• Author / Uploaded
• Bong Padilla

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I.

Problem Solving 1.

An ideal Otto engine with 15% clearance operates on 20 lbm/min of air, initially at 14 psia and 120℉. It uses rf/a = 0.0556 lb fuel/lb air with a lower constant volume heating value at 0°� of qlo = 19,100 Btu/lb fuel. Determine (a) its displacement volume and the mass drawn in during suction for a volumetric efficiency of 75%; (b) the temperatures and pressures at the corners of the cycle; (c) the horsepower and thermal efficiency; and (d) the mep and the brake engine efficiency for a brake thermal efficiency of 24%. Use air table values for the air standard.

2.

(a) Determine the approximate cylinder dimensions of a four-cylinder, four stroke gasoline engine to deliver 100 bhp at 2000 rpm; diameter/stroke ratio = D/L = 1. From experience, it is expected that the bmep ≈ 120 psi, the mechanical efficiency �� ≈ 80%, specific fuel consumption mfb ≈ 0.55 lb/bhp-hr, lower heating value of fuel q ≈ 19,000 Btu/lb. Let rk = 6.5 and use an average k = 1.32 for e. (b) calculate the expected eb and the indicated engine efficiency. Given:

D =1 L

n = 4 cylinders

m fb =0.55

lb bph-hr

q l =19000

BTU lb

4 SCE N

BP=100 bhp

=

200 rpm 2 rev/cycle

Bmep = 120 psia

ηm = 80%

r k =6.5 k = 1.32

Solution: A. Approximate cylinder dimensions

BP= Bmep ( V D )

ft-lb 2 100 bhp s 1 ft V D= × × 2 lb 1 hp 144 in 120 2 in 550

ft V D = 3.1829 s

3

(

)

2

( ) LN n n ft πD 200 rpm 3.1829 =( D( (4 ) ) s 4 2 ( 60 ) ) VD =

πD 4

c

3

p

2

*Shift Solve: D = 0.847 ft or 10.165 in L= 0.847 ft or 10.165 in

B.

e b and the indicated engine efficiency

e tb =

m fb =

BP mf Q H m f ( 3600 ) Pb

0.55 mf =

lb bph-hr

3600

m f = 0.01523

e tb =

BP mf Q H

0.707 BTU/s ( ) 1 bhp = lb BTU 0.01523 (19000 s lb ) 100bhp

e tb

lb s

e tb = 24.43 %

e = 1-

1 r kk-1

e = 1-

1 6.51.32-1

e = 45.06 %

eb =

e tb ×100 = e

0.2443 ×100 0.4506

e b = 54.22 %

3.

An engine designer desires to use a fuel tank that will hold a minimum fuel supply for 1 hr operation for a 6 cylinder 9.21 x 8.89 cm, automotive type gasoline engine which at maximum output develops a brake torque of 267 N-m at 3000 rpm. The corresponding ideal engine showed a thermal efficiency of 56.5%; the brake engine efficiency is expected to be 53%. For the fuel, q = 43,269 kJ/kg and its specific gravity is 0.715. Find the smallest tank that will satisfy. Given:

n c =6 cylinders

N = 3000 rpm

CV=43,269

e t =56.5 %

Time = 1 hr

S.G. = 0.715

T b = 267 N-m

e b = 53.0 %

Solution:

BP=2πTN

BP=2π ( 267 N-m )

BP = 83.88 kW

rpm (3000 ) 60 s

KJ kg

eb et

eeb =

BP m ( cv ) eeb = s et 83.88 m ( 43296 ) 0.53= Sf 0.565 *Shift solve

m Sf = 6.47×10 -3

kg ( 3600 s ) s

m f =23.31 kg

ρf =715

kg m3

mf Vf

ρf =

715=

23.31 Vf

V f = V tank = 0.033 m 4.

