

(800/3600)(2700 - 2000)

Steam Engine

136

Steam Engine

Indicate d Power 93.33 KW 11m = BP/IP 0.90 = BP/93.3 3 BP = 84 KW

137

V 0 = 2[11:/4 (030):' (045) (220/60 )] V o = 0.23326 rnzsec Indicate d Power .~ 392.4(0 .233263 3) Indicate d Power = 91.53/0 .746 Indicate d Power = 122.7 Hp

=

91.53 KW

Steam Engine - 3 g at A steam engine has bore and stroke of 300 mm x 420 mm runnin Kpa. 400 is engine the of e pressur ed 250 rpm has mean indicat Determ ine the indicat ed power. C. 65 KW A. 100 KW D. 99 KW B 50KW SOLUT ION 2 2[11:/4 D L N] = 2[rc/4 (0.3)2 (0.42)(2 50/60)] V::J = 0.247 m' Indicate d Power = P mi X V 0 Indicate d Power = 400 x 0.247 Indicate d Power = 98.96 KW

VD VD

Steam Engine - 5 1034.25 A steam engine develop s 60 Bhp with dry saturat ed steam at 736.36 is ption consum Kpa absolut e and exhaus t at 124.11 Kpa. Steam 90% on based cy efficien engine kg/hr. Calcula te the indicat ed cy. efficien ical mechan A. 34.23% C. 45.23% B. 54.23% D. 66.74%

=

SOLUT ION:

hi = h g at 1.03425 Mpa hi = 2779.4 KJ/kg (interpo lated) S I = Sg at 1.03425 Mpa 5\

6.5748 KJlkg-O K(interp oiated)

=

At 0.12411 Mpa:(B y interpol ation) h f = 443.43 Sf = 1.17165 h fg = 2241.56 Sfg = 5.9152

Steam Engine - 4 220 rpm. The crank shaft of a double acting steam engine rotates at and the mm, 450 x mm 300 is engine steam the of stroke and The bore the Find , mean effectiv e pressur e acting upon the piston is 4 kg/ern" indicat ed horsep ower develop ed in the cylinde r. C. 143.2 hp A. 122.7 hp D. 176.3 hp B. 110.3 hp

SI

=.

S2 =

Sf

+

XSfg

6.5748 = 1.37165 j x(5.915 2) x = 0.8796 h2 = 443.43 + 0.8796( 2241.56 ) h 2 = 241.5.16 Kl/kg Indicate d Power Indicate d Power

SOLUT ION: Indicate d Power

Pm' Pm' VD

=

= =

Pm'

4 kg/em" x 101.325 /1.033 2 3924 KN/m 2[11:/4 [)2 L N]

X

VD

11el

(60/0.9 ) x 07'+6 49.733 KW

Indicate d Power =

m,(h j

11" rtei

= =

=

-0

h2 ) 49.733

(736.36 /3600)( 2779.4 - 24\51) 66.74%

138

Steam Engine

Steam Engme - 6 A steam engine develops 60 Bhp with dry saturated steam at 1034.25 Kpa absolute and exhaust at 124.]] Kpa. Steam consumption is 736.36 kg/hr. Calculate the thermal efficiency of equivalent Rankine engine. A. 15.59% C. 12.45% B 34.23% D. 21.34%

Steam Engine

indicated mean effective pressure is 600 Kpa, determine brake thermal efficiency. A. 23.34% C. 14.66% B. 18.34% D. 27.34% SOLUTION: VD VD

SOLUTION: hi hi

= =

SI

=

SI

=

S 1"= S:

=

Sf

+ XSfg

6.5748 = 1.37165 + x(5.9152) x = 0.8796

h,

~

hz hf2

= =

1,'2

=

Y]R

=

Y]R Y]R

=

2779.4 - 443.43 15.59%

A 350 mm x 450 mm engine running at 260 rpm has an entrance steam condition of 2 Mpa and BO°C and exit at 0.] Mpa. The steam consumption is 2,000 kg/hr and mechanical efficiency is 88%. If

=

At 0.1 Mpa:

s,

=

Sfg

=

hf2

=

1.3026 6.0568

h, = 417.46 h rg = 2258

417.46 KJ/kg Brake Power

Y]tb

m s (h J - h f2 )

m s (h l - h f 2 ) 2779.4 - 2415.1

-----

Prru X V D 600 x 0.37522 225.13 KW

=

=

At 2 Mpa and 230°C(Table 3) h, = 2849.6 SI = 6.4423

m s (hI - h 2)

=

2[71:/4 (0.35)2 (0.45) (260/60)] 0.37522 m 3 /sec

Brake Power = Indicated Power(Y]rn) Brake Power = 225.13(0.88) Brake Power = 198.11 KW

443.43 + 0..8 796(2241.56) 2415.16 KJ/kg h, at 0.12411 Mpa 443.43 KJ/kg

:!/.:: Steam Engine - 7

l

= =

Indicated Power Indicated Power Indicated Power

hg at 1.03425 Mpa 2779.4 KJ/kg (interpolated) Sg at 1.03425 Mpa 6.5748 KJlkg-OK(imerpolated)

At O. ] 2411 Mpa:(By interpolation) Sf = 1.37165 h, = 443.43 Sfg = 5.9152 h rg = 2241.56

139

198.11 Y]tb

=

Y]tb

=

(2000/3600)(2849.6 - 417.46) 14.66%

Steam Engine - 8 A 350 mm x 450 mm engine running at 260 rpm has an entrance steam condition of 2 Mpa and BO°C and exit at 0.1 Mpa. The steam consumption is 2,000 kg/hr and mechanical efficiency is 88%. If

140

Steam Engine

Steam Engine

indicated mean effective pressure is 600 Kpa, determine indicated thermal efficiency. A. 16.66% C. 12.34% B. 34.23% D. 21.23%

SOLUTION: V D = 2[rc/4 (0.35)2 (0.45) (260/60)] V D = 0.37522 m 31sec

Indicated Power Indicated Power Indicated Power

=

Prru

=

600 x 0.37522 225.13KW

=

X

VD

14'1

SOLUTION:

VD

=

piston displacement

VD

=

2[(71:/4)D 2LN]

VD

=

2[(71:/4)(10/12)2(12/12)(300)]

VD

=

327.25 felmin P rru V D

Indicated Power

Indicated Power = (120)(144)(327.25) 33,000

At 2 Mpa and 230°C(Table 3)

hi

=

Sl

=

2849.6 6.4423

Indicated Power

At 0.1 Mpa: Sf = Sfg =

hf2

=

1.3026 6.0568

hf = 417.46 hfg = 2258

417.46 KJ/kg Indicated Power

11tl

m s ( h J - h f2 )

225.13

11 ti 116

-t

(2000/3600)(2849.6 - 417.46) 16.66%

Steam Engine - 9 (ME Bd. Apr. 1997) Steam is admitted to the cylinder of an engine in such a manner that the average pressure is 120 psi. The diameter of thepiston is 10" and the length of stroke is 12". What is the engine when it is making 300 rpm? A. 171.5 C. 173.2 D. 174.4 B. 175

I

=

171.40 hp

1-.+2

.\·1t'1If11

Turbine

STEAM TURBINES

Stca m Turbine - 2 (ME Rd. Apr. 1987) An industrial power plant requires 1,5 kg of dry saturated steam per second at 165°C for heating purposes. This steam may be supplied from an extraction turbine which receives stearn at 4 Mpa and 380°C and 'is exhaust to condenser at the rate of 0.8 kg steam per second at 0.0034 Mpa while rejecting 1400 KW to the cooling water. If mechanical- electrical efficiency is 95% and that the he-at loss in the turbine casing is 10 KW, calculate the power generated by 'he plant. A. 2,126.44 Kw C. 3,123.34 Kw B. 1,556.5 Kw D. 4,344.33 Kw

Steam Turbine - 1 A steam turbine receives 5,000 kglhr of steam at 5 Mpa and 400°C and velocity of 25 m/sec. It leaves the turbine at 0.006 Mpa and 85% quality and velocity of 20 rn/sec. Radiation loss is 10,000 KJ/hr. Find the KW developed. A. 1273.29 C. 1373.29 C. 2173.29 D. 7231.29

SOLUTION:

SOLUTION: At 5 Mpa and 40QoC hi = 3195.7 KJ/kg Sl = 6.6459

---

5MPa.400oC m7=5000kg/h

At 0.006 Mpa h r = 15 I. 53 and hrg = 2415.9 hz = hr +- xhrg hz = 151.53 +- 0.85(2415.9) hz = 2205.045 KJ/kg

w TURBINE

KE 1 = l/2 m yZ KE 1 = 1/2 (5,OOO/3600)(25)z KE 1 = 434.03 W KE 1 = 0.43403 KW KE z = 1/2 m yZ KE z = l!2 (5,000/3600)(20)z KE z = 277.78 W KE z = 0.2778 KW

W

=

W

=

143

Steam Turbine

By energy balance: KE, -t- mh I = KE z -i- mh, +- Q -i- W W = (KE 1 - KE z) +- m(h, - h z) - Q 5000 10,000 (0.43403 - 0.2778) -i- ( - - )(3195.7 - 2205.045) - - 3600 3600 1373.29 KW

From Steam tables: h, = 3165.9 KJ/kg 10i,{'.~ hz = hg at 165°C 4--~Gen hz = 2763.5 KJ/kg ! ~~) OutP':lt h, = h, at 0.0034 Mpa h) = 109.84 KJ/kg ° . ~ @~. By mass balance: ~,~~~5~gIS . 1Tl,=O.8kgls rn, = 1.5 + 0.8 f 1,400kw m, = 2.3 kg/sec-: h By heat balance: m.h, = 1.5h z +- O.Sh) + 10 +- 1400 +- \'1 2.3(3165.9) = 1.5(2763.5) + 0.8(109.84) +- 10 -t- 1400" W W = ]638.45 KW Generator Output = 1638.45(0.95) Generator Output = 1,556.5 KW

r».

't

7=20m/s (x=85%) O.006MPa

-------~------------

Steam Turbine - 3

A steam turbine with 90% stage efficiency receives steam at 7 Mpa and 550°C and exhausts as 20 Kpa. Determine the turbine work. A.. 117 KJ/kg C. 123 K.l/kg . B. 132 KJ/kg D. 143 KJ/kg

SOLUTION: At 7 Mpa and 550°C hi = 3530.9 KJ/kg s\ = 6.9486

)44

Steam Turbine

.~

At 20 Kpa(0.020 Mpa) h, = 251.40 h,g = 2358.3 s I = S2 = Sf + XSfg 6.9486 = 0.8320 + x(7.0766) x = 0.8643 h2 = 251.40 + 0.8643(2358.3) h2 = 2289.78 KJ/kg p hi - h Za llsT = h, - h z j~30.9 - h Za 0.90 = 3530.9 - 2289.78 h2• = 2413.89 Kl/kz WT = hi - h2 • W T == 3530.9 - 2413.89 W T = ]]7.01 KJ/kg

.

