Page 1. bar psi. kPa/MPa. Kg/cm2. Mbar. 0.025. 0.363. 2.5 kPa. 0.025. 25. 0.04. 0.58 5. 72.516. 500 kPa. 5.098. 5000. 6. 87.020. 600 kPa. 6.117. 6000.
pdf pressure card
(a) the specific volume at T = 240°C p = 1.25 MPa
ThermoHmwk Soln
15 kg/min. (3 MW). CGH2. 40 g/L. 69-65 g/L. 1
compressed hydrogen brunner
December 1 2000. J. Murthy. 11.4 A steam power plant operating in an ideal Rankine cycle has a high pressure of 5. MPa and a low pressure of 15 kPa.
as sol
Table 1. Saturation (Temperature) (continued). 5 t °C p
NISTIR Tab
Nov 9 2000 1. 2 s = C. 8.29 A mass and atmosphere loaded piston/cylinder contains 2 kg of water at 5 MPa
extra sol
5. Saturated liquid exits the open feedwater heater and saturated liquid exits State 5: Ps= P2 = 1 MPa (10 bar)
HW
5–7 Air enters a nozzle steadily at 2.21 kg/m until the density in the tank rises to 7.20 kg/m ... steadily at 1 MPa and 500°C with a mass flow rate.
sheet solution
equivalent (200 bar 20 MPa
ms
5-45 A number of brass balls are to be quenched in a water bath at a specified rate. kJ/kg. 3240.9. /kg m. 0.02975. C. 450. MPa. 10. 1. 3. 1. 1. 1.
SOLUTIONS HOMEWORKS F
212031
ME 24-221
THERMODYNAMICS I
Solutions to Assignment 11
December 1, 2000
J. Murthy
11.4A steam power plant operating in an ideal Rankine cycle has a high pressure of 5
MPa and a low pressure of 15 kPa. The turbine exhaust state should have a quality of at least 95% and the turbine power generated should be 7.5 MW. Find the necessary boiler exit temperature and the total mass flow rate.
C.V. Turbine w
T = h 3 - h 4 ; s 4 = s 3
4: 15 kPa, x
4 = 0.95 => s 4 = 7.6458 , h 4 = 2480.4 3: s 3 = s 4 , P3 ? h 3 = 4036.7, T 3 = 758°C w T = h 3 - h 4 = 4036.7 - 2480.4 = 1556.3 m . = W. T /w T = 7.5 × 1000/1556.3 = 4.82 kg/s
11.13A steam power plant has a steam generator exit at 4 MPa, 500°C and a condenser
exit temperature of 45°C. Assume all components are ideal and find the cycle efficiency and the specific work and heat transfer in the components.1 T 3 2 s 4
From the Rankine cycle we have the states:
1: 45°C x = 0/ , v
1 = 0.00101 , h 1 = 188.45
2: 4 MPa
3: 4 MPa, 500°C , h
3 = 3445.3 , s 3 = 7.0901 4: P sat (45°C) = 9.593
C.V. Turbine: s4
= s 3 ? x 4 = (7.0901 - 0.6386)/7.5261 = 0.8572, h 4 = 188.42 + 0.8572 × 2394.77 = 2241.3 w T = h 3 - h 4 = 3445.3 - 2241.3 = 1204 kJ/kg
C.V. Pump: -w
P = v 1 (P 2 - P 1 ) = 0.00101(4000 - 9.6) = 4.03 kJ/kg -w P = h 2 - h1 ? h 2 = 188.42 + 4.03 = 192.45 kJ/kg
C.V. Boiler: q
H = h 3 - h 2 = 3445.3 - 192.45 = 3252.8 kJ/kg
C.V. Condenser: q
L,out = h 4 - h 1 = 2241.3 - 188.42 = 2052.9 kJ/kg TH = w net /q H = (w T + w P )/q H = (1204 - 4.03)/3252.8 = 0.369
11.8Consider the ammonia Rankine-cycle power plant shown in Fig. P11.8, a plant
that was designed to operate in a location where the ocean water temperature is
25°C near the surface and 5°C at some greater depth.
a. Determine the turbine power output and the pump power input for the cycle. b. Determine the mass flow rate of water through each heat exchanger. c. What is the thermal efficiency of this power plant? a) Turbine s 2S = s 1 = 5.0863 = 0.8779 + x 2S
× 4.3269
x 2S = 0.9726 h 2S = 227.08 + 0.9726 × 1225.09 = 1418.6 w ST = h 1 - h 2S = 1460.29 - 1418.6 = 41.69 w T S w ST = 0.80 × 41.69 = 33.35 kJ/kg
ME 24-221
THERMODYNAMICS I
Solutions to Assignment 11
December 1, 2000
J. Murthy
11.4A steam power plant operating in an ideal Rankine cycle has a high pressure of 5
MPa and a low pressure of 15 kPa. The turbine exhaust state should have a quality of at least 95% and the turbine power generated should be 7.5 MW. Find the necessary boiler exit temperature and the total mass flow rate.
C.V. Turbine w
T = h 3 - h 4 ; s 4 = s 3
4: 15 kPa, x
4 = 0.95 => s 4 = 7.6458 , h 4 = 2480.4 3: s 3 = s 4 , P3 ? h 3 = 4036.7, T 3 = 758°C w T = h 3 - h 4 = 4036.7 - 2480.4 = 1556.3 m . = W. T /w T = 7.5 × 1000/1556.3 = 4.82 kg/s
11.13A steam power plant has a steam generator exit at 4 MPa, 500°C and a condenser
exit temperature of 45°C. Assume all components are ideal and find the cycle efficiency and the specific work and heat transfer in the components.1 T 3 2 s 4
From the Rankine cycle we have the states:
1: 45°C x = 0/ , v
1 = 0.00101 , h 1 = 188.45
2: 4 MPa
3: 4 MPa, 500°C , h
3 = 3445.3 , s 3 = 7.0901 4: P sat (45°C) = 9.593
C.V. Turbine: s4
= s 3 ? x 4 = (7.0901 - 0.6386)/7.5261 = 0.8572, h 4 = 188.42 + 0.8572 × 2394.77 = 2241.3 w T = h 3 - h 4 = 3445.3 - 2241.3 = 1204 kJ/kg
C.V. Pump: -w
P = v 1 (P 2 - P 1 ) = 0.00101(4000 - 9.6) = 4.03 kJ/kg -w P = h 2 - h1 ? h 2 = 188.42 + 4.03 = 192.45 kJ/kg
C.V. Boiler: q
H = h 3 - h 2 = 3445.3 - 192.45 = 3252.8 kJ/kg
C.V. Condenser: q
L,out = h 4 - h 1 = 2241.3 - 188.42 = 2052.9 kJ/kg TH = w net /q H = (w T + w P )/q H = (1204 - 4.03)/3252.8 = 0.369
11.8Consider the ammonia Rankine-cycle power plant shown in Fig. P11.8, a plant
that was designed to operate in a location where the ocean water temperature is
25°C near the surface and 5°C at some greater depth.
a. Determine the turbine power output and the pump power input for the cycle. b. Determine the mass flow rate of water through each heat exchanger. c. What is the thermal efficiency of this power plant? a) Turbine s 2S = s 1 = 5.0863 = 0.8779 + x 2S
× 4.3269
x 2S = 0.9726 h 2S = 227.08 + 0.9726 × 1225.09 = 1418.6 w ST = h 1 - h 2S = 1460.29 - 1418.6 = 41.69 w T S w ST = 0.80 × 41.69 = 33.35 kJ/kg