4.24

Question

Calculate the efficiency of a Rankine cycle that is modified from the parameters used in the text in each of the following three ways (one at a time), and comment briefly on the results:

(a) reduce the maximum temperature to 500^oC;

(b) reduce the maximum pressure to 100 bars;

Step-by-Step Solution

Verified

Answer

a) e=0.463

b) e=0.453

c) e-0.490

1Part(a): Step 1 - given information

A Rankine cycle operates between temperatures 20^oC and 500^oC at maximum pressure 300 bars.

2Part(a): Step 2 : Explanation

Efficiency is written as

$e = 1 - \frac{H_{4} - H_{1}}{H_{3} - H_{1}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

Where H_nis enthalpy at point n.

Entropy at point 4 is written as

$S_{4} = x S_{water} + (1 - x) S_{steam}$

where x is fraction of water.

Rearrange and find x.

$x = \frac{S_{4} - S_{steam}}{S_{water} - S_{steam}} . . . . . . . . . . . . . . . . . (2)$

Write the expression of the enthalpy at point 4

$H_{4} = x H_{water} + (1 - x) H_{steam} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)$

Substitute values in equation (2)

S₄= $5.791 kJ \cdot K^{- 1} \cdot {kg}^{- 1}$

S_Water= $0.297 kJ \cdot K^{- 1} \cdot {kg}^{- 1}$

S_Steam= $8.667 kJ \cdot K^{- 1} \cdot {kg}^{- 1}$

$x = \frac{(5.791 kJ \cdot K^{- 1} \cdot {kg}^{- 1}) - (8.667 kJ \cdot K^{- 1} \cdot {kg}^{- 1})}{(0.297 kJ \cdot K^{- 1} \cdot {kg}^{- 1}) - (8.667 kJ \cdot K^{- 1} \cdot {kg}^{- 1})}$

$x = 0.344$

Substitute in Equation 3, we get

x=0.344,

H_Water = $84 kJ \cdot {kg}^{- 1}$

H_Steam = $2538 kJ \cdot {kg}^{- 1}$

$H_{4} = (0.344) (84 kJ \cdot {kg}^{- 1}) + (1 - 0.344) (2538 kJ \cdot {kg}^{- 1}) = 1693.8 kJ \cdot {kg}^{- 1}$

Now substitute these values in equation 1, we get

$e = 1 - \frac{(1693.8 kJ \cdot {kg}^{- 1}) - (84 kJ \cdot {kg}^{- 1})}{(3081 kJ \cdot {kg}^{- 1}) - (84 kJ \cdot {kg}^{- 1})} = 0.463$

3Part(b) : Step 1 : Given information

Pressure reduced to 100 bar.

4Part(b) : Step 2 : Explanation

Substitute Values in equation 2

S₄ = 6.903 kJ.K^-1 kg^-1

S_Water = 0.297kJ.K^-1 kg^-1

S_Steam = 8.667 kJ.K^-1 kg^-1

$x = \frac{(6.903 kJ \cdot K^{- 1} \cdot {kg}^{- 1}) - (8.667 kJ \cdot K^{- 1} \cdot {kg}^{- 1})}{(0.297 kJ \cdot K^{- 1} \cdot {kg}^{- 1}) - (8.667 {kJ}^{- 1} \cdot K^{- 1} \cdot {kg}^{- 1})} = 0.211$

Now substitute in equation (3)

x=0.211

H_Water = 84 kJ. kg^-1

H_Steam =2538 kJ. kg^-1

$H_{4} = (0.211) (84 kJ \cdot {kg}^{- 1}) + (1 - 0.211) (2538 kJ \cdot {kg}^{- 1}) = 2020.2 kJ \cdot {kg}^{- 1}$

Substitute these in the efficiency formula $e = 1 - \frac{H_{4} - H_{1}}{H_{3} - H_{1}}$ , we get

$e = 1 - \frac{(2020.2 kJ \cdot {kg}^{- 1}) - (84 kJ \cdot {kg}^{- 1})}{(3625 kJ \cdot {kg}^{- 1}) - (84 kJ \cdot {kg}^{- 1})} = 0.453$

5Part(c): Step 1: Given Information

minimum temp is reduced to 10^oC

6Part(c): Step 2 : Explanation

Substitute values in equation (2)

S₄ = 6.233 kJ.K^-1 kg^-1

S_Water = 0.151 kJ.K^-1 kg^-1

S_Steam = 8.901 kJ.K^-1 kg^-1

$x = \frac{(6.233 kJ \cdot K^{- 1} \cdot {kg}^{- 1}) - (8.901 kJ \cdot K^{- 1} \cdot {kg}^{- 1})}{(0.151 kJ \cdot K^{- 1} \cdot {kg}^{- 1}) - (8.901 kJ \cdot K^{- 1} \cdot {kg}^{- 1})} = 0.305$