3

A 6-cylinder, 4 stroke cycle, single acting, spark ignition engine with a compression ratio of 9.5 is required to develop 67.14 kW with a torque of 194 Nm. Under the conditions, the mechanical efficiency is 78% and the brake mep is 552 kPa. For the ideal cycle, p1 = 101.35 kPaa, t1 = 35 ℃, and the hot-air standard k = 1.32. If D/L = 1.1 and the fuel rate is mfi = 0.353 kg/kW-hr (q1 = 43,967 kJ/kg), determine (a) the bore and stroke, (b) the indicated thermal efficiency, (c) the brake engine efficiency, and (d) the percentage clearance. Given:

nc =6 cylinders

e m =78 %

4 SCE

Bmep = 552 kPa

r k =9.5

P1 =101.35 kPa

P = 67.14 KW

t 1 =35

Solution:

P=2πTN 67.14 ( 60×106 ) =2π ( 194×10 3 ) N N = 3304.844 rpm

67.14 KW 552 kPa

V D =0.1216

m3 s

A. Bore and Stroke

VD =

πD 2 L N nc n p 4

( )

0.1216

m3 π 3304.844 ( ) = (1.1L)2 (L) 6 s 4 2(60)

(

*Shift Solve: L = 0.0918 m D = 1.1 ( 0.0918 m) = 0.1010 m B. Indicated Thermal Efficiency

e ti =

IP m f QH

: where

IP=

BP em

D/L = 1.1

m f1 =0.353

kg KW-hr

q 1 =43,967

KJ kg

C

T = 194 N-m

V D=

k=1.32

)

67.14 KW =86.077KW 0.78

IP=

e ti =

IP 3600 = m f QH kg KJ 0.353 43,967 kWh kg

(

)

e ti =0.2320 or 23.20 % C. Break Engine Efficiency

eb =

e tb ×100 e

e tb =

BP mf Q H

m f =0.353

kg 1hr ( 86.077kW ) kWh 3600s

(

m f =8.4403×10

e tb = e tb =

-3

kg s

BP mf Q H

67.14kW 8.4403×10-3

kg KJ 43967 s kg

(

)

e tb =0.1809 or 18.09 % e=1−

eb =

1 rk

k−1

=1−

1 9.5

1.32−1

0.1809 ×100 0.5134

e b =0.3524 or 35.24 % D. Percentage Clearance

rk=

C+1 C

=0.5134

)

9.5=

C+1 C

*Shift Solve 5.

C = 0.1176 or 11.76 % A four-stroke single-cylinder engine has a bore of 160 mm and a stroke of 310 mm. When tested at 720 rev/min the net load on the rope brake is 610 N and the drum diameter is 0.98 m. The indicator diagram has a net area of 634 mm2 and the length of 77 mm with a spring rating of 0.9 bar per mm. Determine for the engine: a) Indicated mean effective pressure b) Indicated Power c) Shaft power d) Friction power e) Mechanical efficiency f) Engine capacity g) Mean piston speed

Given: 4 SCE

F = 610 N

l = 77 mm

D = 160 mm

d = 0.98 m k = 0.9

L = 310 mm

A = 634

2

bar mm

mm

N = 720 rpm Solution: A. Indicated mean effective pressure

IMEP =

S × Ai Li

bar ( 634mm 2 ) mm IMEP = 77mm 0.9

IMEP = 7.4 bar B. Indicated power

IP = IMEP×A×L×N π 720rpm IP = 7.4 bar ( 160mm2 ) ( 310mm ) 4 2(60)

()

IP = 27.674 kW C. Shaft power

BP = 2πTN

(

)(

0.1 N/ mm2 1 bar

)(1100m m )(11000kN N )

[ ( )](

BP = 2π 610 N

0.98 2

720

rev min

)( 160mins )(11000kN N )

BP = 22.537 kW D. Friction power

FP = IP - BP

FP = 27.674 - 22.537 FP = 5.137 kW E. Mechanical efficiency

em =

BP IP

em =

22.537 27.674

e m = 0.8144 or 81.44 % F.

