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####### PROBLEMSteam with an enthalpy of 800 kcal/kg enters a nozzle at a velocity of 80 m/s. Find the velocity of the steam at the exit of the nozzle if its enthalpy is reduced to 750 kcal/kg, assuming the nozzle is horizontal and disregarding heat losses. Take g = 9 m/s 2 and J constant = 427 kg m...

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PROBLEM Steam with an enthalpy of 800 kcal/kg enters a nozzle at a velocity of 80 m/s. Find the velocity of the steam at the exit of the nozzle if its enthalpy is reduced to 750 kcal/kg, assuming the nozzle is horizontal and disregarding heat losses. Take g = 9.81 m/s2 and J constant = 427 kg m/kcal. A. 452.37 m/s C. 651.92 m/s B. 245.45 m/s D. 427.54 m/s Solution: Energy Entering = Energy Leaving h1 + KE 1 = h2 + KE 2 KE 2 - KE 1 = h1 - h2 2

2

V 2 −V 1 2 2

= ( 800 - 750)

kcal 4187 J x kg 1 kcal

2

V 2 −(80 ) =¿ 418600 2 Answer: (C) V 2 = 651.92 m/s PROBLEM Steam is expanded through a nozzle and the enthalpy drop per kg of steam from the initial pressure to the final pressure is 60 kJ. Neglecting friction, find the velocity of discharge and the exit area of the nozzle to pass 0.20 kg/s if the specific volume of the steam at exit is 1.5 m3 /kg. A. 346.4 m/s , 879 m2 C. 765.6 m/s , 467 m2 2 B. 356.7 m/s , 278 m D. 346.4 m/s , 866 m2 Solution: The velocity of discharge, V d : V d = √ 2(specific enthalpy drop) V d =√ 2(60,000) thus; V d =346.4 m /s The exit area of the nozzle: Area x Velocity = mass flow x specific volume A x (346.4) = 0.20 (1.5) A = 866 m2 Answer: (D) 346.4 m/s , 866 m2 PROBLEM A 6 MW steam turbine generator power plant has a full-load steam rate of 8 kg/kW-hr. Assuming that no-load steam consumption as 15% of full-load steam consumption, compute for the hourly steam consumption at 75% load, in kg/hr. A. 37,800 kg/hr C. 30,780 kg/hr

B. 38,700 kg/hr

D. 30,870 kg/hr

Solution: Steam consumption at full-load: kg m 2 = (6000kW) 8 kW −hr = 48,000 kg/hr Steam consumption at no-load ( m 1 ) : m 2= (0.15 ) (48,000) = 7200 kg/hr Using two point form: y − y 1 y 2− y 1 = x − x 1 x 2−x 1 where: P1 (x 2 y 2) = P1 (0,7200) P2 ( x 2 y 2 ) =P2 (6000, 48,000)

(

)

P ( x,y ) = P (L, m s ) then; m s−7200 48,000−7200 = 6000−0 L−0 m s−7200 = 6.8 L m s=¿ 6.8 L +¿ 7200 At 75 % load : L = 0.75 (6000) = 4500 kW then Answer: (A) m s=¿ 37,800 kg/hr PROBLEM A 4 kg of air enters a turbine with enthalpy of 600 kJ and velocity of 250 m/s. The enthalpy at exit is 486 kJ and velocity of 170 m/s. What is the work developed if there is a heat loss of 10 kJ? A.128.83 kJ C. 80.2 kJ B.171.2 kJ D. 28.3 kJ Solution: H1 +

KE 1 =

250 ¿ ¿ Jx 1( ) 4 ¿ 2

H 2+Q+W t 170 ¿ ¿

kJ 1 =486+ (4)¿ 1000 J 2

Jx

kJ 1000 J

+ 10 + W t

Answer: (B)

