0.5 kg of water in piston cylinder executes a Camot power cycle as shown in Figu

Question

0.5 kg of water in piston cylinder executes a Camot power cycle as shown in Figure 1. The water initially at exist as saturated mixture at 1500 kPa and quality of 0.25. This saturated mixture was heated until reached as saturated vapor at point 2 during isothermal expansion step. This saturated vapor was then expanded adiabatically to a pressure of 100 kPa while doing 2019 kJ of work. To complete the cycle, the system undergoes isothermal compression from point 3 to point 4 and adiabatic compression from point 4 to point 1. By treating each step as a closed system, determine g of water in piston eylinder executes a a) The amount of heat, Qir and work, W12 involve in step 1-2 b) The quality of saturated mixture at point 3. c) The amount of heat, Qs involve in step 3-4 and the quality of saturated mixture at (10 Marks) (7 Marks) point 4, (8 Marks)

Explanation / Answer

Given data

At 1500 kPa

hf = 844.47 kJ/Kg , hfg = 1947.28 kJ/Kg

h1 = 844.47 + 0.25 (1947.28)

= 1331.29 kJ/Kg

h2 = 2792.15 kJ/Kg

At 1500 kPa

S2 = 6.4448 kJ/Kgk

Given the workdone in process 2-3 = 201.9 kJ

W = m (h2 - h3)

201.9 = 0.5 (2792.15 - h3)

h3 = 2388.35 kJ/Kg

At 100 kPa

2388.35 = 417.44 + x (2258.02)

x = 0.8728 = 87.28%

Since the process 4-1 is adiabatic compression

  S4 = S1 = 3.34745 kJ/Kgk

3.34745 = 1.3025 + x (6.0568)

x4 = 0.3376 = 33.76%

h4 = 417.44 + 0.3376 (2258.02)

= 1179.74 kJ/Kg

(a) Process 1-2 : Isothermal expansion

U = 0

Q = W = 0.5 (2792.15 - 1331.29)

= 730.43 kJ

(b) Quality of steam at point 3 = 0.8728 = 87.28%

(c) Process 3-4: Isothermal compression

U = 0

Q3-4 = W3-4 = m (h3 - h4)

= 0.5 (2388.35 - 1179.74)

= 604.305 kJ

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