Power and Industrial Plant Engineering

1. Find the work possess for a Helium gas at 200C

      *A. 609 KJ/kg     B. 168 KJ/kg     C. 229 KJ/kg      D. 339 KJ/kg

      Solution:

      W = m R T = m (8.314 / M) T

       For helium, M = 4

      W/m =  (8.314/4)(20 + 273) = 609 KJ/kg

 

2. Two kilogram of gas is confined in a 1 m3 tank at 200 kpa and 880C. What type of gas is in the tank?

      A. Helium     *B. Ethane     C. Methane     D. Ethene

      Solution:

      Therefore: the gas is Ethane (C2 H8)

 3. Find the enthalpy of Helium if its internal energy is 200 KJ/kg

  1. 144 KJ/kg   B.  223.42 KJ/kg  *C. 333.42 KJ/kg  D. 168 KJ/kg

Solution:

 

4. Compute the mass of a 2 m3 propane at 280 kpa and 40˚C.

A.      6.47 kg        B. 5.1 kg         C. 10.20 kg       *D. 9.47 kg

Solution:

Propane is C3 H3--------------M = 12(3) + 8(1) = 44

PV = m R T

280(2) = m (8.314/44)(40 + 273)

m = 9.47 kg

 

5. Compute the air flow in ft3/min of mechanical ventilation required to exhaust an accumulation of refrigerant due to leaks of the system capable of revolving air from the machinery room for a mass of 4 lbs refrigerant.

*A. 200           B. 210                  C. 220                  D. 230

Solution:

Q = 100 x G0.5 ft3/min

Q = 100 x (4)0.5 = 200 ft3/min

 

6. Compute the free-aperture cross section in m2 for the ventilation of a machinery room if the mass of refrigerant is 9 kg.

  1. 0.314       *B.  0.414        C.  0.514         D.  0.614

Solution:

F = 0.138 G0.5 m2

F = 0.138 (9)0.5 = 0.414 m2      

 

7. A 29.53” x 39.37” pressure vessel contains ammonia with f = 0.041. Compute the minimum required discharge capacity of the relief device in kg/hr.

A.      106.71 kg/hr   B. 108.71 kg/hr   *C. 110.71 kg/hr        D. 112.71 kg/hr

Solution:

C = f D L, kg/s

C = 0.041(29.53/39.37)(39.37/39.37) = 0.03075 kg/s (3600) = 110.71 kg/hr

 

8. Compute the maximum length of the discharge pipe installed on the outlet of a pressure-relief device in feet for internal pipe diameter of 0.5 inch and rated discharge capacity is 8 lb/min of air. The rated pressure of relief valve is 16 psig.

*A. 0.286 ft   B. 0.386 ft       C. 0.486 ft       D. 0.586 ft

Solution:

P = Pg + Patm = 16 x 1.1  +  14.7 = 32.3 psia

L = 9P2d5/16Cr2 = 9(32.3)2(0.5)5/16(8)2 = 0.286 ft

 

9. A thermal power plant has a heat rate of 11,363 Btu/kw-hr. Find the thermal efficiency of the plant.

  1. 28%         *B. 30%           C. 34%            D. 40%

Solution:

eth = 3412/Heat rate = 3412/11,363 = 30%

 

10. What is the hydraulic gradient of a 1 mile, 17 inches inside diameter pipe when 3300 gal/min of water flow with f = 0.03.

*A. 0.00714          B. 0.00614        C. 0.00234        D. 0.0018

Solution:

v = (3300/7.481)/(π/4)(17/12)2(60) = 4.66 ft/s

L = 1 mile = 5280 ft

hL = fLv2/2_D = 0.03(5280)(4.66)2/2(32.2)(17/12) = 37.7 ft

Hydraulic gradient = 37.7/5280 = 0.007.14

 

11. Find the loss of head in the pipe entrance if speed of flow is 10 m/s.

  1. 5.10 m            B. 10.2 m         C. 17.4 m         *D. 2.55 m

Solution:

Loss at entrance = 0.5 (v2/2g) = 0.5 [102 / 2(9.81)] = 2.55 m

 

12. Wet material, containing 220% moisture (dry basis) is to be dried at the rate of 1.5 kg/s in a continuous dryer to give a product containing 10% (dry basis) . Find the moisture removed, kg/hr

*A. 3543.75 kg/hr          B. 3513.75 kg/hr  C. 3563.75 kg/hr  D. 3593.75 kg/hr

Solution:

Solid in wet feed = solid in dried product

[1/(1 + 2.2)](1.5) = [1/(1 + 0.1)](x)

x = 0.5156 kg/s (total dried product)

Moisture removed = 1.5 – 0.5156 = 0.984 kg/s = 3543.75 kg/hr

 

13. Copra enters a dryer containing 70% moisture and leaves at 7% moisture. Find the moisture removed on each pound on solid in final product.

  1. 6.258 lb    B. 1.258 lb       C. 4.258 lb       *D. 2.258 lb

Solution:

Solid in wet feed = solid in dried product

0.3x = 1

x = 3.333 lbs

1 = 0.93y

y = 1.07527 lb

Moisture removed = x – y = 3.333 – 1.07527 = 2.258 lb

 

 

14. A 1 m x 1.5 m cylindrical tank is full of oil with SG = 0.92. Find the force acting at the bottom of the tank in dynes.

  1. 106.33 x 103 dynes B. 106.33 x 104 dynes        C. 106.33 x 105 dynes    *D. 106.33 x 106 dynes

Solution:

P = w h = (0.92 x 9.81) (1.5) = 13.5378 kpa

F = PA = 13.5378(π/4 x 12) = 10.632 KN = 10,632.56 N x 10,000 dynes/N

F = 106.33 x 106 dynes

 

15. Find the pressure at the 100 fathom depth of water in kpag.

*A. 1,793.96 kpag          B. 1,893.96 kpag  C.  1,993.96 kpag       D. 1,693.96 kpag

Solution:

H = 100 fathom x 6 = 600 ft

P = w h = (600/3.281)(9.81) = 1,793.96 kpag

 

16. Find the depth in furlong of the ocean (SG = 1.03) if the pressure at the sea bed is 2,032.56 kpag.

*A. 1          B. 2        C. 3        D. 4

Solution:

P = w h

2,032.56 = (1.03 x 9.81) h

h = 201.158 m  x  3.281 ft/m  x  1 yd/3ft  x  1 furlong/220yd = 1 furlong

 

17. Find the mass of 10 quartz of water.

  1. 10.46 kg          *B. 9.46 kg       C. 11.46 kg       D. 8.46 kg

Solution:

V = 10 quartz  x  1gal/4quartz  x  3.785li/1gal  x  1m3/1000li

V = 0.0094625  x  10-3m3

w = m/V

1000 = m/0.0094625  x  10-3 

m = 9.46 kg

 

18. Find the mass of carbon dioxide having a pressure of 20 psia at 200˚F with 10 ft3 volume.

  1. 1.04 lbs          B. 1.14 lbs       *C. 1.24 lbs            D. 1.34 lbs

Solution:

PV = m R T

(20  x  144)(10) = m (1545/44)(200  +  460)

m =  1.24 lbs

 

19. Find the heat needed to raise the temperature of water from 30˚C to 100˚C with 60% quality. Consider an atmospheric pressure of 101.325 kpa. Use the approximate enthalpy formula of liquid.

  1. 293.09 KJ/kg            B. 1,772.90 KJ/kg       C. 1,547.90 KJ/kg       *D. 1,647.29 KJ/kg

Solution:

      At 100°C

      hf = Cp t = 4.187(100) 418.7 KJ/kg

      hfg = 2257 KJ.kg

      h2 = hf + xhfg = 418.7 + 0.60(2257) = 1,772.9 KJ/kg

      Q = 1(4.187) (100-30) + 1(1772.9 – 418.7) = 1647.29 KJ/kg

 

20. Find the enthalpy of water at 212˚F and 14.7 psi if the dryness factor is 30%. Use the approximate enthalpy formula of liquid.

  1. 461 Btu/lb        *B. 471 Btu/lb          C. 481 Btu/lb           D. 491 Btu/lb

Solution:

ht = (˚F – 32) = (212 – 32) = 180 Btu/lb

htg = 970 Btu/lb

h = ht  +  x htg

h = 180  +  0.3(970) = 471 Btu/lb

 

21. An air compressor consumed 1200 kw-hr per day of energy. The electric motor driving the compressor has an efficiency of 80%. If indicated power of the compressor is 34 kw, find the mechanical efficiency of the compressor.

  1. 117.65 %          B. 75 %           *C.  85 %         D. 90 %

Solution:

P/m = 1200kw-hr/24 hrs = 50 kw

BP = 50(0.80) = 40 kw

em = 34/40 = 85 %

 

22. A refrigeration system consumed 28,000 kw-hr per month of energy. There are 20 % of energy is lost due to cooling system of compressor and motor efficiency is 90 %. If COP of the system is 6, find the tons of refrigeration of the system.