W t = 171.2 kJ

PROBLEM Calculate drive horsepower for pumping 1703 L/min cold water to a tank suction at 127 mm Hg vacuum, delivery at 5.3 kg/cm2 ga., both measured close to pump, ep = 0.65. A. 31.42 HP C. 35.42 HP B. 20.42 HP D. 23.02 HP Solution: Let; hd = total head at discharge hg = total head at suction h = pump head By Bernoulli’s Equation: h = hd - h s 0 0 2 2 P d−P s V d −V s + +( zd - zs ) or ; h= γ 2g where: Pd = 5.3 kg/cm2 ga. Ps = -127 mm Hg = -0.1727 kg/cm2 2 2 (5.3+0.1727) kg /cm (100 cm/m ) = 54.72 m h= 3 1000 kg/m where: kg 1 m3 L 1000 3 (54.72) = 93,188.16 Pbrake = γ Q h = 1703 min 1000 L m

[

kg− m min

(

(

)] (

)

)

1 kW kg −m =15.24 kW 6116.3 min Pbrake = 20.42 Hp Drive HP for pump: 20.42 HP = 0.65 Answer: (A) HP = 31.42 HP PROBLEM Find the length of a suspension bunker to contain 181 tons of coal without surcharge; width 4.6 m; depth 4.3 m. The level capacity of a suspension bunker is 5/8 wdl where : w = width, d = depth, l = length. Density of coal, 800 kg/m3. A.18.30 m C. 17.61 m B. 13.80 m D. 12.61 m

Solution: 5 Wdl V= 8 Solving for the total volume: 181 (1000 )kg =226.25m 3 V= 3 800 kg / m then; 5 (4.6)(4.3) l 226.25 = 8 Answer: (A) L = 18.30 m PROBLEM 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 0.26 kg/kW-hr with a heating value of 43,912 kJ/kg. Calculate the brake thermal efficiency. A. 31.63 % C. 21.63 % B. 41.63 % D. 35.63 % Solution: etb =

Pb

m f Qh Solving for mf : kg (150 kW ) = 39 kg/hr = 0.0108 kg/s mf =0.26 kW−hr then; 150 etb = 0.0108 (43,912 ) Answer: (A) etb = 0.3163 or 31.63 % PROBLEM The brake thermal efficiency of a 1 MW diesel electric plant is 36 %. Find the heat generated by fuel in kW if the generator efficiency is 89 %. A. 3,121.10 kW C. 4,121.10 kW B. 3,528.64 kW D. 4,528.64 kW Solution: Pb Qs Solving for brake power, Pb : 1000 egen = Pb etb =

Pb =

1000 0.89

Pb=1,123.60 kW then; 1,123.60 etb = Qs Answer: (A) Qs = 3,121.10 kW

PROBLEM In an air–standard Brayton cycle, compressor receives air at 101.325 kPa, 21 ˚ C and leaves at 600 kPa at the rate of 4 kg/s. Determine the turbine work if the temperature of the air entering the tubine is 1000 ˚ C. A. 3000 kW C. 2028 kW B. 2701 kW D. 3500 kW Solution: Wt = m Cp (T3 – T4) Solving for T4: k−1 T4 P4 k = T3 P3 T4 101.325 = 600 1000 + 273 T4 = 765.83 K t4 = 492.83 ˚ C thus; Wt = 4(1)(1000-492.83) Answer: (C) Wt = 2028 kW

( )

(

1.4− 1 1.4

)

PROBLEM Kerosene is the fuel of a gas turbine plant: fuel – air ratio, mf = 0.012, T3 = 942 K, pressure ratio, rp = 4.5, exhaust to atmosphere. Find the available energy in kJ per kg air flow. Assume k = 1.34 and Cp = 1.13. A. 352.64 kJ/kg C. 252.64 kJ/kg B. 452.64 kJ/kg D. 552.64 kJ/kg Solution: The available Energy, Q: Q = (1+ mf) Cp (T3 – T4) Solving for T4: k−1 T3 P3 k = P4 T4

( )

1.34 − 1 972 = ( 4.5 ) 1.34 T4 T4 = 663.63 K then; Q = (1 + 0.012)(1.13)(972 – 663.63) Answer: (A) Q = 352.64 kJ/kg

