SOLVED MECHANICAL ENGINEERING PROBLEMS STEAM POWER PLANT Problem 1 A reheat steam has 13,850 kPa throttle pressure at the turbine inlet and a 2800 kPa reheat pressure. The throttle and reheat temperature of the steam is 540°C, condenser pressure is 3.4 kPa, engine efficiency of high pressure and low pressure is 75%. Find the cycle thermal efficiency. A. 34.46% C. 36.66% B. 35.56% D. 37.76% Solution:
ηth =
2.8 MPa 540°C
Reheat
Wt − Wp QA
T QA
1
Boiler
3
m
13.85 kPa1 540°C
2
H.P.
3
L.P.Turbine. 4
0.0034 MPa 6
2
Condenser
m 5
6
QR H2O out
4
5
S
H2O in
Pump
From Steam Table: At 13.85 MPa and 540°C: h1= 3434.1 kJ/kg (interpolated) At 2.8 MPa and 540°C: h3= 3548.5 kJ/kg h2 at s2 = s1 = 6.5436 kJ/kg : h2 = 2974.9 kJ/kg h4 at s4 = s3 = 7.3810 kJ/kg : h4 = 2204.5 kJ/kg h5 = hf at 0.0034 MPa = 109.84 kJ/kg v5 = vf at 0.0034 MPa = 0.0010032 m3 /kg Solving for h6: h6 – h5 = v5 (P6 – P5) h6 – 109.84 = 0.0010032 (13,850 – 3.4) h6 = 123.73 kJ/kg Wt = [(3434.1 – 2974.9) + (3548.5 – 2204.5)] × 0.75 = 1,352.4 kJ/kg Wp = h6 – h5 = 123.73 – 109.84 = 13.89 kJ/kg QA = (h1 – h6) + (h3 – h2) = (3434.1 – 123.73) + (3548.5 – 2974.9) = 3,883.97 kJ/kg Thus, 1,352.4 − 13.89 ηth = = 0.3446 3,883.97 ηth = 34.46% Answer
Problem 2 In a Rankine cycle, saturated liquid water at 1 bar is compressed isentropically to 150 bar. First by heating in a boiler and then by superheating at constant pressure of 150 bar, the water substance is brought to 750°K. After adiabatic reversible expansion in a turbine to 1 bar, it is then cooled in a condenser to a saturated liquid. How much work is generated in the turbine? A. 967.9 kJ/kg C. 796.9 kJ/kg B. 976.9 kJ/kg D. 769.9 kJ/kg Solution: Wt = h1 – h2 At 150 bar (15 MPa) and 750°K (477°C): h1 = 3240.5 kJ/kg ,
s1 = 6.2549 kJ/kg-°K
At 1 bar (0.10 MPa): h f = 417.46 kJ/kg h fg = 2258 kJ/kg
sf = 1.3026 kJ/kg-°K sfg = 6.0568 kJ/kg-°K
Solving for h2 : at s1 = s2 s1 = (sf + xsfg)2 6.2549 = 1.3026 + x (6.0568) x = 0.8176 or 81.76% h2 = h f + x h fg = 417.46 + (0.8176) 2258 = 2263.6 kJ/kg then, W t = 3240.5 – 2263.6 = 976.9 kJ/kg Answer Problem 3 A Batangas base industrial company operates a steam power plant with a reheat and regeneration. The steam enters a turbine at 300 bar and 900°K and expands to 1 bar. Steam leaves the first stage at 30 bar and part of it entering a closed heater while the rest is reheated to 800°K. Both section of turbine have adiabatic efficiency of 93%. A condensate pump exist between the main condenser and the heater. Another pump lies between the heater and condensate outlet line from the heater (condensed extracted steam). Compute for the extracted fraction of the total mass flow to the heater. A. 0.234 C. 0.765 B. 0.543 D. 0.485 Solution: Let m = fraction of steam extracted By heat balance in the heater: m (h3 – h7) = (1 – m) (h8 – h7) Using Steam Table and Mollier Chart: At 300 bar (30 MPa) and 900°K (627°C): h1 = 3528 kJ/kg s1= 6.31505 kJ/kg -°K (by interpolation) At 30 bar (3 MPa) and s1= 6.31505 kJ/kg -°K : h3 = 2883 kJ/kg (by interpolation)