  1. 43.15 TR          B. 46.15 TR       *C. 49.15 TR            D. 41.15 TR

Solution:

P/m = 28,800/(24  x  30) = 40 kw

BP = 40(0.90) = 36 kw

Wc = 36(1 – 0.20) = 28.80 kw

COP = RE/Wc

6 = RE/28.80

RE = 172.8/3.516 = 49.15 TR

 

23. A 23 tons refrigeration system has a heat rejected of 100 kw. Find the energy efficiency ratio of the system.

  1. 13.42       *B. 14.42         C. 15.42          D. 16.42

Solution:

QR = RE  +  Wc

100 = 23(3.516)  +  Wc

Wc = 19.132 kw

COP = RE/Wc = (23  x  3.516) / 19.132 = 4.32

EER = 3.412 COP = 3.412(4.23) = 14.42

 

24. A 200 mm  x  250 mm, 8-cylinder, 4-stroke diesel engine has a brake power of 150 kw. The mechanical efficiency is 80%. If two of the cylinders were accidentally cut off, what will be the new friction power?

  1. 31.50 kw          B. 33.50 kw       C. 35.50 kw       *D. 37.50 kw

Solution:

em = BP/IP

0.8 = 150/IP

IP = 187.5 kw

FP1 = IP – BP = 187.5 – 150 = 37.50 kw

FP1 = FP2 = 37.50 kw

 

25. If the energy efficiency ratio of the refrigeration system is 12.6, what is the COP of the system?

*A. 3.69       B. 4.23           C. 5.92           D. 6.83

Solution:

EER = 3.412 COP

12.6 = 3.412 COP

COP = 3.69

 

26. An air compressor has a power of 40 kw at 4 % clearance. If clearance will increase to 70 %, what is the new power?

  1. 70 kw       *B. 40 kw         C. 53kw           D. 60 kw

Solution:

The power of compressor will not be affected with the changes in clearance. Therefore the power will still be 40 kw.

 

27. What is the approximate value of temperature of water having enthalpy of 208 Btu/lb?

  1. 138.67 ˚C         *B. 115.55 ˚C           C. 258.67 ˚C            D. 68.67 ˚C

Solution:

h = ˚F – 32

      208 = F – 32

      ˚F = 240 ˚F = 115.55 ˚C

 

28. Convert  750˚R to ˚K

A.      390.33 ˚K         B. 395.33 ˚K            C. 410.33 ˚K            *D. 416.33 ˚K

Solution:

˚R = 1.8 ˚K

750 = 1.8 ˚K

˚K = 416.667

 

29. An otto cycle has a compression ratio of 8. Find the pressure ratio during compression.

*A. 18.38            B. 16.38          C. 14.38          D. 12.38

Solution:

P1V1k = P2V22

(V1/V2)k = (P2/P1)

rkk = rp

rp = (8)1.4 = 18.38

 

30. A diesel cycle has a cut off ratio of 2.5 and expansion ratio of 4. Find the clearance of the cycle.

A.      9.11 %                  B.  5.55 %        *C.  11.11 %            D. 15.15 %

Solution:

rk  = rc re

rk = 2.5(4) = 10

rk = (1  +  c)/c

10 = (1  +  c)/c

c  = 11.11 %

 

31. A dual cycle has an initial temperature of 30 ˚C. The compression ratio is 6 and the heat addition at constant volume process is 600 KJ/kg. If cut-off ratio is 2.5, find the maximum temperature of the cycle.

A.      3638.50 ˚C        *B. 3365.50 ˚C    C. 3565.50 ˚C           D. 3965.50 ˚C

Solution:

T2 = T1 rkk-1 = (30 +273)(6)1.4-1 = 620.44 ˚K

QAV = m cv (T3 – T2)

600 = 1(0.7186)(T3 – 620.44)

T3 = 1455.396 ˚K

rc  = T4/T3

2.5 = T4/1455.396

T4 = 3638.49 ˚K = 3365.50 ˚C

 

32. A three stages air compressor compresses air from 100 kpa to 1000 kpa. Find the intercooler pressure between the first and second stage.

A.      505.44 kpa        B. 108.44 kpa                 C. 316.23 kpa                 *D. 215.44 kpa

Solution:

Px = (P12P2)1/3

Px = [(100)2(1000)]1/3 = 215.44 kpa

 

33. A 10-stages air compressor compresses air from 100 kpa to 800 kpa. Find the intercooler pressure between 1st and 2nd stage.

A.  282.84 kpa        B. 113.21 kpa                 *C. 123.11 kpa          D. 333.51 kpa

Solution:

Px = (P1s-1P2)1/s

Px = [(100)10-1(1000)]1/10 = 123.11 kpa

 

34. A 3-stages air compressor compresses air from 100 kpa to 700 kpa. Find the intercooler pressure between the 2nd and 3rd stage.

*A. 365.88 kpa       B. 375.88 kpa                 C. 385.88 kpa                 D. 395.88 kpa

Solution:

Px = (P12P2)1/3

Px = [(100)2(700)]1/3 = 191.28 kpa

Px/P1 = Py/Px

Py = Px2/P1 = (191.28)2/100 = 365.88 kpa

 

35. Carnot cycle A, B and C are connected in series so that the heat rejected from A will be the heat added to B and heat rejected from B will be added to C, each cycle operates between 30 ˚C and 400 ˚C. If heat added to A is 1000 kw, find the work output of C.

*A. 111.44 kw           B. 549.78 kw            C. 247.53 kw            D. 141.89 kw

Solution:

e1 = e2 = e3 = (400 – 30)/(400 + 273) = 54.98 %

e1 = W1/QA1 = (QA1 – QR1)/QA1

0.5498 = (1000 – QR1)/1000

QR1 = 450.22 = QA2

0.5498 = (450.22 – QR2)/450.22

By heat balance:

      Qgain = Qloss

      mw cp (tb- ta) = mg cpg (t2-t1)

      (0.30)(4.187)( tb – 15) = (0.5)(1.0717)(150 – 80)

      tb = 44.86 oC

 

36. A 350 mm X 450 mm steam engine running at 280 rpm has an entrance steam condition of 2 Mpa and  230 oC and exit at 0.1 Mpa. The steam consumption is 2,000 kg/hr and mechanical efficiency is 85%. If  indicated mean effective pressure is 600 Kpa, determine brake thermal efficiency.

      At 2 Mpa and 230 oC (Table 3):   h1 = 2849.6          s1 = 6. 4423

      At 0.1 Mpa:       sf = 1.3026 hf  = 417.46 sfg = 6.0568      hfg = 2258 hf2 = 417.46 KJ/kg

A. 23.34%   *B. 15.25% C. 14.16%   D. 27.34%

SOLUTION:

      VD = 2[3.1416/4  (0.35) 2 (0.45)(280/60)]= 0.4041 m3/sec

      Indicated Power = Pmi x VD = 600 x 0.4041 =242.45 KW

      Brake Power = IP (em)  = 242.45 (0.85) = 206.08KW

            Brake Power                   206.08                     =  15.25%

      etb =    ms (h1-hf2)     =     (2000/3600)(2849.6 – 417.46)

 

37. A steam turbine receives 5,000 kg/hr of steam at 5 Mpa and 4000oc and velocity of 30 m/sec. It leaves the turbine at 0.06 Mpa and 85% quality and velocity of 15 m/sec. Radiation loss is 10,000 KJ/hr. Find he KW developed.

       At 5 Mpa and 400oC:    h1 = 3195.7 KJ/kg        s1 =6.6459

At 0.006 Mpa:  hf = 151.53          hfg = 2415.9

      A. 1273.29        B. 2173.29        *C. 1373.60       D. 7231.29

SOLUTION:

h2 = hf + xhfg =  151.53 + 0.85(2415.9) = 2205.045 KJ/ kg

      KE1 = ½ m v2 = ½ (5,000/3600)(30)2 = 625 W = 0.625 KW

      KE2 = ½ m v2 =  ½ (5,000/3600)(15)2 = 156.25 W = 0.15625 KW

      By energy balance:

      KE1 + mh1 = KE2 + mh2 + Q + W

      W = (KE1 – KE2) + m(h1-h2) – Q

                                      5000                                  10,000

      W = (0.625 – 0.156) + (3600)(3195.7 – 2205.045) – 3600  = 1373.60 KW

 

38. A steam turbine with 85% stage efficiency receives steam at 7 Mpa and 550oC and exhausts as 20 Kpa. Determine the turbine work.

      At 7 Mpa and 550oC:                 h1 = 3530.9 Kj/kg       s1 = 6.9486

      At 20 Kpa (0.020 Mpa): sf = 0.8320       hf = 251.4 sfg = 7.0766       hfg = 2358.3

A.  1,117 KJ/kg    B. 1,132 KJ/ kg   C. 1,123.34 KJ/ kg      *D. 1,054.95 KJ/kg

SOLUTION:

      s1 = s2 = sf + xsfg

      6.9486 = 0.8320 + x(7.0766)

      x = 0.8643

      h2 = 251.40 + 0.8643(2358.3) = 2289.78 KJ/kg

              h1 - h2a

      eST = h1 – h2

      0.85 =3530.9 – h2a        

             3530.9 – 2289.78

      h2a = 2475.95 KJ/kg

      WT = h1 = h2a = 3530.9 – 2475.95 = 1,054.95 KJ/kg

 

39. A steam turbine with 80% stage efficiency receives steam at 7 Mpa and 550oC and exhaust as 20 Kpa. Determine the quality at exhaust.