PROBLEM: An ideal gas turbine operates with a pressure ratio of 10 and the energy input in the high temperature heat exchanger is 300 kW. Calculate the air flow rate for a temperature limits of 30 ˚ C and 1200 ˚ C. A. 0.25 kg/s ` C. 0.41 kg/s B. 0.34 kg/s D. 0.51 kg/s Solution: Qa = mCp (T3 – T2) Solving for T2: k−1 T2 P2 k = T1 P1 1.4−1 T2 = ( 10) 1.4 30 + 273 T2 = 585 K then; 300 = m(1)(1473 – 585) Answer: (B) m = 0.34 kg/s

( )

PROBLEM: In an air-standard Brayton cycle the inlet temperature and pressure are 20 ˚ C and 101.325 kPa. The turbine inlet conditions are 1200 kPa and 900 ˚ C. Determine the air flow rate if the turbine produces 12 MW. A. 21.41 kg/s C. 19.25 kg/s B. 20.20 kg/s D. 18.10 kg/s Solution: Solving for T4: k−1 T3 P3 k = P4 T4 1.4− 1 900 + 273 1200 1.4 = T4 101.325 T4 = 578.89 K then ; 12,000 = m (1)(1173 – 578.89 )

( )

(

Answer: (B) m = 20.20 kg/s

PROBLEM:

)

A gas turbine power plant operating on the Brayton cycle delivers 15 MW to a standby electric generator. What is the mass flow rate and the volume flow rate of air if the minimum and maximum pressures are 100 kPa and 500 kPa respectively and temperature of 20 ˚ C and 1000 ˚ C. A. 31.97 kg/s, 26.88 m3/s C. 41.97 kg/s, 26.88 m3/s 3 B. 36.98 kg/s, 28.99 m /s D. 46.98 kg/s, 28.99 m3/s Solution: PV = mRT Solving for m: k−1 T3 P3 k = T4 P4 1.4− 1 1000 + 273 500 1.4 = T4 100 T4 = 803.75 K Wt = mCp (T3 – T4) 15,000 = m(1)(1273 – 803.75) m = 31.97 kg/s then; 100V = 31.97(0.287)(20+273)

( )

( )

Answer: (A) V = 26.88 m3/s

PROBLEM: In a hydraulic plant the difference in elevation between the surface of the water at intake and the tailrace is 650 ft when the flow is 90 cfs, the friction loss in the penstock is 65 ft and the head utilized by the turbine is 500 ft. The mechanical friction in the turbine is 110 Hp and the leakage loss is 4 cfs. Find the hydraulic efficiency. A. 87.45 % C. 85.47 % B. 84.57 % D. 78.54 % Solution: Hydraulic Efficiency, eh : 500 eh = 650 −65 Answer: (C) eh = 0.8547 or 85.47 % PROBLEM: A hydro-electric power plant consumes 60,000,000 kW-hr per year. What is the net head if the expected flow is 1500 m3/min and over-all efficiency is 63%. A. 34.34 m C. 44.33 m B. 43.43 m D. 33.44 m Solution: Pw = γ Q h

Solving for water power, Pw: Generator Output enet = Water Power 60,000,000 /8760 Pw Pw = 10,871.93 kW then; substituting: 1500 10,871.93 = 9.81 60 0.63 =

( )

Answer: (C)

h

h = 44.33 m

PROBLEM: A pelton type turbine has a gross head of 40 m and a friction head loss of 6 m. What is the penstock diameter if the penstock length is 90 m and the coefficient of friction head loss is 0.001 (Morse). A. 2040 mm

C. 2440 mm

B. 3120 mm

D. 2320 mm

Solution: 2 f LV2 hL = gD Solving for V : V=

√ 2 gh

=

√ 2( 9.81 )(40−6)

then; substituting: 2 2f LV hL = gD 25.83 ¿ ¿ 6= ¿2 2(0.001 )(90 )¿ ¿

Answer: (A) D = 2.04 = 2,040 mm

PROBLEM:

= 25.83 m/s

The water velocity of a 5 m x 1 m channel is 6 m/s. What is the annual energy produced if the net head is 120 m and the over-all efficiency is 80 %. A. 494, 247, 258 kW-hrs

C. 247, 494, 528 kW-hrs

B. 247, 497, 582 kW-hrs

D. 472, 497, 582 kW-hrs

Solution: Annual Energy Produced = 0.80 Pw (8760) Solving for water power (Pw): Q = AV = (5x1)(6) = 30 m3/s Pw = 9.81(30)(120) = 35,316 kW then; substituting: Annual Energy Produced = 0.80(35,316)(8760) Answer: (C) Annual Energy Produced = 247, 494, 528 kW-hrs

PROBLEM: A hydro-electric impulse turbine is directly coupled to a 24 pole, 60 Hz alternator. It has a specific speed of 60 rpm and develops 3000 Hp. What is the required diameter assuming a peripheral speed ratio of 0.45. A. 0.661 m

C. 0.443 m

B. 0.552 m

D. 0.775 m

Solution: Φ=

лDN √2 g h

Solving for h:

N=

Ns =

120 f P

120 (60) 24

= 300 rpm

N √ HP h

60 =

=

5 4

300 √ 3000 5 4

h

h = 89.13 ft = 27.17 m then substituting: 0.45 =

л D(300 / 60 ) √2(9.81)(27.17)

Answer: (A) D = 0.661 m

PROBLEM: In a hydroelectric power plant the tail water elevation is at 500 m. What is the head water elevation if the net head is 30 m and the head loss is 5% of the gross head? A. 785.25 m

C. 528.57 m

B. 582.57 m

D. 758.25 m

Solution: hg = hhw – htw Solving for hg : h = hg + hL = hg + 0.05 hg = 1.05 hg 30 = 1.05 hg hg = 28.57 m then substituting:

28.57 = hhw – 500 Answer: (C) hhw = 528.57 m

PROBLEM: The tailwater and the headwater of a hydro-electric power plant are 150 m and 200 m respectively. What is the water power if the flow is 15 m3/s and a head loss of 10% of the gross head? A. 6,621.75 kW

C. 5,621.76 kW

B. 7,621.65 kW

D. 4,621.56 kW

Solution: Pw = γ Q h Solving for h: hg = hhw – htw = 200 – 150 = 50 m h = hg + hL = 50 – 0.10(50) = 45 then substituting: Pw = 9.81 (15)(45) Answer: (A) Pw = 6,621.75 kW

PROBLEM: In a hydroelectric power plant, water flows at 10 m/s in a penstock of 1 m 2 cross-sectional area. If the net head of the plant is 30 m and the turbine efficiency is 85%, what is the turbine output? A. 2,501.55 kW

C. 3,626.34 kW

B. 2,100.21 kW

D. 3,124.65 kW

Solution:

Turbine Output = 0.85 Pw Solving for Pw: Q = AV = 1(10) = 10 m3/s Pw = γ Q h = 9.81 (10)(30) = 2,943 kW then substituting: Turbine Output = 0.85(2,943) Answer: (A) Turbine Output = 2,501.55 kW

PROBLEM: A 75 MW power plant has an average load of 35,000 kW and a load factor of 65%. Find the reserve over peak. A.21.15 MW

C. 25.38 MW

B. 23.41 MW

D. 18.75 MW

Solution: Reserve over peak = Plant Capacity – Peak load Solving for Peak Load: Load Factor =

0.65 =

Average Load Peak Load

35,000 Peak Load

Peak Load = 53,846.15 kW = 53.846 MW then; Reserve over peak = 75 – 53.846 Answer: (A) Reserve over peak = 21.15 MW

PROBLEM:

A power plant is said to have/had a use factor of 48.5% and a capacity factor of 42.4%. How many hours did it operate during the year? A. 6,600.32 hrs

C. 8,600.32 hrs

B. 7,658.23 hrs

D. 5,658.23 hrs

Solution: Plant Use Factor =

Annual kW−hrs kW Plant Capacity x No . of hrs Operation

Plant Capacity Factor =

Annual Energy Produced kW Plant Capacity x 8760 hrs

Derived Formula : No. of hrs. Operation = 8760 x

(Capacity Factor ) (Use Factor)