Engine capacity

V D = ALN VD =

[

π 720 rpm ( 0.16m)2 ( 0.31 ) 4 2 ( 60 )

V D = 0.0374

m s

]

3

G. Mean piston speed

MPS =

2LN 60

MPS =

( 2 ) ( 0.310 m )( 720 rpm ) 60

MPS =7.44

6.

m s

A two-cylinder two-stroke engine has a bore of 72 mm and a stroke of 129 mm. the shaft power developed is 12.5 kW at 1600 rev/min and the mechanical efficiency of 84%. If the engine consumes 3.66 kg of fuel with calorific value of 43000 kJ/kg during one hour test, determine: a) Capacity of the engine b) Indicated mean effective pressure c) Brake thermal efficiency d) Specific fuel consumption

Given:

n c =2 cylinders

L = 129 mm BP = 12.5 kW

m f =3.66

kg hr

2 SCE N = 1600 rpm D = 72 mm

e m =84 %

Solution: A. Capacity of the engine

V D =LAN nc

V D = ( 0.129 m )

V D =0.028 B.

m s

[

π (0.072 m) 4

2

][

3

Indicated mean effective pressure

IMEP=

IP VD

12.5 kW 0.84 kW IMEP= m3 0.028 s IMEP=5310463 kP a C. Brake thermal efficiency

e tb = e tb =

BP mf Q H

12.5 kW kg 1 hr 3.66 hr 3600 s

(

)( 43,000 KJkg )

e tb =0.2859 or 28.59 % D. Specific fuel consumption

ISFC=

mf IP

]

1600 rpm (2) 60 s

CV=43,000

KJ kg

kg hr ISFC= 14.881 kW 3.66

ISFC = 0.246

BSFC=

kg kW-hr

mf BP

kg hr BSFC= 12.5 kW 3.66

BSFC= 0.2928

7.

kg kW-hr

A test conducted on a four-stroke four-cylinder petrol engine having a bore of 73 mm and a stroke of 95 mm gave the following data: Engine speed = 3400 rpm Dynamometer torque = 70 N-m Indicated mean effective pressure = 6.7 bar Rate of fuel consumption = 8.5 kg/hr Calorific value of fuel = 42,500 kJ/kg Clearance volume = 56.8 cc Calculate the following for the engine: a) Shaft power, indicated power and mechanical efficiency b) Indicated and shaft thermal efficiencies c) Specific fuel consumption d) Air standard and relative efficiencies Given: 4 SCE

n C =4 cylinders

N = 3400 rpm T = 70 N-m IMEP = 6.7 bar

D = 73 mm L = 95 mm

CV=42,500

m f =8.5

kg hr

Solution: A. Shaft power, indicated power and mechanical efficiency

V C =56.8 cc

KJ kg

BP = 2π ( 70 N-m )

BP = 2πTN

rpm (3400 ) 60

BP = 24,923.3017 W

BP = 24.923 kW

IP=IMEP×A ×L ×N N 2 mm IP= ( 6.7 bar ) 1 bar

(

0.1

)( )

[

IP = 30.1920 kW

em =

BP IP

em =

24.923 kW 30.1920 kW

e m = 0.8255 or 82.55 %

B. Indicated and shaft thermal efficiencies

e ti = e ti =

IP mf QH

30.1920 kW kg 1 hr KJ 8.5 42,500 hr 3600 s kg

(

)(

)

e ti = 0.3009 or 30.09 % e tb = e tb =

BP mf Q H 24.923 kW kg 1 hr 8.5 hr 3600 s

(

)( 42,500 KJkg )

e tb = 0.2484 or 24.84 %

C. Specific fuel consumption

BSFC =

mf BP

] (

π 3400 rpm 1 kN (73 mm )2 ( 95 mm ) ( 4) 4 1000 N 2 ( 60 )

m )(11000 mm )

kg hr BSFC = 24.923 kW 8.5

BSFC = 0.34105

ISFC =

kg kW-hr

mf IP

kg hr ISFC = 30.1920 kW 8.5

ISFC = 0.2815

kg kW-hr

D. Air standard and relative efficiencies

at e a

ea =1rk=

1 ×100 % r k-1 k

V D + VC VC

π 2 ( 7.3 cm) ( 9.5 cm ) +56.8 cc 4 rk= 56.8 cc r k = 8.00 ea =1-

1 ×100 % k-1 rk

ea =1-

1 81.4-1

ea = 0.5731 or 57.31 % at e r e r-ti =

e ti ×100 % ea

e r-ti =

0.3009 ×100 % 0.5731

e r-ti = 0.5250 or 52.50 %

8.

