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• Maria Lourdes Cagungun

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TRANSPORT MECHANICS

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Which of the following denotes the effect of compressibility in fluid flow? a. Reynolds Number c. Mach Number b. Weber Number d. Peclet Number  In a compressible fluid flow, the fluid density varies with its pressure. Compressible flows are usually high speed flows with Mach numbers greater than about 0.3. Which of the following meters will have the highest permanent pressure loss? a. Pitot tube b. orifice c. venture d. rotameter  Orifice plates cause a high permanent pressure drop wherein the outlet pressure will be 60% to 80% of inlet pressure, and also subject to erosion which will later cause inaccuracies in the measured differential pressure. The Fanning friction factor for fluid flow in pipe does not depend upon the a. pipe length c. fluid density & viscosity b. pipe roughness d. fluid’s mass flowrate  Fanning friction factor is a dimensionless number used as a local parameter in continuum mechanics calculations. It is defined as the ratio between the local shear stress and the local flow kinetic energy density given by the local flow velocity and the density of the fluid. The differential height between two points where a fluid has to be transferred is the _____head. a. dynamic

b. velocity

c. pressure

d. potential

 Potential head is given by the difference in elevation between two points in a fluid. 5.

A measure of the resistance of a fluid to shear or angular deformation a. viscosity

b. fanning friction

c. roughness d. surface tension

 Viscosity defines the internal resistance to flow and shear of a fluid. 6.

Pressure level measured with respect at atmospheric pressure are termed as a. absolute pressure b. vacuum

c. gage pressure

d. differential pressure

 Gage pressure is the pressure relative to the atmospheric pressure. In other words, how much above or below is the pressure with respect to the atmospheric pressure. 7.

Blood is an example of a_______fluid a. Pseudoplastic

b. Bingham Plastic

c. Newtonian

d. Dilatant

 Blood is a fluid whose apparent viscosity or consistency decreases instantaneously with an increase in shear rate. Therefore it is a pseudoplastic fluid. 8.

Which of the following fittings exhibits the highest pressure drop for the same flow conditions? a. 900 long radius elbow b. 450 standard elbow C. 900 standard elbow d. square corner elbow  The equivalent length is directly proportional to the pressure drop. Among the choices, the squarecorner elbow has the highest equivalent length therefore exhibiting the highest pressure drop.

9.

For the transfer os solution of corrosive or thick slurry, the pump used is a_______pump. a. reciprocating

b. centrifugal

c. diaphragm

d. gear

 Diaphragm pumps are self-priming and are ideal for viscous liquids. 10.

The head loss in turbulent flow is a pipe varies a. as velocity

c. as 1/(diameter)2

b. 𝐚𝐬 (𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲)𝟐

d. inversely as the velocity

 The head loss in a turbulent flow varies approximately as the square of the velocity. 11.

When the ID of the pipe through which a fluid flows is doubled, the pressure loss due to friction per unit length of pipe and for the same volumetric flowrate is changed by approximately factor of a. 1/8

b. 1/16

c. 1/32

d. 1/64

 In doubling the diameter of a pipe, the head loss is decreased by a factor of 32 thus the energy consumed in moving a given volumetric flow of the fluid is cut down dramatically for a modest increase in capital cost. 12.

This event occurs when vapor pockets form in a liquid flow is doubled, the pressure loss due to friction per unit length of pipe and for the same volumetric flowrate is changed by approximately factor of a. water hammer

c. boiling point elevation

c. cavitation

d. flow separation

 Cavitation happens when the pressure in a liquid suddenly drops. The drop in pressure is caused by pushing a liquid quicker than it can react, leaving behind an area of low pressure often as a bubble of gas. 13.

The general purpose of which is to measure the difference pressures a. manometer

b. venture

c. barometer

d. orifice

 Manometers are devices used to measure the difference in pressure between two points. 14.

According to the affinity laws for dynamic types pumps, this varies with the square of the impeller diameter and speed a. power

b. head

c. capacity

d. flowrate

 The Affinity Laws of centrifugal pumps or fans indicates the influence on volume capacity, head and/or power consumption of a pump or fan due to the change in speed of wheel - revolutions per minute and due to geometric similarity - change in impeller diameter. 15.

Lower BWG means a. lower thickness pump

c. lower cross-section of tube

b. outer diameter of a tube

d. inner diameter of tube

 Birmingham Wire Gauge is a notation for the diameters of metal rods. A cross reference between BWG, imperial sizes and metric equivalents, in terms of tube wall thickness. Therefore the lower BWG is the lower cross section. 16.

Horsepower requirement for a given pump capacity depends upon the a. specific gravity of the liquid

c. discharge head

b. suction lift

d. all of the above

 All these parameters are related by the formula BHP = 17.