      At 7 Mpa and 550oC:                 h1 = 3530.9 Kj/kg       s1 = 6.9486

      At 20 Kpa (0.020 Mpa):        sf = 0.8320             hf = 251.4

      *A. 96.96% B. 76.34%   C. 82.34%   D. 91.69%

SOLUTION:

      sfg = 7.0766      hfg = 2358.3

s1 = s2 = sf + sfg

      6.9486 = 0.8320 + x(7.0766)

      x = 0.8643

      h2 = 251.40 + 0.8643(2358.3) = 2289.78 KJ/kg

      nST =  h1 – h2a

                 h1 – h2

      0.80 = 3530.9 – h2a       

                 3530.9 – 2289.78

      h2a = 2538.004 KJ/kg

      h2a  = hf + x hfg

      2538.004 = 251.40 + x (2358.3)

      x = 96.96%

 

40. A 16,000KW 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 400, 000 kg/hr, determine the number of wells are required to produce if the charge of enthalpy if the change of enthalpy at entrance and exit of turbine is 500KJ/kg.

A.  4 wells       *B. 2 wells      C. 6 wells       D. 8 wells

SOLUTION:

WT = ms(h3 – h4)

      16,000 = ms (500)

      0.9(0.8)

      ms = 44.44 kg/sec

      ms = 160,000 kg/hr

      160,000 = 0.20 mg

      mg = 800,000 kg/hr

      No. of wells = 800,000/400,000 = 2 wells

 

41. A liquid dominated geothermal plant with a single flash separator receives water at 204oC. The separator pressure is 1.04 Mpa. A direct contact condenser operates at 0.034 Mpa. The turbine has a polytropic efficiency of 0.75. For a cycle output of 60 MW, what is the mass flow rate of the well-water in kg/s?

      At 204oC:   hf = 870.51 KJ/kg

      At 1.04 Mpa:      hf = 770.38       hfg = 2009.2      hg = 2779.6       sg = 6.5729

      At 0.034 MPa:  hf = 301.40    hfg = 2328.8      sf = 0.9793     sfg = 6.7463

      *A. 2,933   B. 2,100    C. 1,860    D. 2,444

SOLUTION:

      h3 = hg at 1.04 MPa = 2779.6 KJ/kg

      Solving for h4:

      s3 = s4 = sf + xsfg

      6.5729 = 0.9793 + x4(6.7463)

      x4 = 0.829

      h4 = 301.4 + 0.829(2328.8) = 2232.3 KJ/kg

      WT = ms (h3 – h4)

      60,000 = ms (2779.6 – 2232.3) 0.75

      ms = 146.17 kg/sec

Solving for x2: (h1 = h2)

      h1 = h2 = hf + xhfg

      870.51 = 770.38 + x2(2009.2)

      x2 = 0..049836

      ms = x mg

      146.17 = 0.049836 mg

      mg = 2,933.06 kg/sec

 

42. An engine-generator rated 9000 KVA at 80% power factor, 3 phase, 4160 V has an efficiency of 90%. If overall plant efficiency is 28%, what is the heat generated by the fuel.

A.  18,800 KW      B. 28,800 KW           C. 7500 KW        *D. 25,714 KW

SOLUTION:

      Gen. Output = pf x KVA = 0.8 x 9000 = 7200 KW

      eoverall= Gen. Output

                   Qg

 

      0.28 = 7200/Qg

      Qg = 25,714.28 KW

 

43. The indicated thermal efficiency of a two stroke diesel engine is 60%. If friction power is 15% of heat generated, determine the brake thermal efficiency of the engine.

A.  43%            *B. 45 %          C. 36%            D. 37%

SOLUTION:

      ne = IP/ Qg

      0.60 = IP/Qg

      IP = 0.60 Qg

      BP = IP- FP = 0.60Qg – 0.15Qg = 0.45Qg

      etb = BP/Qg = 0.45Qg/Qg = 45%

 

44. A 305 mm x 457 mm four stroke single acting diesel engine is rated at 150 KW at 260 rpm. Fuel consumption at rated load is 0.56 kg/KW-hr with a heating value of 43,912 KJ/kg. Calculate brake thermal efficiency

A.  10.53%   B. 27.45%   *C. 14.64%  D. 18.23%

SOLUTION:

      mf = 0.56 kg/KW-hr x 150 KW = 84 kg/hr = 0.0233 kg/sec

Brake thermal efficiency =

 

45. A waste heat recovery boiler produces 4.8 Mpa(dry saturated) steam from 104°C feedwater. The boiler receives energy from 7 kg/sec of 954°C dry air. After passing through a waste heat boiler, the temperature of the air is has been reduce to 343°C. How much steam in kg is produced per second? Note: At 4.80 Mpa dry saturated, h = 2796.

A.  1.30     B. 0.92           *C. 1.81    D. 3.43

SOLUTION:

hf = approximate enthalpy of feedwater

      hf = Cpt

      hf = 4.187(104)

      hf = 435.45 KJ/kg

      Heat loss = Heat gain

      m gc p(t 1 - t 2) = m s(h - h f)

      7(1.0)(954 – 343) = ms(2796.0 – 436.45)

      m s = 1.81 kg/sec

 

46. A diesel electric plant supplies energy for Meralco. During a 24-hour period, the plant consumed 240 gallons of fuel at 28°C and produced 3930 KW-hr. Industrial fuel used is 28°API and was purchased at P30 per liter at 15.6°C. What is the cost of the fuel be to produce one KW-hr?

*A. P6.87   B. P1.10    C. P41.07   D. P5.00

SOLUTION:

      SG 15.6C = 141.5/(131.5 + 28) = 0.887

      Density at 15.6°C = 0.887(1kg/li) = 0.887 kg/li

      SG 28C = 0.887[1-.0007(1 – 15.6)] = .879

      Density at 28°C = 0.879(1 kg/li) = 0.879 kg/li

      V28C / V15.6C = SG15.6C / SG28C

240 / V15.6C = 0.887 / 0.879

V15.6C = 237.835 gallons x 3.785 li/gal = 900.21 li

Cost = [(30)(900.21)] / 3930 = P6.87/KW-hr

 

47. In a gas turbine unit, air enters the combustion chamber at 550 kpa, 277°C and 43 m/s. The products of combustion leave the combustor at 511 kpa, 1004°C and 180 m/s. Liquid fuel enters with a heating value of 43,000 KJ/kg. For fuel-air ratio of 0.0229, what is the combustor efficiency of the unit in percent?

A.  70.38%   B. 79.385%  *C. 75.38%  D. 82.38%

SOLUTION:

Heat supplied by fuel = mfQh = 0.0229(43,000) = 984.7 KJ/kg air

Q = heat absorbed by fuel

Q/m = Cp(T2 – T1) + ½(V22 – V12)

      Q/m = (1.0)(1004 – 277) + ½[(180) 2 –(43) 2]/1000 =742.28 KJ/kg air

      Combustor Efficiency =  = 75.38%

 

48. The specific speed of turbine is 85 rpm and running at 450 rpm. If the head is 20 m and generator efficiency is 90%, what is the maximum power delivered by the generator.

A.  450.51 KW            B. 354.52 KW            C. 650.53 KW            *D. 835.57 KW

SOLUTION:

      NS = (N√HP)/h5/4

      85 = (450√HP)/(20 x 3.281) 5/4

      Hp = 1244.52

      Generator Output = (1244.52 x 0.746)(0.9) = 835.57 KW

 

49. In Francis turbine, the pressure gage leading to the turbine casing reads 380 Kpa. The velocity of water entering the turbine is 8 m/sec, if net head of the turbine is 45 m, find the distance from center of spiral casing to the tailrace.

      *A. 3.0 m   B. 3.5 m    C. 4.0 m    D. 4.5m

SOLUTION :

 

      h =  V2/2g

      45 = (380/9.81) + z + [82/(2 x 9.81)]

      z = 3 m

 

50. A turbine has a mechanical efficiency of 93%, volumetric efficiency of 95% and total efficiency of 82%. If effective head is 40 m, find the total head.

A.  48.72 m  B. 40.72 m  *C. 36.22 m D. 34.72 m

SOLUTION:

            eT = emehev

                   0.8 = 0.93(eh)(.95)

            ηh = 0.9055

            Total head = h eh = (40)(0.9055) = 36.22 m

 

51. A Pelton type turbine has 25 m head friction loss of 4.5 m. The coefficient of friction head loss (from Moorse) is 0.00093 and penstock length of 80 m. What is the penstock diameter?