145

Steam Turbine

Sr = 0.8320 Srg = 7.0766

090

3530.9 - h

2a ----

~

3530.9 - 2289.78 h 2> .= 2413.89 KJlkg h2> = h, + xhrg 2413.89 = 25.\.40 + x(2358.3)

x

=

0.9167

x == 91.67%

Steam Turbine - 5 A small steam turbine power plant of 5,000 KW capacity has a full load steam rate of 6 kg steam per KW-hr. No load steam consumption may be taken as 10% of the full load steam consumption. Write the equations of WILLANS LINE of this turbine and at 60% of rated load, calculate the hourly steam consumption of this unit. C. 19200 kg/hr A. 19,100 kg/hr B. 19,300 kg/hr D. 19,400 kg/hr

5

Steam Turbine - 4 SOLUTION: A steam turbine with 90% stage efficiency receives steam at 7 Mpa and 550°C and exhausts as 20 Kpa. Determine the quality at exhaust. C. 82.34% A. 87.45% B. 76.34% D. 91.690/,

Full load steam consumption == 6(5000) Full load steam consumption = 30,000 kg/hr No load steam consumption No load steam consumption

p

0.10(30,000) 3,000 kg/hr

SOLUTION:

By Two point slope formula: At 7 Mpa and 550°C h, = 3530.9 KJIkIl SJ = 6.9486 At 20 Kpa(0.020 Mpa) Sf = 0.8320 h, = 251.40 Sfg = 7.0766 hfg = 2358.3 SI

==

S2 =

sr+ XSrg

6.9486 = 0.8320 + x(7.0766) x = 0.8643 h2 = 251.40 + 0.8643(2358.3) h2 = 2289.78 KJ/kg hl-h Za llsT = hi - h,

m. (kg)

ms

5

-

3,000 ==

30,000 - 3,000

5.4(0.6 x 5000) rn, at 60% load m, = 19,200 kg/hr 2

,(sooo,

(L - 0)

5,000 - 0 ms-3,000 == 5.4L rn, = 5.4 L + 3,000

FL

,

Nl~~30-0-00)-- -+

3000

,, ,

: L(kw)

Steam Turbine - 6 Steam flows into a turbine at the rate of 10 kg/sec and 10 KW of heat are lost from the turbine. Ignoring elevation and kinetic energy effects,

30000)

146

Steam Turhine

147

Steam Turbine

calculate till' power output from the tur hine. Inlet enthalpy is 2739 KJ/kg ami exhaust enthalpy is 2300.5 KJ/kg. A. 4605 KW C 4375 KW B. 4973 KW D. 4000 KW

rn,

=

55(42ll0)

rn,

~

26,100 kg/hr

I

3.000

SOLUTION:

W =

ffi(h 1 -

W

=

10(2739 - 2300.5) - 10

W

=

4375 KW

h2 ) - Q

Steam Turbine - 9 A steam turbine has an entrance enthalpy of 3050 KJ/kg. The exit has 2500 KJ/kg. Determine the actual enthalpy after isentropic expansion if stage efficiency is 90%. A 1255 KJ/kg C. 2500 KJ/kg B. 2555 KJ/kg D. 2000 KJ/kg

Steam Turbine - 7 SOLUTION: Steam entering the turbine has a rate of 10 kg/sec with enthalpy of 3400 KJ/kg and 2600 KJ/kg at the exhaust. If 100 KW of heat is rejected from turbine casing, what is the turbine work? A. 7900 KW 'c. 5600 KW B. 7700 KW D. 5400 KW

h, - h 1 a

"lsr

h, - h 1 3050 - h 1 a

0.90

SOLUTION:

3050- 2500 W

=

W W

= =

W

=

(h, - h-) - Q 10 (3400 - 2600) - 100 8000 -100 7900KW

h1a

ffi s

=

2555 KJ/kg

Steam Turbine - 10 Steam Turbine - 8 A steam turbine of 6 MW capacity has a Willan's line equation of rn, =

5.5L + 3,000, kg/hr. Determine the steam consumption at 70% load. A. 3564 kg/hr C. 26,100 kg/hr B. 3546 kg/hr D. 58,000 kg/hr SOLUTION: At 70% load, L

=

0.7(6,000)

L

=

4200 KW

Steam enters the turbine at the rate of2.5 kg/sec with enthalpy of 3200 KJ/kg and exhaust enthalpy of 1100 KJ/kg. Steam is extracted from the turbine at the rate of 1 kg/sec for heating purposes with enthalpy of 2750 KJ/kg. What is the turbine work. A. 2000 KW C. 3000 KW B. 2500 KW D. 3600 KW

SOLUTION: ffi\

=

2.5 rn-

= =

+ ffi3 1+ m, 1.5 kg/s

ffi2

Stemn Turbine

Ic\X

W = 2~(1:()()) -

W

=

i(2750)-1.5(1100)

Geothermal Power Plant

149

GEOTHERMAL POWER PLANT

36()() 1\W

Geothermal Power Plant - 1 Steam Turbine - 11 A steam turbine has an entrance enthalpy of 3400 KJ/kg and 2500 KJ/kg at exit. If generator generates 2430 KW and has 90% efficiency, what is the mass of steam entering the turbine? A 10,400 kg/hr C. 10,700 kg/hr B. 10,600kglhr D.l0,800kg/hr SOLUTION:

Mass flow rate of ground water in a geothermal power plant is 1,500,000 kg/hr and the quality after throttling is 30%. Determine the brake power of turbine if the change of enthalpy of steam at inlet and outlet is 700 KJ/kg. C. 64.5 MW A. 68.5 MW B. 87.5 MW D. 89.5 MW SOLUTION:

W T = m, (hi - h2 ) 243010.90 = TIl, (3400 - 2500) m, = 3 kg/sec x 3600sec/hr m, = }O,800 kg/hr

ID, =

m,

0.3(1,500,000) 450,000 kg/hr ID, = 125 kg/sec WT = m, (h, - h4 ) W T = 125(700) W T = 87,500 KW W T = 87.5 MW ID, =

Steam Turbine - 12 Steam turbine in Rankine cycle has an exhaust enthalpy of 2650 KJ/kg and delivers 0.8 kg/sec of steam. Determine the heat rejected from the condenser if enthalpy at exit is 200 KJ/kg. A. 1960 KW C. 1995 KW D. 1909 KW B. 1940 KW SOLUTION:

OR

=

mlh 2 - h3 )

OR = 0.8(2650 - 200) QR = 1960 KW

x mg

=

Geothermal Power Plant - 2 Ground water of geothermal power plant has an enthalpy of 700 KJ/kg and at turbine inlet is 2,750 KJ/kg and enthalpy of hot water in flash tank is 500 KJ/kg. What is the mass of steam flow entering the turbine if mass flow of ground water is 45 kg/sec? A. 3.27 kg/sec C. 4.27 kg/sec B. 2.27 kg/sec D. 9.27 kg/sec SOLUTION: h2

=

700

x

=

hi + x(h, - hf)

500 + x(2750 - 500) 0.0888

=

!:'()

Geothermal Power Plant 0.888(45)

The enthalpy entering the turbine of a geothermal power plant is 2750 KJ/kg and mass rate of 1 kg/sec. The turbine brake power is 1000 KW condenser outlet has enthalpy of 210 KJ/kg. If temperature rise of . cooling water in condenser is 8°C, what is the mass of cooling water requirement? C. 46 kg/sec A. 44 kg/sec B. 45 kg/sec D. 47 kg/sec SOLUTION: W T = ill,(h 3 - h 4 ) 1000 = 1(2750 - h4 ) h, = 1750 KJ/kg =

Qw

ills(~

- h s) = ill w Cp(t2 - t.) 1(1750 - 210) = ill w(4.187)(8) m., = 45.97 kg/sec

A 16,000 KW geothermal plant has a generator efficiency and turbine efficiency of 90% and 80%, respectively. If the quality after throttling is 20% and each well discharges 200,000 kg/hr, determine the number of wells are required to produce if the change of enthalpy at entrance and exit of turbine is 500 KJ/kg. A. 4 wells C. 6 wel1s B. 5 wells D. 8 wells SOLUTION: W T = ill s(h3 - h 4 ) 16,000 - - = i l l s(500) 0.9(0.8) ills = 44.44 kg/sec ills = 160,000 kg/hr 160,000 = 0.20 illg illg = 800,000 kg/hr No. of wel1s = 800,000/200,000 No. of wells = 4 wells

Geothermal Power Plant - 6 (ME Bd. Apr. 1988)

Geothermal Power Plant - 4 In a 12 MW geothermal power the turbine is 26 kg/sec. The enthalpy of ground water is efficiency of the plant. A. 7.4% B. 9.6%

plant, the mass flow of steam entering quality after throttling is 25% and 750 KJ/kg. Determine the overall C. 5.4% D. 15.4%

SOLUTION: ills = x illg 26 = 0.25 illg illg = 104 kg/sec 12,000

f]()

Geothermal Power Plant - 5

4 kg/sec

Geothermal Power Plant - 3

f]o

151

x fIl g

ills ill, ill,

QR

Geothermal Power Plant

104(750) 15.38%

A geothermal power plant draws pressurized water from a well at 20 Mpa and 300°C. To produce a steam water mixture in the separator, where the un flashed water is removed, this water is throttled to a pressure of 1.5 Mpa. The flashed steam which is dry and saturated passes through the steam collector and enters the turbine at 1.5 Mpa and expands to 1 atm. The turbine efficiency is 85% at a rated power output of 10 MW. Calculate overall plant efficiency A.7.29% C. 9.34% B. 12.34% D. 19.45% SOLUTION:

At 1.5 Mpa (Table 2) h, = 2792.2 KJ/kg S3 = 6.4448 At I arm (100°C)

152

Geothermal Power Plant Sf Sfg

= 1.3069 =

h, hf.~ -

6.048

4 I9.04 2257

S3 = S4 = Sf -t X4Sf~ 64448 = 1.3069 + X4 (6.0480) X4 ~ 0.8495 h, = h r + xh fg .h, = 419.04 + 0.8495(2257) h, = 2336.4 KJ/kg Wr = ms(h 3 " h4)Tlr 10,000 = m,(2792.2 - 2336.4)(0.85) rn, = 25.81 kg/sec At 20 Mpa and 300°C(Table 4) h,= 1333.3 KJ/kg At 1.5 Mpa: hr = 844.89 hfg = 1947.3 h, = h2 = h, + X2 hfg 1333.3 = 844.89+X2(1947.3) X2 = 0.25 m, = Xl fig (25.81 X 3600) = 0.25(m g \ fig = 37 I ,664 kg/hr 10,000

hr '= 640.23 hrg = 2108.5 m. h, = h g at 0.50 Mpa 1500kPa h, = 2748.7 KJ/kg h, = n2 '= hr + xzhrg . 697~22 = 640.23 + Xz (2~108.5) ® I" I X2 - 0.027 I : I I m.m, = xm g \-' m, = 0.027(29.6) ,29.6 kg/s m, = 0.80 kg/sec -IJG) 6t;"10°C From Mollier Diagram: j ~

14 = 2211 KJ/kg Power produced = m s(h 3 - ~)

Power produced = 0.8(2748.7 - 22 I I) Power produced = 430.16 KW

[d:>

'!® +"""

=

+@

Geothermal Power Plant - 8 (ME Bd. Oct. 1985) A flashed steam geothermal power plant is located where underground hot water is available as saturated liquid at 700 Kpa. The well head pressure is 600 Kpa. The flashed steam enters a turbine at 500 Kpa and expands to 15 Kpa, when it is condensed. The flow rate from the wdl is 29.6 kg/sec. Determine the cooling water flow in kg/sec if water is available at 300 e and a lODe rise is allowed through the

s

(371,644/3600)(1333.3) 110vel'all

153

Geothermal Power Plant

7.26%

condenser. SOLUTION:

Geothermal Power Plant - 7 (ME Bd. Oct. 1985) A flashed steam geothermal power plant is located where underground hot water is available as saturated liquid at 700 Kpa. The well head pressure is 600 Kpa. The flashed steam enters a turbine at 500 Kpa and expands to 15 Kpa, when it is condensed. The flow rate from the well is 29.6 kg/sec. Determine the power produced in KW ! .. 430.13 kg/s C. 370.93 kg/s B. 540.23 kg/s D. 210.34 kg/s

h, h,

=

=

h-at 0.70 Mpa 697.22 KJ/kg (i) e-t>

278

Pumps P2

h

h

=

-

279

Pumps

PI

w 896.071- (-6.773)

T\combtned

0.85(0.7)

11combined

59.50%

-----'----....:...