= 8760 x

(0.424) (0.485)

Answer: (B) No. of hrs. Operation = 7,658.23 hour per day

PROBLEM: A 50,000 kW steam plant delivers an annual output of 238,000,000 kW-hr with a peak load of 42,860 kW. What is the annual load factor and capacity factor? A. 0.634, 0.534

C. 0.634, 0.543

B. 0.643, 0.534

D. 0.643, 0.543

Solution: Load Factor =

Average Load Peak Load

Solving for the Average Load; Ave. Load =

kW −hrs Energy No . of hrs∈one year

Load Factor = 27,168.94 42,860 Load Factor = 0.634

=

238,000,000 kW −hrs / yr 8760 hr / yr

=

27,168.94 kW

Annual Capacity Factor =

Annual Energy Produced kW Plant Capacity x 8760 hrs

=

238,000,000 kW −hr / yrs 50,000 kW x 8760 hrs

Annual Capacity Factor = 0.543 Answer: (C) 0.634, 0.543

PROBLEM: Calculate the use factor of a power plant if the capacity factor is 35 % and it operates 8000 hrs during the year? A. 38.325 %

C. 35.823 %

B. 33.825 %

D. 32.538 %

Solution: No. of hrs Operation = 8760

8000 = 8760

Factor ( Capacity ) Use factor

Factor ( Capacity ) Use factor

Use Factor = 0.38325 Answer: (A) Use Factor = 38.325 %

PROBLEM: If the air required for combustion is 20 kg per kg of coal and the boiler uses 3000 kg of coal per hr, determine the mass of gas entering the chimney. Assume an ash loss of 15 %. A. 40,644 kg/hr B. 70,200 kg/hr Solution: mg + mash = ma + mf where:

mair mf

= 20

C. 62,550 kg/hr D. 50,000 kg/hr

ma = 20 mf mg + 0.15 mf = 20 mf + mf mg = 20.85 mf = 20.85 (3000) Answer: (C) mg = 62,550 kg/hr

PROBLEM: A 15 kg gas enters a chimney at 10 m/s. If the temperature and pressure of a gas are 26°C and 100 kPa respectively, what is the diameter of the chimney? Use R=0.287 kJ/kg-K. A. 1.57 m

C. 2.22 m

B. 2.65 m

D. 1.28 m

Solution: Q = AVactual Solving for Q:

mg R g T g

Q = Vg =

Pg

=

15 (0.287 )(26 + 273 ) = 12.87 m3/s 100

then; substituting: 12.87 =

л D2 4

(10)

Answer: (D) D = 1.28m

PROBLEM: A two-stage air compressor at 90 kPa and 20°C discharge at 700 kPa. Find the polytropic exponent if the intercooler intake temperature is 100°C. A. 1.29

C. 1.4

B. 1.33

D. 1.25

Solution:

Tx Ty

( ) Px Py

=

n−1 n

Solving for Px: Px =

√ P 1 P2

=

√ 90 (700 )

=

(

= 250.40 kPa

then,

(100+273) (20+273)

250.40 90

)

n−1 n

Answer: (A) n = 1.29

PROBLEM: A two stage compressor receives 0.35 kg/s of air at 100 kPa and 269 K and delivers it at 5000 kPa. Find the heat transferred in the intercooler. A. 70.49 kW

C. 90.49 kW

B. 80.49 kW

D. 100.49 kW

Solution: Q = mCp(Tx – T1) Solving for Tx: Px =

Tx T1 Tx 269

√ 100 (5000 )

= 707.11 kPa

( ) Px P1

= =

(

k−1 k

707.11 100

1.4 −1 1.4

)