During the test on the engine in problem 3 it is found that 100 cc of petrol of specific gravity 0.79 is consumed in 22.6 seconds. The air flow is measured using a large surge tank fitted with an orifice plate of 50 mm diameter and having a coefficient of discharge of 0.67. The pressure difference across the orifice is found to be 43 mm of water. If atmospheric conditions in the test room are 1.013 bar and 24℃, determine: a) Mass flow rate of air consumption b) Mass flow rate of fuel consumption c) Air / fuel ratio d) Volumetric efficiency Is the engine operating with a rich or a weak mixture?

9.

The following data were obtained during a test on a diesel engine operating at 2800 rpm: Dynamometer torque measurement = 132 N-m Fuel consumption = 12 kg/hr Intake air flow rate = 168 kg/hr Calorific value of fuel = 44,000 kJ/kg Engine cooling water flow rate = 15.4 kg/min Engine cooling water temperature rise = 48 C Exhaust calorimeter cooling water flow rate = 8.7 kg/min Exhaust calorimeter cooling water temperature rise = 64 C Exhaust gas temperature at exit from calorimeter = 85 C Specific heat of exhaust gas = 1.1 kJ/kg-K Laboratory air temperature = 25 C Draw up a percentage-based energy balance for the engine using the laboratory temperature as the datum.

Given:

T= 132 N-m

m f = 12

kg hr

∆T w = 48 ℃ t exhaust = 64 ℃ rise

kg m a = 168 hr KJ CV= 44,400 kg

m exhaust = 8.7

kg min

Exhaust gas temperature at exit= 85 ℃

Cp exhaust gas= 1.1 m w = 15.4

kg min

KJ kg-K

Laboratory out temperature= 85 ℃ Solution:

Qa = m f Qh

(

Qa = 12

kg hr

Qa = 148

)(44,400 KJkg )(1hr 3600s )

KJ kg

Qcw = 15.4

kg KJ 1min 4.187 ( 48 ) min kg-K 60s

(

Qcw = 51.58384

) (

)

KJ s

Qexhaust = m w Cpw ∆T+ m g Cpg Tg

Qexhaust = 8.7

kg KJ 1min kg 4.187 ( 64 ) + 12+168 min kg-K 60s hr

Qexhaust = 42.155

(

) (

)(

KJ s

BP= 2πTN

BP= 2π ( 132 N-m )( 2800 rpm )

BP= 38.704 KW

1min (1KN )( 1000N 60s )

KJ 1.1 ( 85.25 ) )(1hr )( 3600s kg-K )

Losses

Qa (KJ/s)

Item

Heat Loss (KJ/s)

%

Cooling Water

51.58384 KJ/s

148 KJ/s

34.85%

Exhaust

42.155 KJ/s

148 KJ/s

28.48%

Losses (surrounding)

15.55716 KJ/s

148 KJ/s

9.5133%

BPower

38.704 KJ/s

148 KJ/s

3.8239%

10. A petrol engine using fuel with a calorific value of 42,700 kJ/kg was tested at constant speed. When the air/fuel ratio was varied, the following results were obtained: Fuel rate kg/hr 16.3 16.7 18.3 20.0 21.6 23.0 24.7 Air rate kg/hr 270 269 264 268 273 277 278 Shaft power kW 52.6 54.3 59.8 63.1 65.8 66.8 66.4 Plot suitable characteristic curves for the engine and use them to determine: a) Shaft power and air/fuel ratio at the point of maximum shaft thermal efficiency b) Shaft thermal efficiency, specific fuel consumption and air/fuel ratio at the point of maximum c)

shaft power Minimum specific fuel consumption and maximum shaft thermal efficiency

E. Design & Open-Ended Problems: Exploring Engineering Practice 1.

Automotive gas turbines have been under development for decades but have not been commonly used in automobiles. Yet helicopters routinely use gas turbines. Explore why different types of engines are used in these respective applications. Compare selection factors such as performance, power-to-weight ratio, space requirements, fuel availability, and environmental impact. Summarize your findings in a report with at least three references.

2.

Investigate the following technologies: plug-in hybrid vehicles, all-electric vehicles, natural gas-fueled vehicles, and ethanol-fueled vehicles, and make recommendations on which of these technologies should receive federal research, development, and deployment support over the next decade.