𝑄(𝑇𝐷𝐻)(𝑆𝐺) 3960𝜀

At the plane’s nozzle, the air temperature is 400 K and the plane’s velocity is 450 m/s. The plane’s speed is at a. subsonic

b. sonic

c. supersonic

d. hypersonic

Given: T air = 400K , plane’s velocity = 450 m/s Req’d: speed of the plane Solution: 𝑀=

speed of object 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑠𝑜𝑢𝑛𝑑@400𝐾

𝑀=

450m/s 401.028𝑚/𝑠

𝐌 = 𝟏. 𝟏𝟐𝟐𝟏 Answer: M>1 therefore, supersonic

18.

It offers physical explanations why atmospheric pressure changes with altitude, why wood and oil float on water and why the surface of water is always flat and horizontal whatever the shape of its container. a. Newton’s Law

c. Bernoulli’s Equation

b. Fluid Mechanics

d. Hydrostatic Principle

 The principle of hydrostatic equilibrium is that the pressure at any point in a fluid at rest (whence, “hydrostatic”) is just due to the weight of the overlying fluid. 19.

Fittings that join two pieces of pipe a. elbows

b. flanges

reducers

d. couplings

 In plumbing, a coupling is a pipe fitting that connects two pipes. 20.

Consider two pipes of same length and diameter through which water is passed at the same velocity. The friction factor for rough pipe is f1 and that for smooth pipe is f2 . Pick out the correct statement. a. f1 = f2

b. f1 < f2

c. 𝐟𝟏 > 𝐟𝟐

d. f1 = 2f2

 Since the two pipes have the same length, the same diameter, and the water is passing in the same velocity, pipe roughness will be the deciding factor for friction. 21.

Hydraulic radius of a 3 ft equilateral triangle channel below is a. 0.65 ft

b. 0.51 ft

Given:

2 ft

2 ft

3 ft

c. 0.49 ft

d. 0.38 ft

Req’d: hydraulic radius Solution: 𝐚𝐫𝐞𝐚 𝐨𝐟 𝐭𝐫𝐢𝐚𝐧𝐠𝐥𝐞

rH = 𝐰𝐞𝐭𝐭𝐞𝐝 𝐩𝐞𝐫𝐢𝐦𝐞𝐭𝐞𝐫 = 22.

(𝟎.𝟓)(𝟑)(𝐬𝐢𝐧𝟔𝟎)(𝟑) 𝟏+𝟑+𝟒

= 𝟎. 𝟒𝟖𝟕𝟏 𝐟𝐭 ≈ 𝟎. 𝟒𝟗 𝐟𝐭

Oil with a viscosity of 30 cP and a density of 60 lb/ft 3 flows through a ½ in inside diameter pipe. Determine the velocity in ft/s below which flow will be laminar. a. 87.20

b. 0.63

c. 13.10

d. 16.90

Given: Oil µ = 30 cP = 0.0201168 lb/ft-s ρ = 60 lb/ ft3 d = ½ in = 1/24 ft Req’d: velocity in ft/s below which flow will be laminar (Re = 2100) Sol’n: 𝑅𝑒 =

𝑑𝑣𝜌 µ

𝑣=

𝑅𝑒µ 𝜌𝑑

lb ) 𝑓𝑡 − 𝑠 1 𝑙𝑏 (24 𝑓𝑡) (60 3 ) 𝑓𝑡

(2100) (0.0201168 𝑣=

𝐯 = 𝟏𝟔.898112 ft/s≈ 𝟏𝟔. 𝟗𝟎 𝐟𝐭/𝐬 23.

A Bingham fluid of viscosity µ = 10 Pa. s and yield stress, τ0 = 10 kPa, is shared between flat plate parallel plates separated by a distance of 10−3m. The top plate is moving with a velocity of 1 m/s. The shear stress on the plate a. 10 kPa

b. 20 kPa

Given: viscosity, µ = 10 Pa · s distance, x = 10-3m

c. 30 kPa

d. 40 kPa

yield stress, s = 10KPa velocity, v = 1m/s

Req’d: shear stress on the plate Sol’n: 𝜏 = 𝜏0 + 𝜇

𝑑𝑢 𝑑𝑦

𝟏𝟎 𝐏𝐚 𝟏𝐤𝐏𝐚 𝛕 = 𝟏𝟎𝐤𝐏𝐚 + ( −𝟑 ) ( ) = 𝟐𝟎 𝐤𝐏𝐚 𝟏𝟎 𝐦 𝟏𝟎𝟎𝟎 𝐏𝐚

24.

A piece of glass weighs 258.6 g in air, 152.6 g in water at 40 C and 92 g in sulphuric acid. Calculate the specific gravity of the sulphuric acid. a. 0.85

b. 1.57

c. 2.44

d. 3.12

Given: mass of glass in air = 258.6 g mass of glass in water = 152.6 g mass of glass in sulphuric acid = 92 g Required: SG of H2SO4 Solution: Weight of glass(in air – in H2SO4) = 258.6g – 92g = 166.6g of displaced H2SO4 Weight of plummett (in air - in water) = 258.6 g – 152.6 g = 106 g of displaced water SG of H2SO4 = 166.6 g/ 106 g = 1.5717 25.