Tx = 470.40 K thus, Q = 0.35(1)(470.40 - 269) Answer: (A) Q = 70.49 kW

PROBLEM: A centrifugal pump discharged 20 L/s against a head of 17 m when the speed is 1500 rpm. The diameter of the impeller was 30 cm and the brake horsepower was 6.0Hp geometrically similar pump 40 cm in diameter is to run at 1750 rpm. Assuming equal efficiencies, what brake horsepower is required? A. 51.55 Hp

C. 40.14 Hp

B. 50.15 Hp

D. 45.15 Hp

Solution: New brake horsepower required;

P1 5 1

D N1

3

=

P2 D 25 N 2 3

6 3 ( 0.30) ( 1500) 5

=

P2 5

3

( 0.40) ( 1750)

Answer: (A) P2 = 40.14 Hp

PROBLEM: A pump delivers 20 cfm of water having a density of 62 lb/ft 3. The suction and discharge gage reads 5 in Hg vacuum and 30 psi respectively. The discharge gage is 5 ft above the suction gage. If pump efficiency is 70%, what is the motor power? A. 5.31 Hp

C. 4.31 Hp

B. 3.31 Hp

D. 6.31 Hp

Solution: Pmotor = Pwater / 0.70 Solving for Pwater: H =

P d− P s + γ

V d2−V s2 2g

+ (Zd – Zs)

=

[ 30−(−5)( 14.7 /29.92 )]( 144 ) 62

+ 0 + 5 = 80.38 ft

Pwater = 62 lb/ft3 (20ft3/min)(80.38 ft) = 99674.87 ft-lb/min (

1 Hp ) 33,000 ft −lb / min

Pwater = 3.02 Hp Answer: (C) Pmotor = 3.02 / 0.70 = 4.31 Hp

PROBLEM: Calculate the air power of a fan that delivers 1200 m 3/min of air through a 1 m by 1.5 m outlet. Static pressure is 120 mm WG and density of air is 1.18. A. 20.45 kW

C. 30.45 kW

B. 25.64 kW

D. 35.64 kW

Solution: Pair = γ Q h Solving for h:

Problem: Determine the temperature for which a thermometer with degrees Fahrenheit is numerically twice the reading of the temperature in degrees Celsius. A. -24.6

B. 320

C. 160

D. -12.3

Solution: ˚C = ˚F/2

equation-1

˚C = 5/9 (˚F/2) equation-2 Substitute eq. 1 to eq. 2 ˚F/2 = 5/9 (˚F-32) [˚F/2 = 5/9 (˚F-32)] x 18 9˚F = 10˚F - 320 320 = 10˚F - 9˚F Answer: ˚ F = 320

Problem: During takeoff in a spaceship, an astronaut is subjected to acceleration equal to 5 times the pull of the earth's standard gravity. If the astronaut is 180 lb m and the takeoff is vertical, what force does he exert on the seat? A. 4810.9 N

B. 4414.5 N

C. 8829 N

Solution: ΣFvertical = 0 F = m(5)(g) + m(g) g = 9.8 m/s2 ; k= 9.8 m/s2 F = (180 lbm/2.2)kg (5)(9.8 m/s2) + (180 lbm/2.2)kg (9.8m/s2) Answer: F = 4810.9 N

D. 9620 N

Problem: A pressure cooker operates by cooking food at higher pressure and temperature than is possible at atmospheric conditions. Steam is contained in the seal pot, with vent hole in the middle of the cover, allowing steam to escape. The pressure is regulated by covering the vent hole with a small weight, which is displaced slightly by escaping steam. Atmospheric pressure is 100 kPa, the vent hole area is 7 mm2, and the pressure inside should be 200 kPa. What is the mass of the weight? A. 0.107 kg

B. 1.05 kg

C. 1.75 kg

D. 0.1783 kg

Solution: ΔP = Ppressure cooker - Patm = 250 kPa - 100 kPa = 150 kPa ΔP = F/A [150 kPa = F/7] [1m/1000mm]2 F = 1.05 x 10-3 kN (1000 N/ 1kN) = 1.05 N (1kg / 9.81N ) Answer: F = 0.107 kg

Problem: A barometer can be used to measure an airplane's altitude by comparing the bar...