9.645 h = 93.075 m Q = 100 gal/min x 3.785Ii1lgal x Im 3/1000li x 1/60 Q = 0.006308 m 3/sec P = wQh P = 9.645(0.006308)(93.075) P=5.69KW

Pumps - 7 In a boiler feed pump, the enthalpy at the entrance is 765 KJ/kg. If pump has a head of 900 m, what is the exit enthalpy of the pum p. A 897 KJ/kg C. 774 KJlkg B. 465 KJ/kg D. 864 KJlkg

Pumps - 5 SOlCTION A pump is to deliver 150 gpm of water at ahead of 120 m. If pump efficiency is 70%, what is the horsepower rating of motor required to drive the pump? A. 40.44 Hp C. 38.44 Hp B. 25.66 Hp D.2111Hp

m(h:? - hi) = m x h x 0.00981 h:?-765 = 900 x 0.00981 h:? = 773.83 KJlkg

h,

h,

~~.\ h=900m

SOLUTION:

W, = wQh W, = 9.81 (150gal/min x 0.003785m 3/lgal x 1/60)(120)

w, = 11.139 KW BP = 11.139/0.7 BP = 15.913 KW BP = 21.33 hp

Pumps - 8 A submersible pump delivers 350 gpm of water to a height of 5 ft from the ground. The pump were installed 120 ft below the ground level and a draw down of 8 ft during the operation. If water level is 25 ft above the pump, determine the pump power. C. 7.24 KW A. 7.13 KW B. 4.86 KW D. 864 KW

Pumps - 6 SOLUTION: A motor is used to drive a pump having an efficiency of 85% and 70% respectively. What is the combined efficiency of pump and motor? A. 59.50% C. 62.50% B. 61.50% D. 65.50% SOLUTION: 11combtned =

11p 11m

h = 5+120-(25-8) h = 108/3.281 h = 32.916 m Q = 350 gal/min x 0.003785m 3/gal x Imin/60sec Q = 0.02246 m 3/sec W, = wQ h

DiU

Pumps WI'

Pumps

9.81 (0.02246)(32.916) 725 KW

w,

BP BP

=

=

281

32846/0.65 505.32 KW

Pumps - 9

Pumps -11

Determine the number of stages needed for a centrifugal pump if it is used to deliver 400 gal/min of water and pump power of 15 Hp. Each impeller develops a head of 38 ft. A. 6 C. 8 f}. 4 D.7

What power can a boiler feed pump can deliver a mass of 35 kg/s water at a head of 500 m? C. 456.64 KW A. 356.56 KW D. 171.67 KW B. 354.54 KW SOLUTION:

SOLUTION:

W, 15 x 0.746

=

h h

=

=

P

=

mxhxO.00981

P

=

35 x 500 x 0.00981

p

=

171.675KW

w Q h

9.81 (400 galimin x 0.00785m J /gal x 1/60)h 45.20 m x 3.281 film 148.317 ft

=

Number of stages ~ 148.317/38 Number of stages = 3.903 stages N umber of stages "" 4 stages

Pumps - 12 A pump running at 100 rpm delivers water against a head of 30 m. If pump speed will increased to 120 rpm, what is the increase in head? A. 43.2 m C. 34.6 m B. 13.2 m D. 56.3 m

Pumps - 10 SOLUTION: A boiler feed pump receives 50 Ii/sec of water with specific volume of 0.00112 mJ/kg at ahead of750 m. What is the power output of driving motor if pump efficiency is 65%? A. 505.32 KW C. 785.56 KW B. 643.54 KW D. 356.45 KW SOLUTION:

W,

w, w,

= = =

w Q h (1/0.001 12x 0.00981)(0.050)(750) 32846 KW

h,

N2

2

-=(-) hi N1 h, 120 2 -" = ( - ) 30 100 h 2 = 43.2 m Increased = 43.2 - 30 Increased = 13.2 m

~I II

,.~

·1

.Jv·

-,·r

282

Pumps

283

Pumps

Pumps - 13

Pumps - 15

A pump is used to deliver 50 Ii/sec of sea water at a speed of 120 rpm. If speed will increased to 135 rpm, determine the increase in pump

A certain pump is used to deliver 150 gpm of water having a density of 61.2 Ib/fe. The suction and discharge gage reads 4 in Hg vacuum and 25 psi, respectively. The discharge gage is 2 ft above the suction gage. What is the brake power ofthe motor if pump efficiency is 75%? A. 3.24 Hp C. 5.45 Hp B. 2.67 Hp D. 6.89 Hp

capacity. A. 56.25 li/sec B. 34.56 li/sec

C. 87.54 Ii/sec D. 6.260 IiIsec

SOLUTION: SOLUTION: Q

N

2-2 -

Q1

N\

Q

135

h

=

Pd

-

P,

+z

W

Ps = - 4 in Hg x 14.7/29.92 P, = -1.965 psi Pd = 25 psi 25 - (-1.965) h = [ ](144) + 2 61.2 h = 65.45 ft BP = w Qh (61.2)(150/7.481)(65.45) BP = ...:...----'--'----...:....:...--....:... 33,000(0.75) BP = 3.24 Hp

2- -

50 120 Q2 = 56.25 Ii/sec Increased = 56.25 - 50 Increased = 6.25 Ii/sec

Pumps - 14 A 15 KW motor running at 350 rpm is used to drive a pump. If speed will changed to 370 rpm, what is the increase in power? A. 2.72 KW C. 56.45 KW B. 17.72 KW D. 5.67 KW SOLUTION:

P2

No

3

p\ P2

N, 370

3

]5

350

- = ( - ") -=(-)

P2 = 17.72 KW Increased = 17.72 - IS

Increased = 2.72 KW

Pumps - 16 The discharge pipe of a pump is 400 mm in diameter delivers 0.5 mJ/sec of water to a building which maintains a pressure of 100 Kpa at a height of 30 m above the reservoir. If equivalent head is 2 m, what power must be furnished by the pump? A.21IKW C.340KW B. 480 KW D. 240 KW SOLUTION: Q = Axv 0.5 = (re/4 x 0.4 2) V = 3.9788 mlsec

V

284

A 1265 KW B. 23.54 KW

P

v

h = --+-+Z 2g w

(3.9788) 2 h =

2(8.81) 43 m

285

Pumps

Pumps

100 +-+(30+2) 9.81

r

SOLUTION:

-+Pd

.=~~~_.

110m I

n = W p = wQh W p = 9.81(0.50)(43) W p = 210.92 KW

v,

C. 14.17 KW D. 45.35 KW A

12m

Wp = w Q h h = (72 - 10) + 0.1 h = 62.15 ill W p = 9.81(0.015)(62.15) w, = 9.145 KW Power input = 9.145/0.65 Power input = 14.07 KW

+-

Pumps -17 A centrifugal pump is designed for 1800 Determine the speed if impeller diameter is 254 mm. A. 1000 rpm C. B. 1250 rpm D.

rpm and head of 61 m. reduced from 305 mm to 1500rpm 1600 rpm

SOLUTION:

h,

D2

2

h) h,

D1 254

2

-=(-)

-=(-)

42.3

N2

The elevation of suction reservoir is 5 m above the pump centerline and delivers to 85 m elevation tank which maintain 150 Kpa. If 1.5 mJ/sec of water is used to deliver a total head of 3m, determine the power needed by the pum p. A. 1446 KW C. 4675 KW B. 2567 KW D. 3456 KW

h h P

-=(-) hI N]

-=(-)

Pumps - 19

SOLUTION:

61 305 h 2 = 42.30 m h2 N2 2 2

P P

= =

(85 - 5) + 3 + 150/9.81 98.29 m

=

w Qh

=

9.81(1.5)(98.29) 1446.34 KW

=

61 1800 N 2 = 1499 rpm Pumps - 20 (ME Bd. Oct. 1989) Pumps - 18 Water from a reservoir A 10 m elevation is drawn by a motor driven pump to an upper reservoir B at 72 m elevation. Suction and discharge head loss are 0.15 rn, respectively. For discharge rate of 15 li/sec, find the power input to the motor if overall efficiency is 65%.

I

151i/s

Water from a reservoir is pumped over a hill through a pipe 900 mm in diameter and a pressure of one kg/crrr' is maintained at the pipe discharge where the pipe is 85 m from the pump centerline. The pump have a positive suction head of 5 m. Pumping rate of the pump at 1000 rpm is 1.5 mJ/sec. Friction losses is equivalent to 3 m of head loss. What amount of energy must be furnished by the pump in KW?

I

286

Pumps A. 1372 kw B. 1523 kw

Pumps

C. 1234 kw D. 1723 kw

0.5 Yd =

v. SOLUTION:

•Yd

=

I, Q/A 1.5

(1t 14)(0.45)2

:41

5 II 1.em' 85m

Yd Yd =

(1t/4)(0.9)2 I 2.358 mlsec

':=-=0

Pd = 1 kg/ern" x 101.325/1.033 . P d = 98.088 Kpa P 5 = O(open to atmosphere)

p_p h = (zd -zs)+(

d

w h

=

h

=

Yd

= 3.144 mlsec

Pd

=

Pd

~~

1.0 kg/ern' x 101.325/1.033 98.088 Kpa

h

p_p (zd-zs)+( d s)+(hLs+h Ld)+

h

(30) + (

h

42m

w

98.088-0 9.81

2_y 2

y

s)+(hLs+h Ld)+

----c:-

_------.J

Q=1.5m'/s

5m

s

d

2g

. 98.088 - 0 (2.358)2 - (0)2 (85 - 5) + ( ) + 3 + ...:....-------:---=---..:.9.81 2(9.81) 93.28 m

Water Power = w Q h = 9.81(1.5)(93.28) Water Power = 1372.6 KW

Pumps - 21 (ME Bd. Oct. 1986) . Water from a reservoir is pumped over a hill through a pipe 450 mm in diameter and a pressure of 1 kg/cm 2 is maintained at the summit. Water discharge is 30 m above the reservoir. The quantity pumped is 0.5 m'/sec. Frictional losses in the discharge and suction pipe and pump is equivalent to 1.5 m head loss. The speed of the pump is 800 rpm what amount of energy must be furnished by the pump, KW? A. 202 C. 204 B. 206 D. 208

= Q/A

d

s

2g

(3.144)2 -(0») ) + 1.5 + ...:....------'-------'-2(9.81)

Pumps - 22 (ME Rd. Apr. 1988) Water from an open reservoir A at 8 m elevation is drawn by a motordriven pump to an open reservoir R at 70 m elevation. The inside diameter of the suction pipe is 200 mm and 150 mm for the discharge pipe. The suction line has a loss of head three times that of the velocity head in the 200 mm pipe. The discharge line has a loss of head twenty times that of the velocity head in the discharge pipeline. The pump centerline is at 4 m. Overall efficiency of the system is 78%. For a discharge rate of 10 Ii/sec, find the power input to the motor and the pressure gage readings installed just at the outlet and inlet of the pump in Kpag. A. 3.34 kw C. 6.59 kw B. 5.45 kw D. 7.84 kw

v.;SOLUTION: v,

Yd

2_y 2

y

Water power = w Q h Water power = 9.81(0.5X42) Water power = 206 kw

y, = Q/A

SOLUTION:

287

= 0.010,

v, = 0.31831 mlsec Yd = Q/A

r~ ~ 8m

10lils

® ~J :

--

I

288 0.010

v,

Vd

(re /4 )(0.15)2 = 0.566 m/sec

Yd

h Ls

=

3(

(0.31831)2

Ys

(Zd~Zs)+(

Pd

-

Ps

w

0.1782 (rr / 4)( 5 / 12) 2 1.307 fils 0.1782

Yd =

(n I 4)(4 / 12) 2 = 2.043 fils From Steam Table, at 150 psig(l64.7 psi) and 140°F, w = 61 .424 Ib/fi l p_p y 2_y 2 h=(d s)+(d s) w 2g

)

2(9.81) hLs = 0.01549 m (0.566)2 h Ld = 20( ) 2(9.81) h Ls = 0.32642 m h

289

Pumps

Pumps

Yd

Y/

- Ys2

)+(hLs+hLd)+·~-~-

2g

h =(

150(l44)-(-2xI4.7xI44) 61.424

+[

(2.043)2 -(1.307)2

]

2(32.21)

h = 354 ft h = (66_4)+0+(0.01549+0.3264)+[(0.566)2 -(0.31831)2 2(9.81) ] h = 62.35 m Water Power = w Q h Water Power = 0.010(9.81)(62.353) WaterPower = 6.12KW 20wer Input = 6.12/0.78 Power Input = 7.84 KW

Water Power

(61.424)(80 17.481)(354) 33,000

Water Power = 7.05 hp Brake horsepower = 7.05/0.7 Brake horsepower = 10.07 Hp Power Input of motor = 10.07/0.80 Power Input of motor = 12.59 Hp Power Input of motor = 9.39 KW

Pumps - 23 (ME Rd. Apr. 1982) Pumps - 24 (ME Bd. Apr. 1986) A pump is to deliver 80 gpm of water at 140°F with a discharge pressure of 150 psig. Suction pressure indicates 2 inches of mercury vacuum. The diameter of suction and discharge pipes are 5 inches and 4 inches, respectively. The pump has a efficiency of 70%, while the motor efficiency is 80%. Determine power input to the drive motor A. 7.23 kw C. 8.34 kw B 2.34 kw D. 9.39 kw SOLUTION:

Q Q

= =

(80)/(7.481 x 60) 0.1782 fills

Determine the water horsepower and the mechanical efficiency of a centrifugal water pump which has an input of 3.5 Up if the pump has an 8 inches nominal size suction and 6 inches nominal size discharge if it handles 150 gpm of water at 150°F. The suction line gage shows 4" Ug vacuum and the discharge gage shows 26 psi. The discharge gage is located 2 feet above the center of the discharge pipe line and the pump inlet and discharge lines are at the same elevation. A. 2.52 hp C. 4.23 hp B. 6.33 hp D. 8.34 hp

290

Pumps

291

Pumps

SOLUTION:

Pumps - 26 (ME Bd. Oct.l984) 150 gal/min x I fel7.48gal x IminJ60sec 0.334 ft3/ sec P, = - 4 in Hg x 14.7/29.92 P, = -1.965 psi Vs = velocity at suction v. > Q/A v, = 0.334/[n/4 (8112)2] v, = 0.957 ft/sec Vd = Q/A Vd = 0.334/[n/4 (6!12i] Vd = 1.701 ft/sec From steam table, at 150°F, w = 1/0.01634 w = 61.2 lb/ft'

h=(

Q

=;

Q

=

(26xI44)-(-1.965xI44)

+2+[

=

SOLUTION:

(1.701)2 -(0.957)2

61.2 h

A boiler feed pump receives 40 liters per second at 180°C. It operates against a total head of 900 m with an efficiency of 60%. Determine power output of the driving motor in KW. A. 453.23 C. 983.45 C. 623.34 D. 523.27

From steam table(table 4), at 4 Mpa and 180°C, h, = 764.74 KJ/kg 3/kg VI = 0.00112484 m Density = 1/0.00112484 3/1000kg) Density = 889.015 kg/rrr' (Im Density = 0.889 kg/li Waterpower = (40xO.889015)(900)(0.0098!) Water power = 313.964 kw Power output of motor = 313.964/0.60 Power output of motor = 523.273 kw

]

2(32.2)

67.83 ft

Water Hp

(61.2)(0.334)(67.83) =

33,000 Water Hp

=

2.521 .Hp

Pumps - 27 (ME Rd. Oct. 1984)

Pumps - 25 (ME Rd. Oct. 1984) A boiler feed pump receives 40 liters per second at 4 Mpa and 180°C. It operates against a total head of 900 m with an efficiency of 60%. Determine the enthalpy leaving the pump in KJ/kg A. 783.45 C. 756.23 . B. 773.57 D. 765.23 SOLUTION:

180°C, -4MPa -401ils

----.....'

Pump Work = m(h z - hi) = m x h x 0.00981 (h- - 764.74) = 900 x 0.00981 hz = 773.57 KJ/kg

'"

•

,1 '

p

SOLUTION:

2

~®_

.

A boiler feed pump receives 40 liters per second at 180°C. It operates against a total head of 900 m with an efficiency of 60%. Determine discharge pressure in Kpa for a suction pressure of 4 Mpa. A I J ,850 kpa C. 12,566 kpa B. 13,455 kpa D. 14,233 kpa

"

h, - h.. gOOm h, .

Pump Work = m(hz-h\) = mxhxO.00981 (h z - 764.74) = 900 x 0.00981 h 2 = 773.57 KJ/kg h 2-h l

=

Vl(P 2

- P \)

773.57 - 764.74 P 2 = 11,850 Kpa

=

0.001 12484(P2 - 4000)'

29::'

Pumps

293

Pumps Water power

Pumps - 28 (ME Bd, Apr. 19(0)

(260/7.481)(62.4)(226) 33,000

Water power

A boiler feed pump receives 45li/sec of water at 190°C and enthalpy of 839.33 KJ/kg. It operates against a head of 952 m with efficiency of

Brake Hp = 14.85/0.70 Brake power = 21.21 Hp

70%0 Estimate the water leaving temperature assuming that the temperature rise as due to the inefficiency of the input energy A. 191°C C. 123°C B. 143°C D. 165°C SOLUTION: Let m, mass flow rate. kg/sec Pump work = rn., x h x 0.00981 mw (h 2 - h.) = m; x h x 0.00981 h- - 839.33 = 952 x 0.00981 11 2 = 848.67 KJ/kg m - h ) _,,~.",----_I -m",(h 2 -hl)=m c -t

14.85 Hp

Pumps - 30 (ME Bd. Apr. 1985)

Cc

w

1"1 p

P(t 2

(84867 - 839.32>__ (848.67- 839.33) = (4.187)( t 2 0.70 12 = 191°C

-

l

)

190)

A submersible, multi-stage, centrifugal deep well pump 260 gpm capacity is installed in a well 27 feet below the static water level and running at 3450 rpm. Drawdown when pumping at rated capacity is 10 feet. The pump delivers the water into a 25,000 gallons capacity overhead storage tank. Total discharge head developed by pump, including friction in piping is 243 feet. Calculate the diameter of the impeller of this pump in inches if each impeller diameter developed a head of 38 ft. A 3.28 C. 4.23 B. 5.33 D. 6.34

SOLUTION:

Pumps - 29 (ME Bd, Apr. 1985) A submersible, multi-stage, centrifugal deep well pump 260 gpm capacity is installed in a well 27 feet below the static water level. Drawdown when pumping at rated capacity is 10 feet. The pump delivers the water into a 25,000 gallons capacity overhead storage tank. Total discharge head developed by pump, including friction in piping is 243 feet. Calculate brake horsepower req uired to drive the pump if pump efficiency is 70%. A. 3l.31 C. 21.21 B. 41.41 D. 51.51

Let 0 = diameter of impeller

v

=

nDN

V

=

~2gh

1t

0 (3450/60) == ~2(322)(3'6)

D

~c

D

=

0.2738 ft 3.28 inches

SOLUTION:

Pumps - 31 (ME Bd. Oct. 1981) Total dynamic head Total dynamic head

243 - (27 - 10) 226 ft

A double suction, single stage, centrifugal pump delivers 900 m 3/hr of sea water(SG = 1.03) from a source where the water level varies two meters from high tide to low tide level. The pump centerline is located

294

295

Pumps

Pumps

Water Power

2.6 meters above the surface of the water at high tide level. The pump discharges into a surface condenser, 3 m above pump centerline. Loss of head due to friction in suction pipe is 0.8 m and that in the discharge side is 3 m. Pump is directly coupled to a 1750 rpm, 460 V, 3 phase, 60 Hz motor. Calculate the specific speed of pump in rpm. A. 3131 rpm C. 4141 rpm B. 5151 rpm D. 6161 rpm

wQh 62.4(0.0557)(245)

Water Power

550 1.548 Hp

Water Power

=

Motor size Motor size

1.548/0.64 2.42 Hp

Therefore, use 3 Hp motor SOLUTION:

Total suction head = 2 + 2.6 + 0.8 Total suction' head = 5.4 m Total suction head = 17.71 0=900m'/h Total discharge head = 3 + 3 Total discharge head = 6 m I, Total discharge head = 19.686 ft ,I·: Q = 900/2 (double suction) Q = 450 m 3/hr +1--Q = 1981 gal/min 0/2 Fl~ 012 h = 17.71 + 19.68 h = 37.392 ft 2.6m I750 v"l98! Ns I iHighT'Ide level (37.3 92) 3/4 cbts=.._:r Ns 5151 rpm 2m r I I ilow Tide level

Pumps - 33 (ME Bd. Oct. 1996) Water is pumped at I mJ/sec to an elevation of 5 m through a flexible hose using a 100% efficiency pump rated at 100 KW. Using the same length of hose, what size of motor is needed to pump I mJ/sec of water to a tank with no elevation gain? In both cases both ends of hose are at atmospheric pressure. Neglect kinetic energy. A. 51 KW C. 43 KW B. 18 KW D. 22 KW

n

SOLUTION: At 5 m elevation: Water Power = w Q h 100 = 9.81(1)(h) h = 10.194 m

~

Pumps - 32 (ME Bd. Oct. 1996) A pump driven by an electric motor moves 25 gal/min of water from reservoir A to reservoir B, lifting the water to a total of 245 feet. The efficiency of the pump and motor are 64% and 84% respectively. What size of motqr(HP) is required? A. 5 Hp C. 4 Hp B. 3 Hp D. 7.5 Hp SOLUTION: Q Q

= =

25 gal/min x 1min/60sec x 1ft317.481 gal 0.0557 ft3/sec

-'

Ifthere is no elevation: h = 10.194 - 5 h = 5.194 m

Power Power

9.81(1)(5.194) . 51 KW

Pumps - 34 (ME Bd. Apr. 1996) A vacuum pump is used to drain a flooded mine shaft of 20°C water. The pump pressure of water at this temperature is 2.34 Kpa. The pump 1\ incapable of lifting the water higher than 10.16 m. What is the atmospheric pressure?

L'Jb

Pumps

A. 1o» B. 112

C. 98 D. 101.9

Pumps - 36 (ME Bd. Apr. 1985)

!:.