One of the King Hero’s crowns was found to have weight 13N in air. What is its specific gravity if Archimedes found it weighing 11.5N in water? a. 10.83

b. 11.23

c. 12.43

d. 14.72

Given: W in air = 13N W in H2O = 11.5N Req’d: SG Sol’n: Fb = 13N – 11.5N = 1.5N = 1000(9.81)(Vb) ρcrown =

13𝑁 9.81𝑁

/

1.5 9810

ρcrown = 8666.66667 kg/m3 SG =

8666.6667 1000

SG = 8.6667 26.

An ore sample weighs 15N in air. When the sample is suspended by light chord and totally immersed in water, the tension in the cord is 10.80N. Find the total volume of the sample. a. 4 × 10−4 m3 b. 5.67 × 10−4 m3

27.

c. 3.2 × 10−4 m3

d. 𝟒. 𝟐𝟖 × 𝟏𝟎−𝟒 𝐦𝟑

A wooden cube that is 15 centimeters on each side with a specific weight of 6300 N/m3 is floating in fresh water (γ = 9810N/m3 ). What is the depth of the cube below the surface? a. 8.75 cm

b. 9.12 cm

c. 9.63 cm

d. 10.12 cm

Given: S = 15 cm 6300N3 𝛾𝑐𝑢𝑏𝑒 = m 𝛾𝑠𝑒𝑎 = 9810N/m3 Required: Depth of Cube Solution: 0.15𝑚 (

6300𝑁 9810𝑁 ) = 𝑥( ) 3 𝑚 𝑚3

X = 0.0963m = 9.63 cm 28. A hollow plastic sphere is held below the surface of a fresh water lake by a cable anchored to the bottom of the lake. The sphere has a volume of 0.300 m3 , and the tension on the cable is 900N. Calculate the mass of the sphere. a. 208.25 kg

b. 302.45 kg

c. 435.35 kg

d. 307.65 kg

Given: volume of sphere = 0.300 m3 F = 900N Req’d: mass of the sphere Sol’n: F=

g m gc

N kg(mass) mass = 91.7431 kg m ρ= v kg m 1000 3 = m 0.3 m3 mass = 300 kg ∗∗∗ 𝐦𝐚𝐬𝐬 = 𝟑𝟎𝟎 𝐤𝐠 − 𝟗𝟏. 𝟕𝟒𝟑𝟏 𝐤𝐠 = 𝟐𝟎𝟖. 𝟐𝟓𝟔𝟗 𝐤𝐠 900N = 9.81

29.

In a natural gas pipeline at station 1 the pipe diameter is 2 ft and the flow conditions are 800 psia, 600 F and 50 ft/s velocity. At station 2, the pipe diameter is 3 ft and flow conditions are 500 psia, 600 F. What is the mass flowrate in kg/s? a. 184

b. 175

Given: Condition 1: Diameter: 2ft Pressure: 800 psia Temperature= 60o F Velocity:50 ft/s Req’d: Mass flowrate, m Solution: solution 1:

Condition 2: Diameter: 3ft Pressure: 500 psia Temperature= 60o Velocity:?

c. 198

d. 168

𝜌=

1 atm lb x 16 14.7 psia lbmol ft3 atm 0.7302 x 520R lbmol K

800 psia x

lb ft3

= 2.2932

m1= m2 m1= 𝜌1 A1 V1 m1= 2.2932

lb ft3

π

x 4 (2ft 2 ) x 50

ft s

x

1 kg 2.204 lb

= 163.4369 kg/s

solution 2: 𝜌 1=

𝜌 2=

1 atm lb x 16 14.7 psia lbmol ft3 atm 0.7302 x 520R lbmol K

= 2.2511 ft3

1 atm lb x 16 14.7 psia lbmol ft3 atm 0.7302 x 520R lbmol K

= 1.3911

(800−14.7)psia x

(500−14.7)psia x

lb

lb ft3

Using Bernoulli equation: lbf

(800−14.7) x 144 2 ft lb

2.2511 3 ft

V2 = 49.1645

lbf

−

(500−14.7) x 144 2 ft lb

1.3911 3 ft

ft s

(50 )2 −V22

+

lbm ft

2 x 32.174 2 s lbf

ft s lb

π

lb

π

m1= = 2.2511 ft3 x 4 (2ft 2 ) x 50

ft s

x

1 kg 2.204 lb

m2= = 1.3911 ft3 x 4 (3ft 2 ) x 49.1645 mave = 30.