      *A. 1,355.73 mm   B. 3,476.12 mm          C. 6771.23 mm           D. 1686.73 mm

SOLUTION:

       h =25- 4.5 = 20.5

      v = √(2gh) = [(2 x 9.81 x 20.5)1/2] = 20.55 m/sec

      hL = (2fLv2)/gD

      4.5 = (2)(0.00093)(80)(20.055)2 / 9.81D

      D = 1,355,730 m = 1,355.73 mm

 

52. In an 9,000 KW hydro-electric plant the over-all efficiency is 88% and the actual power received by the customer is 110,000 KW-hrs for that day. What is the secondary power could this plant deliver during the entire day?

A.  58,960 KW-hrs  *B. 80,080 KW-hrs C. 65,960 KW-hrs  D. 70,960 KW-hrs

SOLUTION:

      Plant Capacity = 9,000(0.88)(24) = 190,080 KW-hrs

      Secondary Power = 190,080 – 110,000 = 80,080 KW-hrs

 

53. A Pelton type turbine was installed 30 m below the gate of the penstock. The head loss due to friction is 12 percent of the given elevation. The length of penstock is 100 m and coefficient of friction is 0.00093. Determine the power output in KW. ( Use Moorse equation)

A.  22,273   B. 23,234   C. 32,345   *D. 34,452

SOLUTION:

      hL = 0.12(30) = 3.6 m

      h = 30 – 3.6 = 26.40 m

      v = (2gh)1/2 = [(2)(9.81)(26.4)]1/2 = 22.759 m/sec

      hL= (2fLv2)/gD

      3.6 = (2 x .00093 x 100 x 22.759) / (9.81D)

      D = 2.728 m

      Q = A x v = [2](22.759) = 133.03 m3/sec

      Power = w Q h = 9.81(133.03)(26.4) = 34,452 KW

 

54. Water flows steadily with a velocity of 3.05 m/s in a horizontal pipe having a diameter of 25.4 cm. At one section of the pipe, the temperature and pressure of the water are 21C and 689.3 Kpa, respectively. At a distance of 304.8 m downstream

 

55. A hydro electric plant having 30 sq. km reservoir area and 100 m head is used to generate power. The energy utilized by the consumers whose load is connected to the power plant during a five-hour period is 13.5 x 106 kwh. The overall generation efficiency is 75%. Find the fall in the height of water in the reservoir after the 5-hour period.

A.  5.13 m   B. 1.32 m   C. 3.21           *D. 2.20 m

SOLUTION

      Energy Output = Power x time = (w Q h) x time

13.5 x 106 = 9.81(Q)(100)(0.75)(5)

Q = 3669.725 m3/s

      Volume after 5 hrs = 3669.725(5 x 3600) = 66,055,050 m3

      Volume = A x height

      66,055,050 = (30 x 106) h

      H =2.202 m

 

56. The gas density of chimney is 0.75 kg/m3 and air density of 1.15 kg/m3. Find the driving pressure if the height of chimney is 63.71 m.

A.  0.15 kpa *B. 0.25 kpa      C. 0.35 kpa D. 0.45 kpa

SOLUTION:

hw = H(da – dg) = 63.71(1.15 – 0.75) (0.00981) = 0.25 kpa

 

57. The actual velocity of gas entering in a chimney is 8 m/sec. The gas temperature is 25C with a gas constant of 0.287 KJ/kg-K. Determine the gas pressure for a mass of gas is 50,000 kg/hr and chimney diameter of 1.39m.

A.  95 kpa   *B. 98 kpa  C. 101 kpa  D. 92 kpa

SOLUTION:

      Vg = A x v =  / 4 (1.39)2(8) = 12.139 m3/s

      PgVg = mgRgTg

      P(12.139) = (50,000/3600)(.287)(25 +273)

      P = 97.85 kpa

58. A steam generator with economizer and air heater has an overall draft loss of 25.78 cm of water. If the stack gases are at 177C and if the atmosphere is at 101.3 Kpa and 26C, what theoretical height of stack in meters is needed when no draft fan are used? Assume that the gas constant for the flue gases is the same as that for air.

      A 611.10    B. 631.10   *C.651.10   D.671.10

SOLUTION:  

      w = P/RT

da = (101.325)/[(.287)(26 + 273)] = 1.180 kg/m3

      dg = (101.3)/[(0.287)(177 +273)] = 0.784 kg/m3

          Draft = (0.2578)(1000) = 257.80 kg/m3

      Draft = H(da ­– dw)

257.80 = H(1.18 – 0.784)

H = 651.10 m

 

59. A foundation measures 12 ft x 14 ft x16 ft. Find the number of sacks of cement needed for 1:2:4 mixture.

A.  302            B. 404            C. 356            *D. 598

SOLUTION:

      V = 12 X 14 X 16 = 2,688 ft3 (1 yd3 / 33 ft3) = 99.55 yd3 of concrete

 

60. For every 1 yd3 concrete, it needs 6 sacks of cement

Therefore:

No. of sacks = 6(99.55) = 597.33 sacks or 598 sacks

 

61. A rectangular foundation cross-section has a bed plate dimension of 8 ft x 10 ft. The uniform clearance on each side is 1 ft. The height of foundation is 4.5 ft. If the weight of the steel bar reinforcements needed is 1/2% of weight of foundation, find the weight of steel bars. Use concrete density of 2400 kg/m3 .

A.  173.47 kg      *B. 183.47 kg     C. 163.47 kg      D. 153.47 kg

SOLUTION:

      A = (8 + 2) (10 + 2) = 120 m2

V = Ah = 120(4.5) = 540 ft3 = 15.29 m3

W = wV = (2400)(15.29) = 36,693.25 kg

Weight of steel bars = (1/2%) Wf = 0.005(36,693.25) = 183.47 kg

 

62. A steam pipe having a surface temperature of 250C passes through a room where the temperature is 27 C. The outside diameter of pipe is 100 mm and emissivity factor is 0.8. Calculate the radial heat loss for 3 m pipe length.

A.  1434.7 W B. 37.46 W  *C. 2651.82 W           D. 3545.45 W

SOLUTION:

      A = DL =  = 0.425m2

      Solving for heat due to radiation:

      Tg = 250 +273 = 523K

      T2 = 27 +273 = 300K

      Qa = 20,408.4 x 104 AF(T14 – T24), J/hr = 20,408.4 x 104(0.8)(0.7539)[(523)4 – (300)4]

      Qr = 10,266,539.06 j/hr x 1hr/3600sec = 2851.82 W

 

63. Brine enters a circulating brine cooler at the rate of 60 m3/hr at -*C and leaves at -18C. Specific heat of brine is 1.072 KJ/kg-K and specific gravity of 1.12. Determine the tons of refrigeration.

A.  53.5 TR  B. 65.3 TR  C.33.5 TR   *D. 56.9 TR

SOLUTION:

      Density of brine = 1.12(1000 kg/m3) = 1120 kg/m3

      m = (1120)(60)/3600 = 18.67 kg/sec

      Q = mcp = 18.67(1.072)(-8 + 18) = 200.11 KW

      TR = 200.11/3.516 = 56.91 Tons of refrigeration

 

64. A turbo-charged, 16 cylinder, Vee-type diesel engine has an air consumption of 3,000 kg/hr per cylinder at rate load and speed. This air is drawn in through a filter by a centrifugal compressor directly connected to the exhaust gas turine. The temperature of the air from the compressor is 135C and a counter flow air cooler reduces the air temperature to 45C before it goes to the engine suction heater. Cooling water enters air cooler at 30C and leaves at 40C. Calculate the log mean temperature difference.

A.  47.23C   B. 87.82C   *C. 43.34C  D. 65.24C

SOLUTION:

a = 45-30 = 15C

b = 135 – 40 = 95C

mean = [a - b] / [ln(ab)] = [95-15] / ln(95/15) = 43.34C

 

65. Water is flowing in a pipe with radius of 30 cm at a velocity of 5 m/s at the temperature in pipe. The density and viscosity of the water are as follows: density 997.9 kg/sec viscosity = 1.131 Pa-s. What is the Reynolds Number for this situation?

      *A. 2647    B. 96.2           C. 3100           D. 1140

SOLUTION:

      n = Dvg / v

Where:

      D = 2(0.30) = 0.60 m

      vg = 5 M/SEC

      v = 1.131/997.9 = 0.0011334 m2 / sec

      Nm = 0.60(5)/0.0011334 = 2,647

 

66. Compute the amount of condensate form during 10 minutes warm-up of 180 meter pipe conveys the saturated steam with enthalpy vaporization hfg = 1,947.8 LJ/kg. The minimum external temperature of pipe is 2C. The final temperature of pipe is 195C. The specific heat of pipe material is 0.6 KJ/kg-C. The specific weight is 28 kg/m.