SOLUTION: Using Bernoulli's Theorem:

2 P2 V22 PI VI -+--+Zl =-+--+Z2 W

w

2g

2g

PI P 2 V2 2-V t 2 -=-+ +(Z2- ZI ) w

w

2g

~

2.34 9.81 = 9.81 + 0 + 10.16

The rate of flow of water in a pump installation is 60.6 kg/sec. The intake static gage is located 1.22 m below the pump center line and reads 68.95 Kpa gage; the discharge static gage is 0.61 m below the pump centerline and reads 344.75 Kpa gage. The gages are located close to the pump as much as possible. The area of the intake and discharge pipes are 0.093 m 2 and 0.069 m 2, respectively. The pump efficiency is 70%. Take density of water equals 1000 kg/rrr', What is the hydraulic power in KW? A. 17.0 C. 31.9 D. 15.2 B. 24.5

SOLUTION: ,

Q = 60.6/l 000 Q = 0.0606 m3/sec

PI == 101.9 Kpa

0.0606/0.093 0.652 m/s 0.0606/0.069 0.878 m!s

== = Vct = Yct == Ys Ys

Pumps - 35 (ME Bd. Oct. 1995) It is desired to deliver 5 gpm at a head of 640 m in a single stage pump having specific speed not to exceed 40. If the speed is not to exceed 1352 rpm, how many stages are required? . A. 3 C. 5 B.4 D.2

p_p h

h ==(

SOLUTION:

(

d

s)

2_y 2

y

N == N.jQ

h 3/4

40 = 1352..[5 h 3/4 h = 319.54 ft (head per stage) Number of stages == 640/319.54 Number of stages == 2 stag.es

s

d

+z+(--~-)

w 344.75- 68.95

2g

)+(-0.61+1.22)+[

9.81 h == 28.742 m s

297

Pumps

(0.878)2 - (0.652)2

]

2(9.81)

Hydraulic power == w Q h Hydraulic power == 9.81(0.0606)(28.742) Hydraulic power == 17.10 KW

Pumps - 37 (ME Bd. Apr. 1996) Water in the rural areas is often extracted from underground water source whose free surface is 60 m below ground level. The water is to be raised 5 m above the ground by a pump. The diameter Of the pipe is

298

Pumps

299

Pumps

10 em at the inlet and 15 em at the exit. Neglecting any heat interaction with the surroundings and frictional heating effects, what is the necessary power input to the pump for a steady now of water at the rate of 15 Ii/sec in KW if pu mp efficiency is 85%? A. 9.54 C. 7.82 B. 5.343 D. 11.23

SOLUTION:

H = total head

P

H = z + ­

w

137

H = 8 +­ 9.81 H = 21.96 ill

Power = w Q H

Power = (0.283)(9.81)(21.96)

Power = 61 kw

SOLUTION:

Q Q

15 Ii/sec 0.015 rrr'zsec 0.015

v,

v,

8m Q=28Jlps

(n /4)(0.10)2 1.91 m1s

0.015 Vd Vd

Pumps - 39 (ME Bd. Apr. 1998)

(n /4)(0.15)2

0.85 m/s V

h

)+ (

(Zd- Z s

.

2 _ V 2

d

2g

s)

A pump receives 8 kg/s of water at 220 Kpa and 110·C and discharges it at llOO kpa. Compute for the power required in kilowatts. A. 8.126 C. 7.041 B. 5.082 D. 6.104 SOLUTION:

h

= 5-(-60)+ (0.85)2 -(1.91)2 2(9.81)

h = 6485

Power =

ill

Water Power = w Q h Water Power = 9.81(0.015)(64.85) Water Power = 9.54 KW Power input = 9.54/0.85 Power input = 11.22 KW

h =

ill X

h x 0.00981

1100-220

m=8kgls

9.81

h = 89.704 m

Power = (8)(89.704)(0.00981) Power = 7.04 kw

Pumps - 38 (ME Bd. Apr. 1998)

Pumps - 40 (ME Bd. Apr. 1998)

A pump lifts water at a rate of 283 Ips front a lake and force it into a tank 8 m above the level of the water at a pressure of 137 kpa. What is the power required in kilowatts, ' A. 71 C. 61 B. 41 D. 51

A fuel pump is delivering 10 gallons per minute of oil with a specific gravity of 0.83. The total head is 9.14 m, find how much energy does the pump consumes in KJ per hour. A. 169 C. 189 B. 199 D. 179

301

Fans & Blowers 300

Pumps

SOLUT ION: SO!JJT ION:

3.785(60)

Q = 10x--1000 3/hr Q = 2.271 m

Power Power

= =

20,000

Q

=

Q

=

127 (101.325/760) 16.93 kpa 0.33

PI PI

(0.83 x 9.81)(2.271)(9.14) 169 KJ/hr

Yj

Pumps - 41 (ME Bd, Apr. 1998) of 75 A pump dischar ges 150 liters per second of water to a height 1800 is pump the of speed the and 75% is cy efficien meters. If the ed? subject is shaft drive the which to N-m in torque the is what rpm, C. 791 A. 771 681 D. B 781

YI

(rr 14)(0.40)2 = 2.626 0.33

Y2

=

Y2

=

(rr 1 4)(0.35)2 3.429 m/s

y 2_y 2 p_p s) s)+Z+ (d (d 2g W

h

75-(-1 6.93) h =[

SOLUT ION: h = w Q h Power = (0.150)(9.81)(75) Power = I] 0.4 kw Brake power = 110.4/0.75 Brake power = 147.2 kw

1000(60) 3/s 0.33 m

=

]+(0.45 +0.075 )+[

9.81 10.05 m

(3.429)2 -(2.626 )2

]

2(9.81)

Power

Brake Power

Pumps - 43 (ME Bd. Oct. 1997) Power = wQh

Brake Power = 2 rt T N 147.200 = 2rrT(l8 00/60) T = 781 N-m

.Pumps - 42 (ME Bd. Oct. 1997) er A pump with a 400 mm diamet er suction and a 350 mm diamet water. 15.6°C of minute per dischar ge pipe is to deliver 20,000 liters below Calcul ate the pump head in meters if suction gage is 7.5 em is gage ge dischar and vacuum Hg mm 127 pump centerl ine and reads kpa. 75 reads and ine centerl pump the 45 cm above C. 20 m A. 15 m D. 10m B. 5 m

to a A centrif ugal pump deliver s 300,000 liters per hour of water is 5 water of pressur ized tank whose pressur e is 280 kpa. The source and mm 300 is pipe suction m below the pump. The diamet er of the driving the dischar ge pipe is 250 mm. Calcula te the kw rating of the 72%. be to cy motor assumi ng the pump efficien C. 43.28% A. 41.75 kw D.38.1 6kw B. 35.75k w SOLUT ION:

Q =

300,000

1000(3600) 3/s Q= 0.0833 m

302 0.0833 VI

VI

Pump Efficiency 70

(9.81)(0.8)h =

------'--'-----'-

0.74 h = 66 m

(rr /4)(025)2 1.697 mls p_p

v 2_v 2 s)+Z+(d 5)

h=(d

2g

W

280 - 0

(1.697)2 -(1.178)2 h =(--) + 5 + [ . ] 9.81 2(9.81) h=33.62m Water Power Pump Efficiency Brake Power Water Power = w Q h Water Power = (9.81)(0.0833)(33.62) Water Power = 27.473 kw 27.473

0.72

Water Power Brake Power

(rr /4)(0.3)2 1.178 mls 00833

V2

Vl

303

Pumps

Pumps

Pumps - 45 (ME Bd, Apr. 1997) A pump delivers 500 gpm of water against a total head of 200 ft and operating at 1770 rpm. Changes have increased the total head to 375 ft. At what rpm should the pump be operated to achieve the new head at the same efficiency? A. 2800 rpm C. 3434 rpm B. 3600 rpm D. 2424 rpm

SOLUTION:

=

Brake Power Brake Power = 38.16 kw (motor rating) hI

NI

h, 200

N2 1770

-==(-)

Pumps - 44 (ME Bd. Oct. 1997) A centrifugal pump delivers 80 liters per second of water on test. Suction gage reads 10 mm Hg vacuum and 1.2 meters below pump centerline. Power input is 70 kw. Find the total dynamic head in meters. A. 66 C. 62 B. 60 0.64 SOLUTION:

Q Q

= =

80/1000 0.08 mJ/s

Using the typical pump efficiency of 74%.

2

L

-=(-) 375 N2 N 2 = 2424 rpm

304

SOLUTION:

FANS AND BLOWERS

P

wa

-

Fans & Blowers - 1 (ME Apr. 1997)

wa

=

A fan whose static efficiency is 40% bas a capacity of 60,000 fe/hr at 60°F and barometer of 30 in Hg and gives a static pressure of 2 in of water column on full delivery. What size of electric motor should be used to drive the fan? C. 2 Hp A. 1/2 Hp D. 1 '1/2 Hp B. 1 Hp

Wa

SOLUTION:

h,

RT

101.325 0.287(25 + 273) 1.18 kg/m' h w «;

=

Wa

(00254)( I 000)

hs =

U8 hs = 2 1.52 m 1.42

v =

(n/4)(OJ)2

s

h= s hs

~

h

W

~

(2 112)(62.4) =

wa

1 f ~s:;;:~: ';:{!;)~ =~~~~ 2in

wa

30 in Hg

v,

=

Vd

=

20.09 mls 1.42 1

(n

23.9 mfs (239)-1 - (20.09)"'

=

Vd

4)(0.275)-

h = -----v 2(9.81) h, = 8.5--1 m 300mmO

h, =' 10Alw.

wa Q h W a (60,00 160)(10.4 I

305

Fans & Blowers

Fans & Blowers

Air Power

~

Air Power

=

-''------------=---

Air Power

=

0.315 Hp

W

a)

33,000

h = h, + h, h = ~IS~ + 8.54 h = 30.0C' 111

Therefore: Use I hp motor (standard)

Hi> (1.18 x 0.00981XI.42X30.06)

0.70

A fan draws 1.42 m J per second of air at a static pressure of 2.54 cm of water through a duct 300 mm diameter and discharges it through a duct of 275 mm diameter. Determine tbe static fan efficiency if total fan mechanical is 70% and air is measured at 25"C and 760 mm Hg. A. 50.1 1% C. 65.67% B. 54.34% D. 45.34%

760rnrnHg 25°C

,

BP

= 0.7058 kw

BP

I]

=

wnQ h, = ---

BP

p.1 S:\ O.00981)(l.42X21.52) 1], -= 1],

c

0.7058 50113%

a

L:"'=J-4T~-'----+

w.\Qh

'11

Fans & Blowers - 2 (ME Bd. Oct. 1997)

~

-c

~}54,m

Fans & Blowers

Fans & Blowers

306

Fans & Blowers - 3 Find the air horsepower of an industrial a 900 mm by 1200 mm outlet. ~lIll'r gal-:l' and air density is 1.18 kg/nr', A. 65.35 Hp B. 52.35 Hp Ihruu~h

fan that delivers 25 rolls of air Static pressure is 127 mm of C. 60.35 Hp D. 70.35 Hp

307

b, = 125 rn Air power = wQh Air Power = (1.2 x 0.00981)(90,000/3600)(125) Air Power = 36.78 KW Size of motor = Brake Power Size of motor = 36.78/0.65 Size of motor = 56.59 KW

,t II LIT/ON:

Fans & Blowers - 5 Q=Axv 25 = (0.9 x 1.2) v v = 23.15m1sec h, = v 2/2g h, = (23.15)2/2(9.81) h, = 27.315 m h, = hw(dw/d a ) h, = 0.127(1000/1.18) h, = 107.63 m h = h, + h, h = 107.63 + 27.315 h = 134.94 m Air Power = w Q h Air Power = (1.18 x 0.00981)(25)(134.94) Air Power = 39.052 KW Air Power = 52.35 Hp

At 1.2 kg/m 3 air density operates at 98 Kpa and brake power of the fan? A. 68.4 B. 36.7

SOLUTION:

For Standard air, w = 1.2 kg/m ' h, = hwCdw/d.) h, = 0.15(10001l.2)

C. 67.5 KW D. 93.3 KW

KW KW

SOLUTION:

WI W2 W2 W2

1.2 kg/rn' P/RT 98/(0.287)(32 + 273) 1.1195 kg/m'

= =

= =

w2

BP 2

----

BPI WI BP1 J.I195 ---_.--

Fans & Blowers - 4 A boiler requires 90,000 ml/hr of standard air. The mechanical efficiency of fan to be installed is 65%. Determine the size of driving motor assuming fan can deliver a total pressure of 150 mm of water gage. A. 56.6 KW C. 45.5 KW B. 78.5 KW D. 23.5 KW

a fan develops a brake power of 100 KW. If 32°C with the same speed, what is the new

100 BP2

=

1.2 93.29 KW

Fans & Blowers - 6 A fan has a suction pressure of 30 mm water vacuum with air velocity of 3 m/sec. The discharge has 150 mm of water gage and discharge velocity of 7 m/sec, Determine the total head of fan if air density is 1.2 3 . kg 1m . C. 154 rn A. 150 m D.156m B. 152 m

I * ",;

301l

Falls & Blowers SOLUTION:

Fans & Blowers A. 150 mm water gage B. 180 mm water gage

h, hs

(h w 2

-

h W1 )d w

da [0.15 - (-0.03)J(I 000) 150 m v

hv

WI

2g

h,

7

2

_

32

2(9.81) h, = 2.038 m h = 150 + 2.038 h = 152.038 m

1.2 (standard air density) Wz = PIRT 735( 10 1.325/ 760)

Wz

0.287(93 + 273) 0.933 kg/rn '

2 _ V 2

_ct_ _ s

C. 24 J mm water gage D. 456 mm water gage

SOLUTION:

1.2 hs

309

Wz

h2

w2

hi h,

WI

0.933

- - - -

310 1.20 11 sz = 24 J mm water gage

Fans & Blowers - 9 Fans & Blowers - 7 A 50 KW motor is used to drive a fan that has a total head of 110m. If fan efficiency is 70%, what is the maximum capacity of the fan using standard density of air? J A. 27 m /sec C. 3 I rrr'/sec B. 29 mJ/sec D. 33 m'zsec

The total head offan is 185 m and has a static pressure of 210 mm of water gage, what is the velocity of air flowing if density of air is 1.15 kg/rn''? A. 6.85 m/sec C. 4.76 m/sec B. 3.45 m/sec D. 8.54 m/sec SOLUTION:

SOLUTION:

h, =0.21(1000/].J) h, = 182.6J m h = h, + h, J85 = 182.61 + h, h, = 2.39J m hv = y 2/2g 2.3J = YZ/2(9.8J) v =6.85 m/sec

Air power = 50(0.7) Air power = 35 kw Air power = w Q h 35 = (1.2 x 0.00981)(Q)(110) Q = 27.02 m'zsec

Fans & Blowers - 8

Fans & Blowers - 10

A fan using standard air condition can developed a static pressure head of 310 mm water gage. If fan will operate at 93°C and 735 mm of Hg, find the new static pressure required.

The volume flow of air delivered by fan is 20 mJ/sec and 180 mm water gage. The density of air is 1.185 kg/rn'' and the motor power needed to drive the fan is 44 KW. What is the fan efficiency?

Fan.\ .e Blowers

j/u

A. 70.26% B. 80.26%

C. 75.26% O. 90.26%

Fans & Blowers Static pressure Static pressure

31 ]

(0.9329/J .2)(310) 24] mm water gage

SOLUTION:

h, = 0.18(1000/1.185) h, = ]51.89m Air power = w Q h Air power = (1.185 x 0.00981)(20)(151.89) Airpower = 35.316 KW T] = 35.3 ]6/44 T] = 80.26%

Fans & Blowers - 11 (ME Bd. Apr. 1984) A fan is listed as having the following performances with standard air: Volume discharged - 120 m3/sec Speed - 7.0 rps Static pressure - 310 mm water gage Brake power required - 620 KW The system duct will remain the same and the fan will discharge the 3/sec same volume of 120 m of air at 93°C and a barometric pressure of 735 mm Hg when its speed is 7.0 rps. Find the brake power input and the static pressure required. A. 23] rum water C. 24] mm water B. 251 mm waterO. 261 mm water

Air enters a fan through a duct at a velocity of 6.3 mls and an inlet static pressure of 2.5 em of water less than atmospheric pressure. The air leaves the fan through a duct at a velocity of 11.25 rn/s and a discharge static pressure of 7.62 cm of water above the atmospheric pressure. if the specific weight of the air is 1.20 kg/m' and the fan delivers 9.45 m3;sec. what is the fan efficiency when the power input to the fan is 13.75 KW at the coupling? C. 81.34% A 7181% D. 61.34~·{' B. 91.23%

SOLUTION:

h = h, + h, p_p h=(d

For standard air, w

=

1.2 kg/rrr'

Solving for the density at 93°C and 735 mm Hg w = P/RT

73S(101.J25 I 760)

y 2 _ y2

s)+(d

w

h =[

h

SOLUTION:

w

Fans & Blowers - 12 (ME Bd. Oct. 1994)

=

0.0762 - (-0.025)

s)

2g

](1000)+[

1.20 88.76 m

(11.25) 2

-

(6.3)2

]

2(9.81)

Air power ,~ w Q h Air power = (1.2 x (\.0098])(9.45)(88.761) Air power = 9.874 KW Fan Efficiency = 9.874/13.75 Fan Efficiency c= 71.81 %

0.287(93 + 273) w

= 0.9329 kg.nr'

Using fan laws: Brake power input = (0.9329/] .2)(620) Brake power input = 482 kw

Fans & Blowers - 13 (ME Bd. Apr. 1995) A fan delivers 4.7 m 3/sec at a static pressure of 5.08 em of water when operating at a speed of 400 rpm. The power input required is 2.963 KW. If 7.05 m3/sec are desired in the same fan and installation, find the pressure in cm of water.

l i

'J

312

Fans & Blowers

Fans & Blowers A. 7.62 B. 17.14

BPI

C. 11.43 D. 5.08

BPz w2 T] 6.5 65 + 273

N,

P2 P2

-1 - Q2 N 2 4.7 400 7.05 N z N 2 = 600 rpm h, NJ z -==(-) h, N2 5.08 400 2 -==(-) hz 600 h1 = 11.43 em of water

Fans & Blowers - 14 (ME Bd. Apr. 1995)

=

2 1+ 273 5.68 KW

Fans & Blowers - 15 (ME Bd. Oct. 1996) Air is flowing in a duct with velocity of 7.62 m/s and static pressure of 2.16 em water gauge. The duct diameter is 1.22 m, the barometric pressure 99.4 Kpa and the gage fluid temperature and air temperature are 30°e. What is the total pressure against which the fan will operate in em of water? A. 3.25 C. 3.75 B. 2.5 D. 1.25 SOLUTION: V

A fan described in a manufacturer's table is rated to deliver 500 m3/m in at a static pressure (gage) of 254 cm of water when running at 250 rpm and requiring 3.6 KW. If the fan speed is changed to 305 rpm and air handled were at 65°C instead of standard 21°C, find the power in KW. A. 3.82 C. 4.66 B. 5.08 D. 5.68 SOLUTION: At 305 rpm and 21 0C: PI/P z = (N,/N 2 ) 3 3.6/P2 = (250/305)3 P1 = 6.5 KW At 305 rpm and 65°C: w = P/RT WI PI RT, ---\V P/RT2 2 \V 1 T2

---

w2

T]

T2

--=-=-

SOLUTION:

Q

WI

313

hv

2

fa

~b

h

(7.62)2 =

---

2(9.81) h, = 2.959 m w = P/RT w = 99.4/(0.287)(30+273) w = 1.143 kg/nr' Solving for the velocity head in terms of em of water h, = 2.959(1.14311000) h, = 0 0034 m of water h, = 0 34 em of water h = h, + h, h = 216 + 0.34 h = 2.5cmofwater v

'[I

I

11

314

Fluid Mechanics

Fluid Mechanics

315

FLUID MECHANICS Fluid - 2 (Math-ME Bd Oct. 1997} . What is the expecten nead loss per mile of a closedcircular pipe (17 in inside diameter), friction factor of 0.03 when 3300 gal/min of water now under pressure? A. 38 ft C. 0.007 ft B.3.580ft D. 0.64 ft

Fluid - I (Math-ME Bd Oct. 1997) A perfect venturi with throat diameter of 1.8 inches is placed horizontally in a pipe with a 5 in inside diameter. Eighty (80) Ib of water now through the pipe each second. What is the difference between the pipe and venturi throat static pressure? z A. 29.91b/in C. 5020lb/in z z B. 34.81b/in D. n.3lb/in z

SOLUTION:

Velocity

~-

QIA 3300/7.481

SOLUTION

Velocity (n ! 4 )(17 / 12) 2 (60) ~

VI

Velocity = 4.66 ft/s L = 5280 ft (I mile)

QIA I 80/62.4

VI

CD

-..::c..~":.:c _ _

®

I --."""::"-.... -:.~J.~::-:'-=--(rr 14)(5/12)2 m=80lb/s V 9.4 fils --+ \ 5" ................8"1 P ~

VI

QIA z 80/62.4

Vz

V,

p,(,L~i~:,:.~~.,.;"=·~l~_~< "-~ c

_

Head loss =

f L v

2

..Y.J

12:] Q=33~Ogal/min L=1mile

2 g 0 )

Head loss

0.03(5280)(4.66) 2

2(32.2)(17/12) Head loss = 37.7/1

Vz

(n 14)(1.8/12)2

= 72.549 fils

Vz

Fluid - 3 (Math-ME Bd Apr. 1997)

Apply 109 Bernoulli's Equabon: 2

VI P2 V/ PI -+--+Zt =-+--+Z w 2g . w 2g For horizontal, z,

= Zz

PI - P2

2_V 2 V_2_ _1

w

2g

PI -

P2

62.4

(72.549)2 _(9.4)2

2(32.2) PI - Pz = 5014.281b/tY x I1144 PI - P2 = 34.82 psi

A rigid container is closed at one end and measures 8 in diameter by ] 2 in long. The container is held vertically and is slowly moved downward until the pressure in the container is 15.5 psia. What will be the depth of the water surface measure from the free. water surface'? A. 22 in C. 12 in B 9.2 in D. 9.8 in Water Surface

SOLUTION: Pabs

=

P~age + Patn.