=0

160.4365+219.3465 2

ft s

= 160.4365 kg/s 1 kg

x 2.204 lb = 219.3465 kg/s

kg/s = 189.8915 kg/s ≈ 184 kg/s

A hose shoots water straight up a distance of 2.5 m. The end opening on the hose has an area of 0.75 cm2. How much water comes out in minute? a. 13.5 L

b. 31.5 L

c. 41.5 L

d. 41.5 L

Given: Z=2.5 m

0.75 𝑐𝑚2 Req’d: volume of water coming out of the hose Sol’n: 1m 2 m q = 0.75 cm2 ( ) √2 (9.8 2 ) (2.5 m) 100 cm s 𝟑 𝐦 𝟏𝟎𝟎𝟎𝐋 𝐪 = 𝟓. 𝟐𝟓 × 𝟏𝟎−𝟒 × × 𝟔𝟎𝐬 = 𝟑𝟏. 𝟓𝐋 𝐬 𝟏𝐦𝟑

31.

Water is flowing in a pipe of varying cross-sectional area and at all points the water completely fills the pipe. The cross-sectional area at point 1 is 0.80m2 , and the velocity is 3.5 m/s. Compute the fluids velocity at point 2 where the cross-sectional area is 0.60m2 . a. 5.22 m/s

b.4.67 m/s

c.3.25 m/s

d.2.57 m/s

Given:  

@point 1: area= 0.8 m2, velocity = 3.5 m/s @point 2: area= 0.6 m2

Required: 

Velocity @ point 2

Solution: MASS FLOWRATE @ 1 = MASSFLOWRATE @ 2 V1P1S1 = V2P2S2 V2 =

𝑉1 𝑆1 𝑆2 𝟎.𝟖∗𝟑.𝟓 𝟎.𝟔

V2 = 32.

= 6.67 m/s

A cylindrical pipe with water flowing downward at 0.02 m3 /s having top diameter of 0.08 m, bottom diameter of 0.04 m and height of 1.5 m. Find the pressure between the pipe. a. 94 kPa

b. 104 kPa

c. 124 kPa

Given: top diameter of cylinder = 0.08m Bottom diameter of cylinder = 0.04m Height of cylinder = 1.5m Velocity of flowing water = 0.02 m3 /s Req’d: pressure between pipe Sol’n: Given: Q= 0.02 m3/s Dtop= 0.08 m Dbottom= 0.04 m Req’d: pressure between the pipe, ∆P Sol’n: P1 αV12 P2 αV22 + Z1 + + ηWp − Hex − Hloss = + Z2 + ρg 2g ρg 2g P1 − P2 αV22 αV12 = − − Z1 ρg 2g 2g

d. 134 kPa

V1 = V1 =

0.02 m3 /s π (0.08 m)2 4

0.02 m3 /s π (0.04 m)2 4

∆P= (

= 3.9789 m/s = 15.9155 m/s

m 2 m 2 ) − (3.9789 ) s s N 2(9.81 ) kg

(15.9155

=104,020.7475 Pa x

N

kg

N

kg

) (9.81 kg) (1000 m3 ) − (1.5 m) (9.81 kg) (1000 m3 )

1 𝑘𝑃𝑎 1000 𝑃𝑎

∆P= 104.0207 kPa

33.

A 60-cm water pipes carries a flow of 0.1m3 /𝑠. At point A the elevation is 50 meters and the pressure is 200 kPa. At point B, 1200 meters downstream of A, the elevation is 40 meters and pressure is 230 kPa. The head loss, in feet, between A and B is a. 6.94

b. 15.0

c.22.77

d.100.2

Given: Diamter of pipe = 60 cm m3 𝑠

Velocity of water = 0.1 At point A: Za = 50 m

P = 200kPa

At point B: L = 1200 m downstream of A

Zb = 40 m

P = 230 kPa

Req’d: head loss between A and B Sol’n: 𝑣𝑎 2 𝑃𝐴 𝑣𝐵 2 𝑃𝐵 + = 𝑍𝐵 + + + ℎ𝑓 2𝑔 𝛾 2𝑔 𝛾 𝑃𝐴 𝑃𝐵 𝑍𝑎 + = 𝑍𝐵 + + ℎ𝑓 𝛾 𝛾 𝑍𝑎 +

𝐡𝐟 = 34.

𝟐𝟎𝟎𝐤𝐏𝐚 − 𝟐𝟑𝟎𝐤𝐏𝐚 + 𝟓𝟎𝐦 − 𝟒𝟎𝐦 = 𝟔. 𝟗𝟒𝐦 ≈ 𝟐𝟐. 𝟕𝟕𝐟𝐭 𝟗. 𝟖𝟏𝐦/𝐬

A rectangular duct 4m x 1.5 m in cross section carries conditioned air, its equivalent diameter is a. 2.18 m

b. 1.5 m

c. 2.75 ft

Given: Area = 4m x 1.5 m Req’d: equivalent diameter Sol’n: 4(1.5) 𝐷𝑒𝑞 = 4 [ ] 2(4) + 2(1.5) 𝑫𝒆𝒒 = 𝟐. 𝟏𝟖𝟏𝟖 𝒎

d. 2.18 ft

35.