h1 − h 2 = 0.93 h1 − h 3 3528 − h 2 = 0.93 3528 − 2883 h2 = 2928.15 kJ/kg h5 = 417.46 kJ/kg (hf at 0.10 MPa) v5 = 0.0010432 m3/kg (vf at 0.10 MPa) h6 – h5 = v5 (P6 – P5) h6 – 417.46 = 0.0010432 (3000 – 100) h6 = 420.48 kJ/kg h7 = 1008.42 kJ/kg (h f at 3 MPa) Thus, m ( 2928.15 – 1008.42 ) = (1 – m) (1008.42 – 420.48 ) m = 0.234 Answer Problem 4 A drum containing steam with 2.5 m in diameter is 7.5 m long. Of the total volume, 1/3 contains saturated steam at 800kPa and the other 2/3 contains saturated water. If this tank should explode, how much water would evaporate? Consider the process to be constant enthalpy. A. 2,948.11 kg C. 2,651.24 kg B. 2,424.62 kg D. 2,123.76 kg Solution: Volume of saturated steam and water: 1 π 2 3 Vs = ( 2.5 ) ( 7.5 ) = 12.27 m 34 2 π 2 3 Vw = ( 2.5 ) ( 7.5 ) = 24.54 m 34 From Steam Table 2: At 800 kPa (0.80 MPa) : v f = 0.0011148 m3/kg v g = 0.2404 m3/kg
hf = 721.11 kJ/kg hfg = 2048.0 kJ/kg hg = 2769.1 kJ/kg
Initial mass of steam, mS: mS =
12.27 = 51.04 kg 0.2404
Initial mass of water, mW: mW =
24.54 = 22, 012.92 kg 0.0011148
Initial quality, x1: 51.04 = 0.00231 51.04 + 22, 012.92 h = hf + x hfg = 721.11 + 0.00231(2048) h = 725.84 kJ/kg x1 =
After explosion, final condition will be atmospheric At 100°C and 101.3 kPa hf = 419.04 kJ/kg hfg = 2257.0 kJ/kg v f = 0.0010435 m3/kg hg = 2676.1 kJ/kg v g = 1.6729 m3/kg Quality of the final state: h2 = h1 h1 = hf + x2 hfg 725.84 = 419.04 + x2 (2257) x2 = 0.13593 Mass of steam at atmospheric pressure: mS = 0.13593 (51.04 + 22,012.92) mS = 2,999.15 kg The amount of vapor evaporated: mV = 2,999.15 – 51.04 = 2,948.11 kg
Answer
Problem 5 A back pressure steam turbine of 100,000 kW serves as a prime mover in a cogeneration system. The boiler admits the return water at a temperature of 66°C and produces the steam at 6.5 MPa and 455°C. Steam then enters a back pressure turbine and expands to the pressure of the process, which is 0.52 MPa. Assuming a boiler efficiency of 80% and neglecting the effect of pumping and the pressure drops at various locations, what is the incremental heat rate for electric? The following enthalpies have been found: turbine entrance = 3306.8 kJ/kg, exit = 2700.8 kJ/kg, Boiler entrance = 276.23 kJ/kg, exit = 3306.8 kJ/kg A. 22,504 kJ/kW-hr C. 25,200 kJ/kW-hr B. 24,500 kJ/kW-hr D. 30,506 kJ/kW-hr Solution: Wt = m (h1 – h2) = m ( 3306.8 – 2700.8 ) = 606 m kW 3600 × m ( h1 − h 2 ) 3600 × m ( 3306.8 − 2700.8 ) = = 13, 637,565m kJ/hr QA = ηb 0.80 13, 637,565m kJ = 22,504 Heat Rate = Answer 606m kW-hr Problem 6 A coal-fired power plant has a turbine-generator rated at 1000 MW gross. The plant required about 9% of this power for its internal operations. It uses 9800 tons of coal per day. The coal has a heating value of 6,388.9 kcal/kg and the steam generator efficiency is 86%. What is the net station efficiency of the plant in percent? A. 33.07% C. 37.05% B. 35.70% D. 42.05%