15.5 = Pgage + 14.7 Pgage = 0.8 psi Pgage = w h 0.8(144) = 62.4(h) h = 1.846 ft x 12 h = 22.15 in

- - ---- -- --- -_ -.h

8"16 Air 15.5psia -

12"¢

316

Fluid Mechanics

Fluid Mechanics

PI -P2

Fluid - 4 (Math-ME Bd Apr. 1997)

C. 1.05 hp D. 2.54 hp

h h

V

2

)

'+(Z2- Zj

V/-O

0.075

2(32.2)

+0

V z = 73.2 ft/sec (3600/5280) V 2 = 49.9mph P/w 25 x 144

~TUrblne~gpm

62.4 57.69 ft

P,-P,=25psi

Turbine power = w Q h 7.35( 62.4 )(57 .69) Turbine power = 33,000 Turbine power = 0.802 hp

BBN

Let V = volume of stone

D. 62 mph

Rear

For the liquid PI-Pz=wh PI - P2 = (0.6 x 62.4)(2/12) PI - Pz = 6.24 Ib/ft 2 By applying the Bernoulli's equation: 1

PI VI P2 V2 - + - + Z , =-+--+Z2 W 2g w 2g

w.s.

~

Two tubes are mounted to the roof of a car. One tube points to the front of the car while the other point to the rear. The tube are connected to a manometer filled with fluid of specific gravity 0.60. When the height difference is 2 inches, what is the car's speed? A. 46 mph C. 50 mph

SOLUTION:

What is the density of a stone that weighs 19.9 Ib (88 N) in air and 12.4 Ib (55 N) in water?

--../::

SOLUTION:

Fluid - 5 (Math-ME Bd Apr. 1997)

B. 43.8 mph

Fluid - 6 (Math-ME Bd, Apr. 96)

A 2,651.2 kg/rn' C. 2,578.2 kg/rrr' 3 3D. B. 2,612.5kg/m 26,700kg/m

Q = 5517.4ISl Q = 7.35 ft 3/min

2

_

6.24

--=

SOLUTIOW

h

22

w 2g w = density of air 3 w = 0.075 Ib/ft (standard)'

The pressure drop across a turbine is 25 psi. The flow rate is 55 gallons per minute. Calculate the power output of the turbine. A. 0.802 hp B. 0.41 hp

V

317

rront

r

CD

+v;-

BBN

t

If body floats. then the weight of the object is equal to the Bouyant force. W = BF 55N BF is also the difference in weight of object in air and in water. BF = 19.9 - 12.4 BF = 7.51bs BF = wV 7.5 = (62.4) V V = 0.1202 ft3 Density of stone = 19.91b/0.1202 ft3 Density of stone = 165.561b/ft3 (1000/62.4) Density of stone = 2,653.2 kg/m'

2"

Fluid

Fluid - 7 (Math-ME Bd. Oct. 1995) The now rate of water through a cast iron pipe is 5000 GPM. The pipe is 1 ft and the coefficient of friction f = 0.0173. What is the pressure drop over a 100 ft length of pipe?

318

fluid Mechanics A. 3] 7 26 lblin z B 21078 lb/in z

C. 337.261b/ft z D. 23.7801b/ft z

. Turbine power

8.02(62.4)(69.23) =

-----'--'------'--

33,000 Turbine power = 1.05 hp

SOLUTION

5000

o o

=

7481(60) 11.139 ft3 Is

Fluid - 9 (ME Bd. Oct. 1996)

~:.D1ft

v =

,

)

a=50~Ogpm

L=100ft

v =

v = h,

319

Fluid Mechanics

SOLUTION:

=

v

0.0173(100)(14.183)2 =

hi

=

@

P J - w h P, PJ-P Z = wh 6.2(144) = (62.4 x 0.8)(h)

2gD hr

The fluid in a manometer tube is 60% water and 40% alcohol (SC 0.8). What is the manometer fluid height difference if a 6.2 psi pressure is applied across the two ends of a manometer? A 15.5 in C. 36 in B.186in D.215in

T _

2(32.2)(1) 5.404 ft of water

h

17.88 ft

=

=

SG=O.80

214.6 in

Fluid - 10 (ME Bd. Apr. 1996) Fluid - 8 (Math-ME Bd. Oct. 1995) Pressure drop across a turbine is 30 psi, the flow rate is 60 gpm. Calculate the power output of the turbine. A 0.41 hp C. 6,30 hp D. 2.54 hp B. 105 hp

An air bubble rises from the bottom ofa well where the temperature is 25°C, to the surface where the temperature 27°C. Find the percent increase in volume of the bubble if the depth of the well is 5 m. Atmospheric pressure is 101.528 Kpa, A. 49.3 C. 56.7 D. 38.6 8413

SOLUTION: SOLUTION:

h

=

o o

t,=27"C

30 x 144

h h

P/w

= =

=

P,~~..... ~

~=60gpm

624 69.23 ft 60/7.481 8.02 ft3 /min

Turbine power

=

P,·P,=30psi

P2 T2

PlY]

T2

T]

----

, J:

wh+ 101.528 PI = 9.81(5) + 101.528 PI = 150.378Kpa 150.378(Y 1 ) 101.528(Y2 ) PI

wOh 33,000

·_·-~'V2

=

(25 + 273)

(27 + 273)

5m

.. _ _' V, t,=25°C

320

fluid Mechanics. ~

V2

321

Fluid Mechanics

1491V 1 Fluid - 12 (Math-ME Bd Oct. 1998)

%lncrease in volume

=

V2 -V1

---

'\ 24 inches long rod floats vertically in water. It has a I sq. in. cross section and has a specific gravity of 0.6. What length L is submerged? A. 14.4 in C. 9.6 in B. 24 in D. 18.0 in

V1 %lnerease in volume = %lnerease in volume

=

1.491Vj

VI

-

VI 49.10%

~w

SOLUTION:

Fluid - II

For floating object,

An empty, open can is 30 cm high with a Io-cm diameter. The can, with the open end down, is pushed under water with a density of 1000 kg/m ', Find the water level in the can when the top of the can is 50 cm below the surface.

W

=

w., v,

C. 4.20 em D. 5.87 em

24"

vc;

Consider the air pressure: r'IV, = P ZV 2 10 1.325(A x 0.3) 30.3795 Pz = (0.3 - x) Pw = Pz

109.173 - 9.81x 2 .

A

= =

0.02l18m 2.12 em

':]m

I,. ~-- -.-----,1 O .

80

I

P"P,

=

Pz[A (0.3 - x)]

80-x

bo . ,- __=- x !

=

14.4 in

=

1000 [I xL)

--. ·~U t BF

10

I

-----=-------

Pw

Flow of water taking over in a pipe having a velocity of 10 m/s. Determine the velocity head of the water. A 50.1 m C. 8.2 m B. 5.1 ill D. 100 m

SOLUTION: h = vZ/2g h = (10)z 12(9.81) h = 5.1 ill

J., II . , . . ~10mJs

30.3795

(0.3- x) 112.1 16x + 2.3705 = 0

By qua' Iratie formula:

x

..A

v,

Fluid - 13 (Power-ME Bd Oct. 1999)

Consider the water pressure: t'w = wh + 101.325 P w = (0.8 - x)(9.81) + 101.325 P; = 109.173 - 9.81x

9.81x

W.~L t= I_: J ,I __ ____ =!j=__

L

L

SOLUTION:

A=1 in'---.

BF

(0.6 x 1000)( 1 x 24) A. 17.20 em B. 2.12 em

R~

Fluid - 14 (Power-ME Bd Oct. 1999) The length of pipe is 168 meters. If the pressure drop is 50 Kpa for every 30 meters, what is the total oressure drop? A. 260 kpa C. 280 kpa B. 300 kpa D. 100 kpa

,')')

Fluid Mechanics

.)-­

"OU: IJON:

PI-p"

Total Pressure Drop Total Pressure Drop

=

SO kpa (168 m!30 m)

=

280 kpa

VI

VI

Hvd ra ulic clficicncv S5%, find Q in Ii/s, Power developed 10,500 kw C 327,1 D 3623

2g

(re / 4)(0.08)2 3.98 m/s

V2

z,

(re I 4)(0.04)2 15.91 mJs

(15.91)

p,-p)

9.81 PI - P? 11

=

I

15 m -1.5 m 0.02

2

- - - ' = SOU lION

-z) 2

0.02 \/2

Ill.

A 3935 B 3e.j.52

, ? V7'-V t ­ =--~--+(z

w Z, - 7.1 Z2 - ZI

Fluid - 15 (I'O\~er-ME Bd Apr. 1999)

under head of 320

323

Fluid Mechanics

=

-(3.98)

2

+ (-1.5)

1-

Reference

2(9.81) 104.016 kpa

Brake Power ~-

Water Power 10,500 0.85 = Water Power Water Power = 12,352.94 kw Water Power = w Q h 12,35294 = 981 (Q) (320) Q = 3935 m 3/s (1000) Q = 3935 li/s

Fluid - 17 (Power-ME Bd Apr. 1999) A fluid that has a velocity of 18 m/s will have an equivalent head of:

A. 16.51 m B. 12.44 rn

C 18.34 m D. (0.34 m

SOLUTION:

h

v 2g

FIlJi,[ - 16 (l)o\\cr-i\lE Bd Apr. 1999)

(18)2

A c vl iud rira l pipe with water flowing downward at 0.02 mJ/s having lOp d ia me t cr "I' O.OS, bottom diameter of 0.04 m and height of 1.5 m. Find rhc prcxsur« between the pipe. A 104 kpa C 120 Kpa

B 97 kpa 0 143 kpa

h 2(9.81)

h

16.51 m

SUU.' IIO"-J.

Pi

VI"

P1

V,"

w

2g

-+--+zl = -+--+Z) \V

~g

­

Fluid - 18 Determine the size of pipe which will deliver 5 liters of medium oil (v = 6.10 X 10,6 m 2/s) assuming laminar now conditions.

324

522 111m

A

B 454

C. 550 mm

D. 650 mm

!TIfT!

Fluid - 20

The type of flow occupying in a 30 em diameter pipe which water 6 2/s flows at a velocity of 2.10 m/s. Use v = 1.13 X 10. m for water. A turbulent C. laminar B. constant D. none of these

SOLUTION

Q

v

A

0005

v

SOLUTION: 2

4)d V ~c o 006366 d)

dV Re ~(T[

dV

Re

v

v

For laminar flow, Re

2000

~

325

Fluid Mechanics

Fluid Mechanics

2000

=

d(0.006366 I d

2

_~

(0.30)(2.1 0)

)

Re

113

610 x 10

d 0522 m

d- 522 mm

Re

=

X

10-

0

577,52212

Since it is greater than 2000, then it is turbulent flow Fluid - 19

The power available in ajet having a cross-sectional area of II.OOS m Z with a velocity of 25.80 rn/s, A. 34 kw C 43 kw B 49 k w D. 23 kw

SOL!)TION

h

=

'/12g

(2580)2 h

----

~~' ~"" tIJijiiiI1J ..."

2(98 J) h = 33926 m

Q .~ Av Q ~ (0.005)(2580)

Q= 0129 mJ/s P = wQh P = (981)(0129)(33926) P = 42.93 kw

..