Oil with a viscosity of 30 cP and a density of 60 lb/ft 3 flows through a ½ inch inside diameter pipe. Determine the velocity in ft/s below which flow will be laminar a. 87.20

b. 0.63

c. 13.10

d. 16.90

Given: oil µ = 30cP ρ = 60lb/ft3 ID = 0.5in Req’d: velocity in ft/s Sol’n: 𝐷𝑣ρ Re = µ ; Re = 2100 (laminar flow) 2100 =

36.

0.5 𝑙𝑏 𝑓𝑡)(𝑣)(60 ) 12 𝑓𝑡3 6.72𝐸−4 𝑙𝑏 (30• ) 1 𝑓𝑡•𝑠

(

; v = 16.9344ft/s

Water flows through a horizontal coil heated by steam condensing on the outside. If the inlet pressure and temperature are 2 atm and 1600 F and at the exit 1 atm and 2200 F, calculate the heat added (Btu) to the coil per pound mass of water. The entering velocity is 5 ft/s and the leaving is 500 ft/s. Data: H1 = 127.9

Btu and H2 lb

= 1154.4 Btu/lb

a. 1277

b. 1042

c. 1031

Given: P1 = 2 atm P2 = 1 atm T1 = 160°F T2 = 220°F V1 = 5 ft/s V2 = 500 ft/s H1 = 127.9 Btu/lb H2 = 1154.4 Btu/lb Required: Q Solution: ∆H + ∆PE + ∆KE = Q + Ws Q = ∆H + ∆KE 1

Q = 2 𝑚(𝑣22 − 𝑣12 ) + 𝑚(𝐻2 − 𝐻1) 1

1 𝐵𝑡𝑢

Q = 2 (1)(5002 − 52 )𝑓𝑡 − 𝑙𝑏𝑓(778𝑓𝑡−𝑙𝑏𝑓) + (1)(1154.4 − 127.9)𝐵𝑇𝑈 Q = 1031.6032 BTU Q =1031 BTU 37.

Hydraulic radius of a 4 ft equilateral triangle channel below is a. 0.31 ft b. 0.54 ft c. 0.65 ft d. 0.91 ft

d. 1022

Given: 4 ft equilateral triangle Req’d: hydraulic radius Sol’n: rH =

𝐴𝑟𝑒𝑎 𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟

38.

0.5 (4𝑠𝑖𝑛60)(4) 4+2.667+2.667+1.333

=

= 0.65 ft

A cast iron pipe of equilateral triangular cross section with side of 20.75 inch has water flowing through it. The flowrate is 6000 gpm, and the friction factor is 0.017. what is the pressure drop in a 100 ft pipe section? a. 1800 lb/ft 2

b. 310 lb/ft 2

c. 1661 lb/ft 2

d. 704 𝐥𝐛/𝐟𝐭 𝟐

Given: sides of equilateral triangle = 20.75 inch Flowrate = 6000 gpm Friction factor = 0.017 Req’d: pressure drop

Sol’n: Deq=

1 (20.75)(17.97 𝑖𝑛) 2

𝑎𝑟𝑒𝑎 4(𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟)=

4

20.75(3) 1𝑓𝑡2 )= 144 𝑖𝑛

A∆= 20.75 (17.97)(0.5)( 1 𝑓𝑡3

1 𝑓𝑡

(12 𝑖𝑛)= .9983 ft

1.2947 ft2

1𝑚𝑖𝑛

Q= 6000 gal/min (7.481 𝑔𝑎𝑙)(60 𝑠𝑒𝑐)= 13.3673 ft3/s V=

13.3673 = 1.2947

∆P= 39.

10.32 ft/s

𝟐(𝟔𝟐.𝟒)(.𝟎𝟏𝟕)(𝟏𝟎𝟎)(𝟏𝟎.𝟑𝟐)𝟐 = 𝟑𝟐.𝟏𝟕𝟒(.𝟗𝟗𝟖𝟑)

𝐥𝐛𝐟

𝐥𝐛𝐟

703.49𝐟𝐭𝟐 ≈ 𝟕𝟎𝟒 𝐟𝐭𝟐

Air flows past a 2-inch diameter sphere at 100 ft/sec. What is the drag force experienced by the sphere given that it has a coefficient of drag of 0.5 and that the density of the air is 0.0753lb/ft 3? a. 0.064

b. 0.137

c. 0.244

Given: Dsphere = 2 in. V = 100 ft/sec CD = 0.5 Density of air = 0.0753 lbm / ft3 Required: Drag force, FD Solution: 𝐹𝐷 =