A.  249.69 kg            B. 982.45 kg            *C. 299.64 kg           D. 423.45 kg

SOLUTION:

      mp = mass of pipe = 28(180) = 5,040 kg

      Heat Loss by steam = Heat loss from pipe

      m(hg - hf) = mpcp (t2 – t1)

      m(1947.8) = (5040)(0.6)(195-2)

      m = 299.64 kg

 

 

67. The discharge pressure of an air compressor is 5 times the suction pressure. If volume flow at suction is 0.1 m3/sec, what is the suction pressure if compressor work is 19.57 kw? (use n=1.35)

A.  97 kpa   *B.98 kpa   C. 99 kpa   D.100 kpa

SOLUTION:

W = [(P2/P1)n-1/n – 1]

      19.57 = 1.35(P1)(0.1)/(1.35-1)[(5)1.35-1/1.35 – 1]

      P1 = 98 KPa

 

68. The initial condition of air in an air compressor is 98 KPa and 27C and discharge air at 450 KPa. The bpre and stroke are 355 mm and 381 mm, respectively with percent cleared of 8% running at 300 rpm. Find the volume of air at suction.

A.  541.62 m3/hr   B. 551.62 m3/hr         C. 561.62 m3/hr         *D. 571.62 m3/hr

SOLUTION:

      ev = 1 + c – c(P2/P1)1/n = 1 + 0.08 - 0.08(450/98)1/1.4 = 0.842

      VD =  D2 LN =  (0.355)2 (0.381)(300/60) = 0.1885 m3/sec

      V1 = 0.1885(0.842) = 0.15878 m3/sec = 571.62 m3/hr

69. An air compressor has a suction volume of 0.35 m3/sec t 97 KPa and discharges to 650 KPa. How much power saved by the compressor of there are two stages?

A.  18.27 KW       B. 16.54 KW       C. 13.86 KW       *D.  11.58 KW

SOLUTION:

      W = [(P2/P1)n-1/n – 1] = (1.4 x 97 x 0.35)/(1.4 -1) [(650/97)1.4-1/1.4 – 1] = 85.79 KW

      For two stages :

      Px = (P1P2)1/2 = (97 x 650)1/2 = 251.097 KPa

      W = [(Px/P1)n-1/n – 1] = 2(1.4)(97)(0.35)/(1.4 – 1) [(251.0.97/97)1.4-1/1.4 – 1] = 74.208 KW

      POWER SAVED = 85.79 – 74.208 = 11.582 KW

 

70. A two stage air compressor has an intercooler pressure of 4 kg/cm2. What is the discharge pressure if suction pressure is 1 kg/cm2?

A.  3 kg/cm2 B. 9 kg/cm2 C. 12 kg/cm2      *D. 16 kg/cm2

SOLUTION:

      Px = (P1P2)1/2

      Px2 = P1(P2)

      42 = 16 kg/cm2

 

71. A two stage air compressor compresses air at 100 KPa and 22C discharges to 750 KPa. If intercooler intake is 105C. Determine the value of n.

A.  1.400    *B. 1.325   C. 1.345    D. 1.288

SOLUTION:

      P­­­x =  = 273.86 Kpa

1.281 =

      n = 1.326

 

72. A single acting compressor has a volumetric efficiency of 89%, operates at 500 rpm. It takes in air at 900 KPa and 30C and discharges it at 600 KPa. The air handled is 8 m3/min measured at discharge condition. If compression is isentropic, find mean effective pressure in KPa.

      *A. 233.34  B. 973.17   C. 198.34   D. 204.82

SOLUTION:

      P1V1K = P2V2K

      100(V11.4) = 600(6)1.4

      V1 = 28.768 m3/min

      VD = 28.768/0.89 = 32.32 m3/min

      W = n P1V1/n-1 x  [(P2 / P1)n-1/n – 1] = [(1.4 x 100 x 32.32)/(1.4 – 1)] x [(600/100)1.4-1/1.4 – 1]

      W = 7562.19 KJ/min

      W = Pm x Vd

7562.19 = Pm x 32.32

Pm = 233.34 KPa

 

73. A water-jacketed air compressed handles 0.343 m3/s of air entering at 96.5 KPa and 21C and leaving at 460 KPa and 132C; 10.9 kg/h of cooling water enters the jacket at 15C and leaves at 21C. Determine the compressor brake power.

A.  26.163 KW      *B. 62.650 KW           C. 34.44 KW       D. 19.33 KW

SOLUTION:

      T2/T1 = (P2/P1) n-1/n

          (132+273) / (21+273) = (480/96.5)n-1/n

      n = 1.249

      W = (1.249 x 96.5 x 0.343) / (1.249-1) [(480 / 96.5)1.249-1/1.249 – 1]

      W = 62.57 KW

      Q = heat loss = mcp(t2 – t1) = (10.9/3600)(4.187)(21 – 15) 0.075 KW

      Brake power = W + Q = 62.57 + 0.076 = 62.65 KW

 

74. A double suction centrifugal pumps delivers 20 ft3/sec of water at a head of 12 m and running at 650 rpm. What is the specific speed of the pump?

A.  5014.12 rpm    B. 6453.12 rpm          *C. 2770.73 rpm   D. 9966.73 rpm

SOLUTION:

      N = N(Q)1/2 / h3/4

      Q = 20/2 ft3/sec x 7.481 gal/ft3 x 60 sec/1min = 4,488.6 gal/min

      h = 12 x 3.281 = 39.37 ft

      N = (650 x (4,488.6)1/2)/(39.37)3/4

          N = 2,770.73 rpm

 

75. 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 30 ft.

A.  6        B. 4        *C. 5       D. 7

SOLUTION:

      Wp = w Q h

      15 x 0.746 = 9.81(400 gal/min x 0.00785m3/gal x 1/60)h

      h = 45.20 m x 3.281 ft/m = 148.317 ft

      Number of stages = 148.317/40 = 4.94 stages = 5 stages

 

76. The suction pressure of a pump reads 3 in. of mercury vacuum and discharge pressure reads 140 psi is use to deliver 120 gpm of water with specific volume of 0.0163 ft3/lb. Determine the pump work.

A.  4.6 KW   B. 5.7 KW   *C. 7.4 KW  D. 8.4 KW

SOLUTION:

      P1 = -3 in Hg x 101.325/29.92 = -10.16 KPa

      P2 = 140 psi x 101.325/14.7 = 965 KPa

      w = 1/v = 1/0.163 = 61.35 lb/ft3 x 9.81/62.3 = 9.645 KN/m3

      h = (P2 – P1)/w = (965 +10.16)/9.645 = 101.105 m

      Q = 120 gal/min x 3.785/1gal x 1m3/1000li x 1/60 = 0.00757 m3/sec

      P = w Q h = 9.645(0.00757)(101.105) = 7.38 KW

 

77. A submersible pump delivers 350 gpm of water to a height of 5 ft from the ground. The pump were installed 150 ft below the ground level and draw down of 8 ft during the operation. If water level is 25 ft above the pump, determine the pump power.

A.  7.13 KW        B. 4.86 KW        C. 7.24 KW        *D. 9.27 KW

SOLUTION:

      h = 5 + 150 – (25 – 8) = 138/3.281 = 42.06 m

      Q = 350 gal/min x 0.003785 m3/gal x 1 min/60sec = 0.02246 m3/sec

      Wp = w Q h = 9.81(0.02246)(42.06) = 9.27 KW

 

78. A vacuum pump is used to drain a flooded mine shaft of 20 water. The pump pressure of water at this temperature is 2.34 KPa. The pump is incapable of lifting the water higher than 16 m. What is the atmospheric pressure?

      *A. 159.30        B. 32.33          C. 196.22         D. 171.9

SOLUTION:

      Using Bernoulli’s Theorem:

      P1/w + V12/2g + z1 = P2/w + V2/2g + z2

          P1/w = P2/w + (V22 - V12)/2g + (z2 - z1)

      P1/9.81 = 2.34/9.81 + 0 + 16

      P1 = 159.30 KPa

 

79. 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 3000 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           B. 5.33                 *C. 3.71          D. 6.34

SOLUTION:

      V =  D N

      V =

       D (3000/60) = (2(32.2)(38))1/2

      D = 0.315 ft = 3.708 inches

80. A fan pressure of 2.54 cm of water t 1.42 m3 per second of air at static pressure of 2.54 cm of water through a duct 300 mm diameter and discharges it through a duct 275 mm diameter. Determine the static fan efficiency if total fan mechanical is 75% and air measured at 25 and 60 mm Hg.

A.  50.11%         *B. 53.69%        C. 65.67%         D. 45.34%

SOLUTION:

      wA = P/RT = 101.325/(0.287)(25 + 273) = 1.18 kg/m3

      hA = hwww/wA = (0.0254)(1000)/1.18 = 21.52 m

      vA = 1.42/(/4)(0.3)2 = 20.09 m/s

      Vd = 1.42/(/4)(0.275)2 = 23.9 m/s

      hv = (23.9)2 – (20.09)2 / 2(9.81) = 8.54 m

      h = ha + hv = 21.52 + 8.54 = 30.06 m

      eT = wa Q h/BP

      0.75 = (1.18 x 0.00981)(1.42)(30.06) / BP

      BP = 0.6588 KW

      ep = wa Q hs/BP = (1.18 x 0.00981)(1.42)(21.52) / 0.6588 = 53.69%

 

81. A water cooler uses 50 lb/hr of melting ice to cool running water from 80 to 42.  Based on te inside coil area, U1 = 110 Btu/hr-ft2-. Find the gpm of water cooled.