Fluid - 21 A man weighing 64 kg causes a flat area 30 em thick to be just fully submerged in a sea water (SG = 1.03). Neglecting weight the area must be:

C. 0.173 m 2 A. 0.085 m ' 2 D. 0.062 m 2 B. 0.756 m

S

p

---+ 25.80m/s

A"'O.005m

SOLUTION:

M

1

For floating object:

W W 64 A

= = = =

BF

wV

(1.03 x 1000)(0.3 x A) 0062 m)

~;i,~·

·f••

r

SF

326

Fluid Mechanics

327

Fluid Mechanics

o 778

V,

Fluid - 22

rn/s

0.055 V2

What force is exerted by water jet 50 mm diameter if it strikes a wall at the rate of 15 m/s? A. 342 N C. 764 N D. 5113 N B. 442 N

,

(rt . 4)(0.075)~

V,= 12456 mis o

)

(12456f - (0.778r P, - P, --- +-0 2(981) w P, -. P, 79 111

SOLUTION:

w

F = wQv

Q = A v

Q = [(11/4)(0.05)2](15) Q = 0.02945 m3/s

F

F

Flu id - 2-t

A jet of water 50 mm diameter with a velocity of 35 mls is being eli,charged in a horizontal direction from a nozzle mounted on a fire truck. The force required to hold the nozzle stationary is: A 1.34 KN C. 4.23 KN D. 323 KN B 240 KN

(1000)(0.02945)( I 5) 441.75 N

Fluid - 23 SOLUTION:

A 300 mm x 75 mm venturi meter is inserted in a 300 mm diameter pipeline where water flows at 55 liters/s. Neglecting friction loss, compute the drop in pressure head from the inlet to the throat. A8m C.6m B. 10 m D. 12 m

F = wQ v Q = A x v Q = [(11/4)(0.05/](35) Q = 0.0687 m 3/s

\61

SOLUTION:

F

"

V 75mm

,

v, 2

PI

Vj

w

2g

P2

V/

w

2g

-+--+Zl = -+--+Z2

PI-PO V) ---- w

2

-VI

2g 0.055

VI

(11 /4)(0.3)2

F F

Z,

Z,

(1000)(0.0687)(35) 2404.5 ~

2.404 KN

Reference

Ftuid-25

2

+(Z2- Z j )

An open storage vessel has 3 m of oil (SG = 0.82) and 6 m of water. The pressure at the bottom is: C. 83 kpa A 45 kpa D. 92 kpa B 65 kpa

328

Fluid Mechanics

Fluid Mechanics.

329

SOLlfTiON

Fluid - 27

F=--C:=-Oil--­ P = pressure at the bottom p = h, + Ww h., P = (082 x 9.81)(3) + 981(6) P = 83 kpa

3m=h.

""0

Water

6m=h w

p

A 300 0101 diameter pipe discharges water at the rate of 200 Ii/s. Point I on the pipe has a pressure of 280 kpa and 3.4 m below point 1 is point 2 with a pressure of 300 kpa. Compute the head loss between points 1 and 2. A 4.2 m C. 6.3 m B. 2.5 m D. 1.4 m

Fluid - 26 SOLUTION:

A 200 0101 diameter pipe gradually reduces to a 100 0101 diameter. The 200 0101 diameter pipe is connected to another pipe having a pressure of 600 kpa horizontally with a flow of 0.04 m 3/s. Find the pressure at the 100 0101 diameter. A 588 kpa C 566 kpa

B 642 kpa D. 598 kpa

PI

V

2

PI -

hL = ;:,OLUT10N: hL hL

PI V\2 P2 V2 2 -+--+ZI =-+--+Z2

W 2g w 2g

Ffuid -

0.04 VI

VI

(n: /4)(0.2/ 1.2732 m/s 0.04

Q=0.04m'/s

-----+

100mm

P2

j

v/

- + - + Z \ =-+--+Z2 +h L W 2g w 2g P2

+(Z\-Z2)

w

280- 300 ----+3.40 9.81 1.36 m

CD

I

p ,=280K Pa

3.4m

P,=300Kpa

2~

An object weighs 90 N in air and when immersed in water it weighs 50 N. Compute the specific gravity of the object. A. 1.25 C. 2.25 B. 3.25 D. 4.25

V2

V2

(n: 14)(0.1/

509 m/s

SOLUTION:

-W'$''--T~~~-c.:

P7 (5.09)2

600 (12732)2 -+ +0=-- + +0

9.81 2(9.81) 9.81 2(9.81)

BF = 90 - 40 BF = 50 N

Po = 587.85 kpa

BF = W Va 50 = 9810 v; V o = 0.00408 rrr'

11F 56N

Wo

weight of object in air

330

Fluid Mechanics Wo 90 w., SG SG

c.

= =

=

Past Board Examination Elements (1994-1999)

331

w, v;

"0 (0.00408) 22,058.82 N/m J 22,058.82/9810 2249

PAST BOARD EXAMINATION

ELEMENTS

Fluid - 29

Elements - 1 (ME Rd. October 1994)

A rectangular open box 7.6 m by 3 m in plan and 3.7 m deep, weighs J5U KN and is launched in fresh water. If water is 3.7 m deep what weight of stone placed in the box will cause it to rest at the bottom? /\ 35034 KN C. 498.34 KN B 65345 KN D. 477.57 KN

When a substance is gaseous state is below the critical temperature it is called: A. Vapor B cloud C. moisture

Do steam

Sl)J.l iTlON:

Elements - 2 (ME Rd. October 1994)

w.s.

Total weight

=

J."Oj W,

w V

=

BF

Is the condition of pressure and temperature at which a liquid and its vapor are indistinguishable: A. critical point B. dew point C. absolute humidity D. relative humidity

Elements - 3 (ME Rd. October (994) 3"i() W,

f

~

W, = 9.111 [(7.6)(3)(3.7)] 477.57 KN

If the temperature is held constant and the pressure is increased beyond the saturation pressure, we have a: A. saturated vapor C. saturated liquid B. compressed liquid D. subcooled liquid

Elements - 4 (ME Rd. October (994) A Francis turbine has what flow: A. inwardflow reaction B. outward flow impulse C. outward flow reaction D. inward flow impulse

332

Past Board Examination

/:'/1'1111'/11\

(1994-1999)

Past Board Examination Elements (1994-1999)

333

A. velocity of now only B. pressure only C. height above a chosen datum, density, internal energy,

Elements - 5 (ME Rd. October 1994)

pressure and velocity ojflow

The latent heat 01 vaporization in joules per kg is equal to: 2 A 5.40 x 10 3 B.4.13xl0 5 C. 22.6 X 10 5 D. 335 X 10

D. pressure, height above a chosen datum, velocity of f1ow, density of fluid

Elements - 10 (ME Rd. October 1994) Elements - 6 (ME Rd. October 1994) Form of energy associated with the kinetic energy of the random motion of large number of molecules: A. internal energy B. kinetic energy C. heat of fusion

o

A type of water turbine: A. Parson B. Hero C. Pelton D. Banki

Elements - 11 (ME Rd. October 1994)

heat

Elements - 7 (ME Rd. October 1994) In a P- T diagram of pure substance, the curve separating the solid phase from the liquid phase is: A. vaporization curve B. fusion curve C. boiling point D. sublimation point

Elements - 8 (ME Rd. October 1994) The number of protons in the nucleus of an atom of the number of electrons in the orbit of an atom: A. atomic. volume B. atomic number C. atomic weight D. atomic mass

If the pressure of the confined gas is constant, the volume is directly proportional to the absolute temperature: A. Boyle B. Joule C. Charles D. Kelvin

Elements - 12 (ME Rd. October 1994) A theoretical body which when heated to incandescence would emit a continuous light-ray spectrum: A black body radiation B. black body C. blue body D. white body

Elements - 13 (ME Rd. October 1994) Elements - 9 (M E Bd. October 1994) The energy off1uid flowing at any section in a pipeline

IS;I

Iuncuon of

Ignition of the air fuel mixture in the intake of the exhaust manifold: A. backlash B. backfire

334

Past Board Examination Elements (1994- / 999) C. exhaust pressure D. back pressure

Elemer..ts - 14 (ME Rd. October 1994)

Is the condition of pressure and temperature at which a liquid and its vapor are indistinguishable: A. relative humidity B. absolute humidity C. critical point D. dew point

Elements - 15 (ME Rd. October 1994)

When a substance in gaseous state is below its critical temperature it is called: A. steam B. cloud C. moisture D. vapor

Elements - 16 (ME Rd. October 1994)

.

Which of the following a set of standard condition A. 1 atm, 255k, 22.41 m 3/kg mole B. 101.325,273 'k, 22.4 m'rkg mole C. 101.325, 273°k, 23.66 m3/kg mole D. 1 arm, lOoC, 22.41 m 3/kg mole

Elements - 17 (ME Rd. October 1994)

Number of molecules in a mole of any substance is a constant called: A. Rankine cycle B. Avogadro's number C. Otto cycle D. Thompson constant

Pas: Board Examination Elements (/994-1999)

335

Elements - 18 (ME Rd. October 1994)

A simultaneous generation of-electricity and steam (or heat) in a single power plant: A. gas turbine B. steam turbine-gas turbine plant C. waste heat recovery D. cogeneration

Elements - 19 (ME Rd. October 1994)

Is one whose temperature is below corresponding to its pressure: A. compression B. condensation C. constant volume process D. subcooled liquid

the

saturation' temperature

Elements - 20 (ME Rd. October 1994)

Pump used to increase air pressure above normal, air is then used as a motive power: A. air cooled engine B. air compressor C. air condenser D. air injection

Elements - 21 (ME Rd. October 1994)

If the temperature is held constant and the pressure is increased beyond the saturation pressure, we have a: A. compressed liquid

B. subcooled liquid C. saturated %por D. saturated liquid Elements - 22 (ME Rd. October 1994)

The locus of elevations:

336

Past Board Examination Elements (1994-/999) A. B. C. D.

337

critical point

hydraulic gradient energy gradient friction gradient

Elements - 23 (ME Ed. October 1994) In sensible cooling process, the moisture content: A. does not change B. decreases C. indeterminate D. increases

Elements - 24 (ME Bd. April 1995) What is the A. B. C. D.

Past Board Examination Elements (1994-1999)

process that has no heat transfer? reversible isothermal polytropic

adiabatic

Elements - 25 (ME Ed. April 1995) The internal combustion engines never work on_ _cycle: A. Rankine B. diesel C. dual combustion D. Otto

Elements - 27 (ME Bd. April 1995) What is the force which tends to draw a body toward the center about which it is rotating? A. centrifugal force B. centrifugal in motion C. centrifugal advance D. centripetal force

Elements - 28 (ME Bd. April 1995) A simultaneous generation of electricity and steam (or heat) in a single power plant: A. steam turbine - gas turbine B. cogeneration C. gas turbine plant D. waste heat boiler

Elements - 29 (ME Bd. April 1995) Percent excess air is the difference between air actually supplied and theoretically required divided by: A. tlle theoretically air supplied B. the deficiency air supplied C. gas turbine plant D. waste heat boiler

Elements - 30 (ME Bd. April 1995) Elements - 26 (ME Ed. April 1995) The dividing point between the high-pressure arid low pressure sides of the refrigeration cycle occurs at the: A. expansion 'valve

B. compressor C. condenser D. cooling oil

What amount of air is required in a low bypass factor? A. greater B. lesser C. indeterminate D. does not change Elements - 31 (ME Bd, April 1995) Work done per unit charge when charge is moved from one point to another:

331
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B. cd-cl C. cd D. cd

Elements - 228 (ME Board April 1999) If the exhaust lowered or the boiler is raised the moisture content of steam: A. vaponzes B. decreased C. liquifies D. increase

Elements - 229 (ME Board April 1999) The relative humidity become [00% and where the water vapor starts to condensate. A. critical point B. dew point C. saturated point D. cold point

Elements - 230 (ME Board April 1999) For reciprocating compression slip at positive displacement. A. cd = ]

=

0

= [

Elements - 233 (ME Board April 1999) When the number of reheat stages in a reheat cycle is increased, the average temperature: A. increases B. constant C. decreases D. zero

Elements - 234 (ME Board April 1999) When the boiler pressure increases or when the exhaust pressure decreases, the amount of moisture: A. increases B. constant C. decreases D. zero

Elements - 235 (ME Board April 1999)

B. cd