1 𝜌 𝑣 2 𝐶𝐷 𝐴 2

d. 0.256

𝐴=

2 𝜋 2 𝜋 2 1 𝐷 = ( 𝑓𝑡) = 𝜋 𝑓𝑡 2 4 4 12 144

1 𝑙𝑏𝑚 𝑓𝑡 2 1 𝑙𝑏𝑚 . 𝑓𝑡 (0.0753 ) (100 ) (0.5) ( 𝜋 𝑓𝑡 2 ) = 4.106977896 3 2 𝑓𝑡 𝑠𝑒𝑐 144 𝑠𝑒𝑐 2 𝒍𝒃𝒎. 𝒇𝒕 𝟒. 𝟏𝟎𝟔𝟗𝟕𝟕𝟖𝟗𝟔 𝒔𝒆𝒄𝟐 = 𝟎. 𝟏𝟐𝟕𝟔 𝒍𝒃𝒇 𝑭𝑫 = 𝒍𝒃𝒎 . 𝒇𝒕 𝟑𝟐. 𝟏𝟕𝟒 𝒍𝒃𝒇 . 𝒔𝒆𝒄𝟐 𝐹𝐷 =

40.

A venturi meter is to be installed in a 100 mm line to measure the flow of water at 15℃. The maximum flowrate is

75 𝑚3 . ℎ

The manometer reading is 1.25m of Hg. What will be the power in Kw required to

operate the meter at full load? a. 1.26

b. 1.07

c. 0.55

d. 0.32

Given: Density of water at 15°C = 998.1231 kg/m3 Q = 75 m3/hr P = 1.25 mHg Required: Power, kW Solution: Assume: permanent loss in pressure is 10% of the venturi differential N 1000 mmHg 101325 m2 m3 1 hr 1 kW Power = 1.25 mHg ( )( ) (75 x ) (0.1) )( 1 mHg 760 mmHg hr 3600 s 1000 W = 𝟎. 𝟑𝟒𝟕𝟐 𝐤𝐖

41.

A solution of specific gravity 1.84 is being discharged from a tank through an orifice to the atmosphere. Level of the liquid in the tank is 20 ft above the centerline of the exit pipe. Frictional and contraction losses in the pipe amount to 12 ft head of solution. Under these conditions; the discharge velocity in ft/s is a. 22.7

b. 13.8

Given: orifice SG=1.84 Hliquid = is 20ft above the centerline of the exit pipe Frictional + contraction losses = 12ft Required: Discharge velocity, ft/s Sol’n:

c. 25.5

d. 28.8

𝑣 = √2𝑥32.174𝑥(20 − 12) = 𝟐𝟐.

42.

𝟔𝟖𝟖𝟗𝒇𝒕 ≈ 𝟐𝟐. 𝟕𝒇𝒕/𝒔 𝒔

Water at 60℉ is flowing through a 3-inch diameter smooth horizontal pipe. If the Reynolds number is 35,300, calculate the ratio of the maximum velocity to the average velocity a. 0.81

b. 1.05

c. 1.22

d. 1.72

Given: Temp = 60oF µ = 1129.2652 x 10-6 Pa-s ρ = 998.0622 kg/m3 Req’d: Vmax/Vave ratio Solution: Dvρ µ µRe Vmax = ρD Vmax = 0.5242 m/s Re =

Get f by Blasius equation: F = (100 Re)-1/4 = 0.0231 Vave =

V

= 0.4361 1 + 1.33√f 𝐕𝐦𝐚𝐱 = 𝟏. 𝟐𝟎𝟐𝟎 𝐕𝐚𝐯𝐞

43.

If a centrifuge is 3 ft diameter and rotate at 1000 rpm, the speed (rpm) of a laboratory centrifuge of 6 inch diameter be run if it is to duplicate the plant conditions. a. 2449

Given: r1 = 3 ft N1 = 1000 rpm r2 = 6 inches Reqd: N2 Soln: 2πN1 2 2πN2 ) = r2( 60 )2 60

r1(

r1N12 = r2N22 3(1000)2 = 0.5N22 N2 = 2449.4897 rpm

b. 2500

c. 2469

d. 2000

44.

Engine oil with kinematic viscosity of 0.00024 m2 /s is flowing inside an annulus at 0.5 m/sec. Annulus is made up of 3 in. Sch. 40 and 1 in. Sch 40 concentric pipes 10 m long. Pressure drop expressed in meters of oil is a. 2.33

Given:

b. 4.57

c. 1.98

d. 0.65

Kinematic viscosity= 0.00024 m2/sec Annulus: 3 in Sched 40 & 1 in Sched 40 L=10 m

Req’d: Pressure drop Sol’n: 3 in Sched 40 (OD 3.5 in, ID=3.068 in= 0.0779 m) 1 in Sched 40 (OD=1.315=0.0334 m ID=0.133) 𝜋 ( 𝐼𝐷)2 𝜋 ( 0.0779)2 = = 4 4 2 𝜋 ( 𝑂𝐷) 𝜋 ( 0.0334)2 = = 4 4