Solution: Net Output = 1000 (1 – 0.09) = 910 MW = 910,000 kW 9800 ( 907 ) ( 6,388.9 × 4.187 ) = 2, 752, 001 24 ( 3600 ) Net Output Net Station Efficiency = Heat Input 910, 000 = 0.3307 × 100% = 33.07% Answer = 2, 752, 001 Heat Input = mf Qh =
Problem 7 A superheat steam Rankine cycle has turbine inlet conditions of 17.5 Mpa and 530°C expands in a turbine to 0.007 Mpa. The turbine and pump polytropic efficiencies are 0.9 and 0.7 respectively. Pressure losses between pump and turbine inlet are 1.5 Mpa. What should be the pump work in kJ/kg? A. 17.3 kJ/kg C. 37.3 kJ/kg B. 27.3 kJ/kg D. 47.3 kJ/kg Solution: Wp =
v3 ( P4 − P3 ) ηp
where: 3 1 = 0.001 m kg 1000 P4 = 17.5 + 1.5 = 19 MPa = 19, 000 kPa
v3 =
P3 = 0.007 MPa = 7 kPa
ηp = 0.70 Wp =
0.001( 19,000 − 7 ) = 27.1 kJ kg 0.70
Answer
Problem 8 A steam plant operates with initial pressure of 1.70 MPa and 370°C temperature and exhaust to a heating system at 0.17 MPa. The condensate from the heating system is returned to the boiler at 65.5°C and the heating system utilizes from its intended purpose 90% of the energy transferred from the steam it receives. The turbine efficiency, ηt is 70%. If the boiler efficiency is 80%, what is the cogeneration efficiency of the system in percent? Neglect pump work. Steam Properties: At 1.7 MPa and 370°C: h = 3187.1 kJ/kg s = 7.1081 kJ/kg-K At 0.170 MPa: hf = 483.20 kJ/kg sf = 1.4752 kJ/kg-K hfg = 2216.0 kJ/kg sfg = 5.7062 kJ/kg-K At 65.5°C: hf = 274.14 kJ/kg A. 78% B. 102.10%
C. 91.24% D. 69%
Solution: h1 = 3187.1 kJ/kg Solving for h2 : s1= s2 = (sf + x sfg)2 7.1081 = 1.4752 + x2 (5.7062) x2 = 0.9871 h2 = (hf + x2 hfg) = 483.20 + 0.9871 (2216.0) = 2670.6 kJ/kg h3 = h4 = 274.14 kJ/kg Wt = (h1 – h2) ηt = (3187.1 – 2670.6) (0.70) = 361.55 kJ/kg QR = 0.90 (h2 – h3) = 0.90 (2670.6 – 274.14) = 2156.81 kJ/kg h1 − h 4 3187.1 − 274.14 = QA = =3641.2 kJ/kg ηb 0.80 Wt + Q R 361.55 + 2156.81 = = 0.6916 = 69.16% Cogeneration Efficiency = QA 3641.2
Answer
Problem 9 In a Rankine cycle, steam enters the turbine at 2.5 Mpa and condenser pressure of 50 kPa. What is the thermal efficiency of the cycle? Steam Properties: At 2.5 Mpa, hg = 2803.1 kJ/kg, Sg = 6.2575 kJ/kg-K At 50 kPa, Sf = 1.0910 kJ/kg-K, Sfg = 6.5029 kJ/kg-K, hf = 340.49 kJ/kg, hfg = 2305.4 kJ/kg, vf = 0.0010300 m3/kg Solution: W W − WP ( h1 − h 2 ) − ( h 4 − h 3 ) eth = NET = T = QA QA h1 − h 4 where: h g at 2.5 MPa = h1 = 2803.1 kJ/kg h f at 50 kPa = h 3 = 340.49 kJ/kg Solving for h 2 and h 4 : S = Sf + xSfg 6.2575 = 1.0910 + x ( 6.5029 ) x = 0.7945 h 2 = h f + xh fg = 340.49 + 0.7045 ( 2305.4 ) = 2172.13 kJ/kg h 4 = h f + vf ( P2 − P1 ) = 340.49 + 0.00103 ( 2500 − 50 ) = 342.98 kJ/kg thus; eth =
WNET ( 2803.1 − 2172.11) − ( 342.98 − 340.49 ) = QA 2803.1 − 342.98
eth = 25.55%