A. 0.10 GPM B.  0.21 GPM*C. 0.38 GPM            D. 0.45 GPM

SOLUTION:

            Q = mf L = mwcpw(t1 – t2)

            50 (144) = mW(1)(80-42)

            mw = 189.474 lb/hr

            V = (189.474/62.4) (7.48/60) = 0.38 GPM

 

82. The charge in a Diesel engine consists of 18.34 grams of fuel, with lower heating value of 42,571 KJ/kg, and 409 grams of fuel and products of combustion. At the beginning of compression, t1 = 60. Let rk = 14. For constant cP = 1.11 KJ/kg-C, what should be the cut-off ratio in the corresponding ideal cycle?

A.  2.05           B. 2.34     C. 5.34           *D. 2.97

SOLUTION:

            QA = mfQh = 0.01283(42,571) = 780,752 KJ

            T2/T1 = rkk-1

            T2 = (60 + 273)1.4-1 = 956.964K

            mt + mg = 409

            mt + ma + mf = 409

            ma = 409 – 2(18.34) = 372.32 grams

            QA = macp(t3 – t2)

            780.752 = 0.37232(1.11)(T3 – 956.964)

            T3 = 2846,146 

            rC = T3/T2 = 2846.146/956.964 = 2.97

 

83. The gain of entropy during isothermal nonflow process of 5 lb of air at 60 is 0.462 Btu/R. Find the V1/V2.

A.  3.85           *B. 0.259         C. 1.0            D. 0.296

SOLUTION:

            s = m R T ln(V2/V1)

            0.462 = 5 (53.33/778) ln (V2/V1)

            V2/V1 = 3.85

            V1/V2 = 1/3.85 = 0.259

 

84. An auditorium seating 1500 people is to be maintained at 80 dry bulb and 85 wet bulb temperature when outdoor air is at 91 dry bulb and 75 wet bulb. Solar heat load is 110,000 Btu/hr and supply air at 60 determine the amount of supply air.

*A. 93,229.17 lb/hr     B. 83,229.17 lb/hr      C. 73,229.17 D. 63,229.17 lb/hr

SOLUTION:

            Sensible heat per person = 225 Btu/hr

            Qa = 225(1500) + 110,000 = 447,500 Btu/hr

            Qa = m cp(t1 – t2)

            447,500 = ma(0.24)(80 – 60)

            ma = 93,229.17 lb/hr

 

85. In a Brayton cycle that operates between temperature limits of 300K and 1773K wit k = 1.4, determine the temperature at the end of the compression (isentropic) for maximum work of the cycle.

A.  700K           B. 590.5K         *C. 730K    D. 350K

SOLUTION:

            For maximum work: T2 = (T1T3)1/2

            T2 = (300 x 1773)1/2 = 730K

 

86. At 35% solution leaves the absorber and 30% solution enters the absorber. The heat removed from the absorber by cooling water is 547.6 Btu and ammonia is superheated by 10. Find the pound per pound of ammonia gas from the evaporating coils.

A.  11       B. 12       *C. 13            D. 14

SOLUTION:

            n = lb/lb of ammonia gas from the coils

            n = (1 - x2) / (x1 - x2) – 1

            n = (1- 0.3) / (0.35 – 0.3) – 1 = 13

 

87. A Carnot refrigeration system operates at Tmax / Tmin = 1.5. Find the KW per tons of refrigeration.

A.  1.91     B. 2.15           *C. 1.76    D. 1.55

SOLUTION:

            Wo / TR = 3.516 / COP = 3.516 / (Tmin / (Tmax – Tmin)

            Wo / TR = [3.516(Tmax – Tmin)] / Tmin

            Wo / TR = 3.516[(Tmax / Tmin) – 1] = 3.516(1.5-1) = 1.758 KW/TR

 

88. Assume 8 ft3 of air at 100 psi, 100 are compressed isothermally to a volume of 2 ft3. For each of end states of the process, find the bulk modulus.

*A. 400 and 100 psi B. 400 and 110 psi C. 400 and 120 psi D. 400 and 130

 

89. An empty open can is 30 cm high with a 15 cm diameter. The can, with the open end and down, is pushed under water with a density of 1000kg/m3. Find the water level in the can when the top of the can is 50 cm below the surface.

A.  17.20 cm *B. 2.12 cm C. 4.20 cm  D. 5.87cm

SOLUTION:

      Consider the water pressure

      Pw = w h + 1010.325 + (0.8-x)(9.81) + 101.325 = 109.173 – 9.81x

      Consider the air pressure

      P­­­­­1V1 = P2V2

      101.325(Ax0.3) = P2[A(0.3-x)]

      P2 =

      Pw = P2

     

      109.173 – 9.81x =

      9.81x2 – 112.116x + 2.3705 = 0

      By quadratic formula:

      X = 0.02118 m = 2.12 cm

 

90. A cylindrical pope with water flowing downward at 0.03 m3/s having top diameter of 0.08, bottom diameter of 0.04 m and a height of 1.5m. Find the pressure between the pipe

A.  154.63 Kpa     B. 197.93 Kpa     *C. 252.44 Kpa    D. 243.92 Kpa

SOLUTION:

        + Z1 =  + Z2

     

     

 

      Z1 – Z­2 = 1.5 m

      Z2 – Z1 = -1.5 m

 

      V1 =  = 5.968 m/s

 

      V2 =  = 23.87 m/s

     

 

      P1 – P2 = 252.44 Kpa

 

 

91. Determine the size of pipe which will deliver 8 liters of medium oil (v= 6.10 x 10-6 m2/s) assuming laminar flow conditions:

A.  622 mm   B. 754 mm   C. 950 mm   *D. 835 mm

SOLUTION:

 

      V =

 

      Re =

      For laminar flow, Re = 200

 

      2000 =

      d = 0.835 m = 835 mm

 

92. The type of flow occupying in a 1 cm diameter pipe which water flows at a velocity of 2.50 m/s. Use v = 1.13 x 10-6 m2/s for water.

      *A. turbulent     B. constant       C. laminar D. none of the above

 

SOLUTION:

     

      Re =

      Re =

           

      Re = 22,124

      Since it is greater than 2000 then it is turbulent flow

 

93. What is the force is exerted by water jet 60 mm diameter if it strikes a wall at the rate of 15 m/s?

      *A. 636.17 N            B.442.62 N C. 764.23 N       D. 563.34 N

SOLUTION:

      F = w Q v

      Q = A v =  = 0.0424 m3/s

      F = (1000)(0.0424)(15) = 636.17 N

 

94. A 300 mm diameter pipe discharges water at the rate of 200 li/s. Point 1 on the pipe has a pressure of 260 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.  4.29 m   B. 2.59 m   C. 6.32 m   *D. 1.87 m

SOLUTION:

      hL

      hL =  

 

95. Water flowing at the rate of 10 m/s from an orifice at the bottom of a reservoir. Find the pressure at the bottom of the reservoir.

A.  30 kpag B. 40 kpag *C. 50 kpag       D. 60 kpag

SOLUTION:

      h = V2/ 2g = 102/ 2(9.81) = 5.0968 m

      P = w h = 9.81(5.0968) = 50 kpag

 

96. Steam flows through a nozzle at 400oC and 1 Mpa (h = 3263.9 KJ/kg) with velocity of 300 m/s. Find the stagnation enthalpy.

A.  3300 KJ/kg     B. 3290 KJ/kg     *C. 3320 KJ/kg    *D. 3309 KJ/kg

SOLUTION:

      ho = h + v2/2000 = 3263.9 + 3002/2000 = 3309 KJ/kg

 

97. Air flows through a nozzle at a speed of 350 m/s. Find the stagnation temperature if entrance temperature is 200oC.

A.  241.25oC B. 251.25oC *C. 261.25oC      D. 271.25oC

SOLUTION:

      To = T1 + v2/2000Cp = (2000 + 273) + 3502/2000(1)

      To = 534.25oK = 261.25

 

98. Carbon dioxide flows through a nozzle with a speed of 400 m/s. Compute the dynamic temperature.

A.  92.56oK *B. 94.56oK C. 96.56oK D. 98.56oK

SOLUTION:

      For CO2: Cp = 0.846 KJ/kg-K

      Dynamic temperature = v2/2000Cp = 4002/2000(0.846) = 94.56oK

 

99. Carbon dioxide flows through a nozzle with a speed of 380 m/s. The entrance condition of nozzle is 250oC and 1200 kpa. Find the stagnation pressure.