Inner cross section are of the larger pipe =

4.7661 𝑥 10−3

Outer cross section are of the smaller pipe = 8.7616𝑥10−4 Cross sectional area of the annulus= Inner –Outer=3.8899 x 10-3 Inner circumference of the larger pipe = 𝜋 ( 𝐼𝐷) = 𝜋 ( 0.0779) = 0.2447 m Outer circumference of the smaller pipe = 𝜋 ( 𝑂𝐷) = 𝜋 ( 0.0334) = 0.1049 𝑚 Wetted perimeter = Cinner + Couter=0.2447 + 0.1049 = 0.3496 m Dh= 4 x

Area Wetted perimeter

=4x

3.8899 x 10−3 0.3496

= 0.0445 m

Flow velocity=0.5 m/sec Relative roughness= 0.0457 mm=4.57x10-5 m Dh 0.5 x ( 0.0445) Reynolds number = (v) ( ) = = 92.7083 u 0.00024 16 16 Fanning friction factor f = Re = 92.7083 = 0.1726 𝐋 𝐕𝟐 𝟏𝟎 𝟎. 𝟓𝟐 𝐡𝐟 = 𝟐𝐟 ( ) 𝐱 ( ) = 𝟐(𝟎. 𝟏𝟕𝟐𝟔) ( )𝐱( ) = 𝟏. 𝟗𝟕𝟔𝟗 𝐦 ≈ 𝟏. 𝟗𝟖𝒎 𝐃𝐡 𝐠 𝟎. 𝟎𝟒𝟒𝟓 𝟗. 𝟖𝟏 45.

A perfect venturi with throat diameter of 1.8 inches is placed horizontally in a pipe with a 5 in inside diameter. Eighty pounds of water flow through the pipe each second. What is the difference between the pipe and venture throat static pressure? a. 29.9 psi

b. 34.8 psi

c. 5020 psi

d. 72.3 psi

Given: venturi throat pipe inside mass flow rate,m= 80 lbs/s

diameter,D2= diameter,

Req’d: difference between the pipe and venturi throat static pressure Sol’n:

1.8 D1=5

inches inches

Q=

m 80 lbs/s 3 = = 1.2821 ft ⁄s 3 ρ 62.4 lbs/ft

v1 =

(4)(1.2821) 4Q = 9.4027 ft⁄s 2 = πD1 5 2 (π) ( ) 12

v2 =

(4)(1.2821) 4Q = = 72.5520 ft⁄s πD22 1.8 2 (π) ( ) 12 P1 − P2 v22 − v12 = ρ 2g

(P1 − P2 )(1728) (72.5520)2 − (9.4027)2 12 =( )( ) (62.4) (2)(32.174) 1 𝐏𝟏 − 𝐏𝟐 = 𝟑𝟒. 𝟖𝟓𝟐𝟏 𝐩𝐬𝐢~𝟑𝟒. 𝟖 𝐩𝐬𝐢 46.

A 4m3/h pump delivers water to a pressure tank. At the start, the gauge reads 138 kPa until it reads 276 kPa and then the pump was shut off. The volume of the tank is 150 L. At 276 kPa the water occupied 2/3 of the tank volume. Determine the volume of water that can be taken out of until the gauge reads 138 kPa. a. 14.5L

Given:

b. 29.5L

c. 44.0L

d. 58.0L

3

4m V̇ = h

P1 = 138 kPa

P2 = 276 kPa

P3 = 138 kPa

Vtank = 150 L @ P2 = 276 kPa : V = 23(150 L) = 100 L @ Pair = 276 kPa : V = 13(150 L) = 50 L Req’d: volume of water removed Sol’n: P2 V2 = P1 V1 (138 + 101.325)(V2) = (276 +101.325)(50 L) V2 = 78.8311 L VH2O to be removed = 100 L – (150 - 78.8311) L = 28.8311 L ≈ 29.5 L 47.

Water is pumped at a constant rate of 10m3 /h from a large reservoir resting on the floor to the open top of an absorption tower. The point of discharges is 6m above the floor, and the frictional losses in the 40 mm pipe from the reservoir to the tower amount to 4J/kg. At what height in the reservoir must the water level be kept if the pump can develop only 0.12 kW? a. 2.2 m

b. 6.4m

c. 1.9m

Given: pump rate = 10m3 /h point of discharge = 6m above the floor D = 40mm

frictional loss = 4 J/kg

d. 2.7 m

Pump work = 0.12 kW Req’d: height of water Solution: αV22 − nWp + HL 2g m3 1h 10 x 3600s m h V= = 2.2105 Π s x 0.042 m2 4 z1 = z2 +

Re =

0.04 x 1000 x 2.2105 = 88420 (Turbulent: α = 1) 0.001 120

nWp = 10

J s

kg m3 1h N x x 1000 3 x 9.81 h 3600s kg m

= 4.4037 m

J 2.21052 kg z1 = 6 m − 4.4037 m + + N 2x9.81 9.81 kg 4

𝐳𝟏 = 𝟐. 𝟐𝟓𝟏𝟎 𝐦 48.

Cooling water at 22 ℃ flows thru a smooth pipe of unknown diameter resulting to a pressure drop of 2Kn/m2 for a length of 5 meters. It’s mass flow rate is 1590 kg/h. Determine the pipe diameters in meters. a. 200

b. 171

c. 330

Given: T of cooling water = 22 ℃ ; Length of pipe = 5 m ∆𝑃 = 2Kn/m2 ; mass flowrate of water = 1590 kg/h Req’d: pipe diameter Sol’n: properties of water at 22 ℃ µ =9.6892 x 10-4 Pa.s ρ =996.795 kg/m3 Q=

𝜋𝐷 4 ∆ 𝑃 128 µ ∆𝑋

Q=

1590

𝑘𝑔 1 𝑥 ℎ𝑟 3600 𝑠 𝑘𝑔 996.795 𝑚3

= 4.4309 x 10-4 m3/s

4.4309 x 10-4 m3/s = d = 0.025 m

𝜋 𝐷 4 𝑥2000 𝑃𝑎 5 𝑚 𝑥 128 𝑥 9.6892 𝑥 10−4

d. 245

49.

A venturi meter having a throat diameter of 38.9 mm is installed in a line having an inside diameter of 102.3 mm. It meters water and the measured pressure drop across the venturi is 156.9 KPa. Calculate the flow rate in gpm if the coefficient (Cv) is 0.98 a. 200

b. 171

c. 330

d. 245

Given: Throat diameter=38.9 mm Pipe ID = 102.3 mm ΔP = 156.9 KPa Cv=0.98 Req;d: Q, gpm

Sol’n: 𝛽=

Throat diameter 38.9 mm = = 0.3803 Pipe ID 102.3 mm Vb =

Vb =

Cv

2gcΔP √ ρ √1 − β4

2(1)(156.9x103 ) m √ = 17.5446 1000 s √1 − 0.38034 0.98

𝛑 𝟑𝟖. 𝟗 𝟐 𝟏𝟎𝟎𝟎𝐋 𝟏 𝐠𝐚𝐥 𝟔𝟎𝐬 𝐐 = (𝟏𝟕. 𝟓𝟒𝟒𝟔) ( ( 𝐱 𝐱 = 𝟑𝟑𝟏. 𝟐𝟕𝟐𝟖 𝐠𝐩𝐦 )) 𝐱 𝟑 𝟒 𝟏𝟎𝟎𝟎 𝟏𝐦 𝟑. 𝟕𝟖𝟓𝟒𝐋 𝟏 𝐦𝐢𝐧𝐮𝐭𝐞 50.

A cylindrical tank 1 ft in diameter discharges through a nozzle connected to the base. Find the time needed for the water level in the tank to drop from 6 ft to 3 ft above the nozzle. The diameter of the nozzle is 1 inch and its discharge maybe taken as unity. a. 15s

b. 42s

c. 21s

d. 30s

Given:

D of cylinder tank = 1 ft

6 ft

D of nozzle = 1 in

Req’d: time needed for the water level in the tank to drop from 6 ft to 3 ft Sol’n:

ρAV − 0 = −

d 0 ∫ ρdv dt cv

d [ρAT (h + x)] dt dh ρAv = −ρAT dt −AT dh ∆v = A√2gh −AT dh dt = A√2gh −AT = A√2gh 1 −AT = dh(h)−2 A√2g =−

1

t=

−2AT (h)2

When

√2g A

+c

t=0, c=

h=6ft

1 2AT (6 ft)2

√2g A 1

t=

−2AT (2.4495 − h2 ) √2g A

When h=3 ft

𝐭=

𝟏 𝛑 𝟐 (𝟒 ) (𝟏 𝐢𝐧)𝟐 (𝟐. 𝟒𝟒𝟗𝟓𝐟𝐭 − 𝟑𝐟𝐭 𝟐 )

𝟒

****

𝟏 𝟐 𝟏𝟐)

𝟏 𝛑 [𝟐(𝟑𝟐. 𝟐 𝐟𝐭/𝐬)]𝟐 ( ) (

When

= 𝟐𝟓. 𝟕𝟒𝟕𝟖𝐬

t=0, c=

1 2AT (6 ft)2

√2g A 1

t=

−2AT (2 − h2 ) √2g A

When h=2 ft

𝐭=

𝟏 𝛑 𝟐 (𝟒 ) (𝟏 𝐢𝐧)𝟐 (𝟐𝐟𝐭 − 𝟐𝐟𝐭 𝟐 ) 𝟏 𝛑 [𝟐(𝟑𝟐. 𝟐 𝐟𝐭/𝐬)]𝟐 ( ) (

𝟒

𝟏 𝟐 𝟏𝟐)

= 𝟐𝟏. 𝟎𝟐𝟐𝟕𝐬 ≈ 𝟐𝟏𝐬

h=4ft