      *A. 2,136.34 kpa B. 2,146.34 kpa   C. 2,156.34 kpa   D. 2,166.34 kpa

   SOLUTION:

      T1 = 250 + 273 = 523oK

      To = T1 + v2/2000 = 523 = 3802/2000 = 595.2oK

      P1 = 1200 kpa

      T1/To = (P1/Po)k-1/k

      For CO2: k = 1.289

      523/595.2 = (1200/Po)1.289-1/1.289

      P0 = 2,136.34 kpa

 

100. Air enters a diffuser with a velocity of 200 m/s. Determine the velocity of sound if air temperature is 30oC.

      *A. 349 m/s       B. 359 m/s C. 369 m/s D. 379 m/s

SOLUTION:

      C =

         

 

101. Air flows through a nozzle with temperature of entrance of 420oK stagnation temperature of 468oK. Find the mach number.

A.  0.744    *B. 0.754   C. 0.764    D. 0.774

SOLUTION:

      To = T1 + v2/2000Cp

      468 = 420 + v2/2000

      v  = 309.838 m/s

      C =

      M = v/C = 309.838/410.8 = 0.754

 

102. Air at 300oK and 200 kpa is heated at constant pressure to 600oK. Determine the change of internal energy.

A.  245.58 KJ/kg   B. 235.58 KJ/kg   C. 225.58 KJ/kg   *D. 215.58 KJ/kg

SOLUTION:

      ΔU = mCv (T2 – T1) = 1(0.7186)(600 -300) = 215.58 KJ/kg

 

103. An insulated rigid tank initially contains 1.5 lb of helium at 80oF and 50 psia. A paddle wheel with power rating of 0.02 hp is operated within the tank for 30 min. Determine the final temperature.

A.  159.22oF B. 169.22oF *C. 179.22oF      D. 189.22 oF

SOLUTION:

      W = ΔU = m Cv (T2 – T1)

      0.02 hp (0.50hr)(2545Btu/hr/hp) = 1.5(0.171)(t2 – 80)

      t2 = 179.22oF

 

104. A 4m2 asphalt pavement with emissivity of 0.85 has a surface temperature of 50oC. Find the maximum rate of radiation that can be emitted from the surface.

A.  2,068.32 watts       B. 2,078.32 watts       C. 2,088.32 watts       *D. 2.098.32 watts

SOLUTION:

      Qr = e kev A Ts4

      Kev = 5.67 x 10-8 ( Stefan Boltzman constant)

      Qr = 0.85(5.67 x 10-8)(4)(50 +273)4 = 2,098.32 watts

 

105. Air at 10oC and 90 kpa enters a diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area diffuser is 0.40 m2. Determine the mass flow rate of air.

A.  72.79 kg/s     B. 74.79 kg/s     C. 76.79 kg/s     *D. 78.79 kg/s

SOLUTION:

      W = P/RT = 80/0.287(10 + 273) = 0.985 kg/m3

      m  =  w v A = 0.985(200)(0.40) = 78.79 kg/s

 

106. Consider a refrigeration whose 40 watts light bulb remains on continuously as a result of a malfunction of the switch. If the refrigerator has a COP of 1.3 and the cost of electricity is 8 cents per kw-hr., determine the increase in the energy consumption of the refrigerator and its cost per year if the switch is not fixed.

      *A. P49.59 B. P47.59   C. P45.59   D. P43.59

SOLUTION:

      COP = RE/Wref

      1.3 = 40/Wref

      Wref = 30.769 watts

      W = Wb + Wref = 40 + 30.769 = 70.77 watts

      W = 0.07077 Kw

      Cost = 0.07077(8760)(P0.08) = P49.59

 

107. A 75 hp motor that has an efficiency of 91% is worn out and is replaced by a high-efficiency motor that has an efficiency of 95.4%. Determine the reduction in heat gain of the room due to higher efficiency under full-load conditions.

A.  2.24 KW       *B. 2.44 KW       C. 2.64 KW D. 2.84 KW

SOLUTION:

      P01 = (75 x 0.746)(0.91) = 50.91 KW

      P02 = (75 x 0.746)(0.954) = 53.376 KW

      Qreduced = 53.376 – 50.91 = 2.44 KW

 

108. A household refrigerator that has a power input of 450 watts and a COP of 2.5 is to cool five large watermelons, 10 kg each, to 8oC. If the watermelons are initially at 20oC, determine how long will take for the refrigerator cool them. The watermelons can be treated as a water whose specific heat is 4.2 KJ/kg-oK.

A.  2220 seconds   B. 2230 seconds   *C.2240 seconds   D. 2250 seconds

SOLUTION:

      COP = RE/Wc

      2.5 = RE/450

          RE = 1,125 watts

      RE = m cp (t2 – t1)

      450 t = (10 x 5)(4.2)(20-8)

      t = 2240 seconds

 

109. When a man returns to his wall-sealed house on a summer day, he finds that the house is at 32oC. He returns on the air conditioner which cools the entire house to 20oC in 15 minutes, if COP is 2.5, determine the power drawn by the airconditioner. Assume the entire mass within the house is 800 kg of air for which cv = 0.72 KJ/kg-K, cp = 1.0KJ/kg-K.

A.  1.072 KW             B. 2.072 KW             *C. 3.072 KW      D. 4.072 KW

SOLUTION:

      RE = m cv (T2 –T1) = (800/15x60)(0.72)(32-20)

      RE = 7.66 KW

      Wc = 7.68/2.5 = 3.072 KW

 

110. A heat source at 8000K losses 2000 KJ of heat to a sink at 500oK. Determine the entropy generated during this process.

       *A. 1.5 KJ/K           B. 2.5 KJ/K             C. -2.5 KJ/K            D. 4 KJ/K

SOLUTION:

       ΔSsource = -2000/800 = -2.5

       ΔSsink = 2000/500 = 4

       ΔSgen. = -2.5 + 4 = 1.5 KJ/K

 

111. Helium gas is compressed in an adiabatic compressor from an initial state of 14 psia and 50oF to a final temperature of 320oF in a reversible manner. Determine the exit pressure of Helium.

A.  38.5 psia           *B. 40.5 psia           C. 42.5 psia            D. 44.5 psia

SOLUTION:

       T2/T1 = (P2/P1)n-1/n

           (320 + 460)/(50 +460) = (P2/14)1.587-1/1.587

       P2 = 40.5 psia

 

112. Air  pass thru a nozzle with efficiency of 90%. The velocity of air at the exit is 600 m/s. Find the actual velocity at the exit.

A.  382 m/s       B. 540 m/s        C. 458 m/s *D. 568 m/s

SOLUTION:

       e = (v2/v3)2

       0.9 = (v2/600)2

       v2 = 568.21 m/s

 

113. A 50 kg block of iron casting at 500K is thrown into a large lake that is at a temperature of 258oK. The iron block eventually reaches thermal equilibrium with the lake water. Assuming average specific hear of 0.45 KJ/kg-K for the iron, determine the entropy generated during this process.

       *A. -12.65 KJ/k B. 16.97KJ/K            C. 4.32 KJ/K            D. 6.32 KJ/K

SOLUTION:

       ΔSiron = m c ln (T2/T1) = 50(0.45)ln(285/500) = -12.65 KJ/K

       ΔSlake = Q/T = [50(0.45)(500-285)]/285 = 16.97 KJ/K

       ΔSgen. = -12.65 + 16.97 = 4.32 KJ/K

 

114. A windmill with a 12 m diameter rotor is to be installed at a location where the wind is blowing at an average velocity of 10 /s. Using standard conditions of air (1 atm, 25oC), determine the maximum that can be generated by the windmill.

A.  68 KW         *B. 70 KW         C. 72 KW          D. 74 KW

SOLUTION:

       w = P/RT = 101.325/(0.28)(25+ 273) = 1.1847 kg/m3

       m = w A v = 1.1847(π/4 x 122)(10) = 1,1339.895 kg/s

       KE = v2/2000 = 102/2000 = 0.05 KJ/kg

       Power = m KE = 1,339.895(0.05) = 70 KW

 

115. Consider a large furnace that can supply heat at a temperature of 2000oR at a steady rate of 3000Btu/s. Determine the energy of this energy. Assume an environment temperature of 77oF.

A.  2305.19 KW    *B. 2315.19 KW          C. 2325.19 KW     D. 2335.19 KW

SOLUTION:

       e =  = 0.7315

       W = e Q = 0.7315(3000) = 2194.5 Btu/s = 2315.19 KW

 

116. A heat engine receives hat from a source at 1200oK at a rate of 5000KJ/s and rejects the waste heat to a medium at 3000oK. The power output of the heat engine is 180 KW. Determine the irreversible rate for this process.

A.  190 KW        *B. 195 KW        C. 200 KW         D. 205 KW

SOLUTION:

       e = (1200 – 300) / 1200 = 0.75

       W = 0.75(500) = 375 KW

       Irreversibilities = 375 – 180 195 KW

 

117. A dealer advertises that he has just received a shipment of electric resistance heaters for residential buildings that have an efficiency of 100 percent. Assuming an indoor temperature of 21oC and outdoor temperature of 10oC, determine the second law efficiency of these heaters.

A.  8.74%         B. 6.74%          *C. 3.74%         D. 4.74%

SOLUTION:

       COP1 = 100% efficient = 1

       COP2 = (21 + 273) / (21 – 10) = 26.72

       e = COP1 /COP2  = 1/ 26.72 = 3.74 %

 

118. A thermal power plant has a heat rate of 11,363 Btu/KW-hr. Find the thermal efficiency of the plant.

A.  34%           B. 24%            C. 26%            *D. 30%

SOLUTION:

       e = 3412 / Heat rate = 3412 / 11363 = 30 %

 

119. A rigid tank contains 2 kmol of N2 and 6 kmol of CO2 gasses at 300oK and 115 Mpa. Find the tank volume us ideal gas equation.

A.  7.33 m3        B. 5.33 m3        C. 3.33 m3        *D. 1.33 m3

SOLUTION:

       PmVm = Nm R Tm

       15,000 Vm = (6 + 2)(8.314)(300)

       Vm = 1.33 m3

 

120. A spherical balloon with a diameter of 6 m is filled with helium at 20oC and 200 kpa. Determine the mole number.

       *A. 9.28 Kmol    B. 10.28 Kmol     C. 11.28 Kmol           D. 13.28 Kmol

SOLUTION:

       P V = N R T

       (200)[(4/3)(π)(6/2)3] = N (8.314)(20 + 273)

       N = 9.28 Kmol

 

121. The air in an automobile tire with a volume of 0.53 ft3 is at 90oF and 20 psig. Determine the amount of air that must be added to raise the pressure to the recommended value of 30 psig. Assume the atmospheric to be 14.7 psia and the temperature and the volume to remain constant.

       *A. 0.026 lb     B. 0.046 lb       C. 0.066 lb             D. 0.086 lb

SOLUTION:

       P V = m R T

       (20 + 14.7)(144)(0.53) = m1 (53.3)(90 + 460)

       m1 = 0.09034 lb

       (30 + 14.7)(144)(0.53) = m2(53.3)(90 + 460)

       m2 = 0.11634 lb

       madded = m2 – m1 = 0.11634 – 0.09034 = 0.026 lb

 

122. A rigid tank contains 20 lbm of air at 20 psia and 70oF. More air is added to the tank until the pressure and temperature rise to 35 psia and 90 oF, respectively. Determine the amount of air added to the tank.

A.  11.73 lb      *B. 13.73 lb      C. 15.73 lb             D. 17.73 lb

SOLUTION:

       P1V1 = m1 R1T1

       (20 x 144)(V1) = 20 (53.3)(70 + 460)

       V = 196.17 ft3

       P2V2 = m2R2T2

       (35 x 144)(196.17) = m2 (53.3)(90 + 460)

       m2 = 33.73 lbs

       madded = m2 –m1 = 33.73 – 20 = 13.73 lb

 

123. A rigid tank contains 5 kg of an ideal gas at 4 atm and 40oC. Now a valve is opened, and half of mass of the gas is allowed to escape. If the final pressure in the tank is 1,5 atm, the final temperature in the tank is:

       *A. -38oC         B. -30oC          C. 40oC     D. 53oC

SOLUTION:

       PV = m R T

       (4 x 9.81)(V) = 5(0.287)(40 + 273)

       V = 11.446 m3

       PV = mRT

       (1.5 x 9.81)(11.446) = (5/2)(0.287)(T)

       T = 234.74oK = -38.26oC

 

124. The pressure of an automobile tire is measured to be 200 kpa(gage) before the trip and 220 kpa(gage) after the trip at a location where the atmospheric pressure is 90 kpa. If the temperature of the air in the tire before the trip is 25oC, the air temperature after the trip is:

       *A. 45.6oC        B. 54.6oC         C. 27.5oC         D. 26.7oC

SOLUTION:

       T2 / T1 = P2 / P1

       T2 / (25+ 273) = (220 +90) / (200 + 90)

       T2  = 318.55 K

       t2 = 45.55oC

 

125. Water is boiling at 1 atm pressure in a stainless steel pan on an electric range. It is observed that 2 kg of liquid ater evaporates in 30 mins. The rate of heat transfer to the water is:

A.  2.97 KW       B. 0.47 KW        *C. 2.51 KW       D. 3.12 KW

SOLUTION:

       Q = mL = (2257) = 2.51 KW            

 

126. Consider a person standing in a breezy room at 20oC. Determine the total rate of heat transfer from this person if the exposed surface area and the average outer surface temperature of the person are 1.6 m2 and 29oC, respectively, and the convection heat transfer coefficient is 6 W/m2 with emissivity factor of 0.95.

A.  86.40 watts   B. 61.70 watts    C. 198.1 watts          *D. 168.1 watts

SOLUTION:

       Qc = h A (t2 – t1) = (6)(1.6)(29.20) = 86.40 watts

       Qr = (0.95)(5.67 x 10-6)[(1.6)(29 + 273)4 – (20 + 273)4] = 81.7 watts

       Q = Qc + Qr = 86.40 + 81.7 = 168.1 watts

 

127. Water is boiler in a pan on a stove at sea level. During 10 minutes of boiling, it is observed that 200 grams of water has evaporated. Then the rate of heat transfer to the water is:

A.  0.84 KJ/min   *B. 45.1 KJ/min   C. 41.8 KJ/min    D. 53.5 KJ/min

SOLUTION:

       Q = m L = (0.2 / 10) (2257) = 45.1 KJ/min

 

128. An aluminum pan whose thermal conductivity is 237 W/m-C has a flat bottom whose diameter is 20 cm and thickness 0.4 cm. Heat is transferred steadily to boiling water in the pan through its bottom at a rate of 500 watts. If the inner surface of the bottom of the pan is 105oC, determine the temperature of the surface of the bottom of the pan.

A.  95.27 oC       *B. 105.27oC      C. 115.27oC       D. 125.27oC

SOLUTION:

       A = π / 4 ( 0.20)2 = 0.0314 m2

       Q =

 

       500 =

 

      T2  = 105.27oC

 

129. For heat transfer purposes, a standing man can be modeled as a 30 cm diameter, 170 cm long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of 34oC. For a convection heat transfer coefficient of 15 W/m2- oC, determine the rate of heat loss from this man by convection in an environment at 20oC.

A.  316.46 watts   B. 326.46 watts   *C. 336.46 watts            D. 346.46 watts

SOLUTION:

      Qc = k A (t2 – t1) = 15 (π x 0.30 x 1.7) (34 – 20) = 336.46 watts

 

130. A 5cm diameter spherical ball whose surface is maintained at a temperature of 70oC is suspended in the middle of a room at 20oC. If the convection heat transfer coefficient is 15 W/m2 – C and the emissivity of the surface is 0.8, determine the total heat transfer from the ball.

A.  23.56 watts    *B. 32.77 watts   C. 9.22 watts           D. 43.45 watts

SOLUTION:

      A = 4 π r2 = 4 π (0.05)2 = 0.0314 m2

      Qc = h A (t2 – t1) = 15 (0.0314)(70 – 20) = 23.56 watts

      Qr = (0.80)(5.67 x 10-6)(0.0314)[(70 + 273)4 – (50 + 273)4] = 9.22 watts

      Q = Qr + Qc = 23.56 + 9.22 = 32.77 watts

 

131. A frictionless piston-cylinder device and rigid tank contain 1.2 kmol of ideal gas at the same temperature, pressure, and volume. Now heat is transferred, and the temperature of both system is raised by 15oC. The amount of extra heat that must be supplied to the gas in the cylinder that is maintained at constant pressure.

 

SOLUTION:

A.  0        B. 50 KJ    C. 100 KJ   *D. 150 KJ

      Q = m cp (t2 – t1) = (1.2 x 8.314)(1)(15) = 150 KJ

 

132. A supply of 50 kg of chicken needs at 6oC contained in a box is to be frozen to -18oC in a freezer. Determine the amount of heat that needs to be removed. The latent heat of chicken is 247 KJ/kg, and its specific heat is 3.32 KJ/kg-oC above freezing and 1.77 KJ/kg-oC below freezing. The container box is 1.5 kg, and the specific heat of the box material is 1.4 Kj/kg-oC. Also the freezing temperature of chicken is -2.8oC.

      *A. 15,206.4 KJ   B. 50.4 KJ  C. 15,156 KJ            D. 1,863 KJ

SOLUTION:

      Qchicken = 50 [3.32(6 + 2.8) = 247 1.77(-2.8 + 18)] = 15,156 KJ

      Qbox = 1.5(1.4)(6 + 8) = 50.4 KJ

      Q = 15,156 + 50.4 = 15, 206.4 KJ

 

133. Water is being heated in a closed pan on top of a range while being stirred by a paddle wheel. During the process, 30 KJ of heat is transferred to the water, and 5 KJ of heat is lost to the surrounding air. The paddle wheel work amounts to 500 N-m. Determine the final energy of the system if its initial energy is 10 KJ.

      *A. 35.5 KJ       B. 45.5 KJ        C. 25.5 KJ        D. 14.5 KJ

SOLUTION:

      Final energy = Qa + ΔU – Qloss + W = 30 + 10 – 5 + 0.50 = 

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