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;

(c) reduce the minimum temperature to 10^oC.

Short answer

a) e=0.463

b) e=0.453

c) e-0.490

Step-by-step solution

Part(a): Step 1 - given information 

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

Part(a): Step 2 : Explanation

Efficiency is written as

$e=1-\frac{H_4-H_1}{H_3-H_1}..............................\left(1\right)$

Where H_nis enthalpy at point n.

Entropy at point 4 is written as

$S_4=x{S}_{\text{water}}+(1-x)S_{\text{steam}}$

where x is fraction of water.

Rearrange and find x.

$x=\frac{S_4-S_{\text{steam}}}{S_{\text{water}}-S_{\text{steam}}}.................\left(2\right)$

Write the expression of the enthalpy at point 4

$H_4=x{H}_{\text{water}}+(1-x)H_{\text{steam}}......................................\left(3\right)$

Substitute values in equation (2)

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

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

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

$x=\frac{\left(5.791\mathrm{kJ}·{\mathrm{K}}^{-1}·{\mathrm{kg}}^{-1}\right)-\left(8.667\mathrm{kJ}·{\mathrm{K}}^{-1}·{\mathrm{kg}}^{-1}\right)}{\left(0.297\mathrm{kJ}·{\mathrm{K}}^{-1}·{\mathrm{kg}}^{-1}\right)-\left(8.667\mathrm{kJ}·{\mathrm{K}}^{-1}·{\mathrm{kg}}^{-1}\right)}$

$x=0.344$

Substitute in Equation 3, we get

x=0.344,

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

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

$$\begin{gathered} H_4=(0.344)\left(84\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)+(1-0.344)\left(2538\mathrm{kJ}·{\mathrm{kg}}^{-1}\right) \\ =1693.8\mathrm{kJ}·{\mathrm{kg}}^{-1} \end{gathered}$$

Now substitute these values in equation 1, we get

$$\begin{gathered} e=1-\frac{\left(1693.8\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)-\left(84\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)}{\left(3081\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)-\left(84\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)} \\ =0.463 \end{gathered}$$

Part(b) : Step 1 : Given information

Pressure reduced to 100 bar.

Part(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

$$\begin{gathered} x=\frac{\left(6.903\mathrm{kJ}·{\mathrm{K}}^{-1}·{\mathrm{kg}}^{-1}\right)-\left(8.667\mathrm{kJ}·{\mathrm{K}}^{-1}·{\mathrm{kg}}^{-1}\right)}{\left(0.297\mathrm{kJ}·{\mathrm{K}}^{-1}·{\mathrm{kg}}^{-1}\right)-\left(8.667{\mathrm{kJ}}^{-1}·{\mathrm{K}}^{-1}·{\mathrm{kg}}^{-1}\right)} \\ =0.211 \end{gathered}$$

Now substitute in equation (3)

x=0.211

H_Water = 84 kJ. kg^-1

H_Steam =2538 kJ. kg^-1

$$\begin{gathered} H_4=(0.211)\left(84\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)+(1-0.211)\left(2538\mathrm{kJ}·{\mathrm{kg}}^{-1}\right) \\ =2020.2\mathrm{kJ}·{\mathrm{kg}}^{-1} \end{gathered}$$

Substitute these in the efficiency formula $e=1-\frac{H_4-H_1}{H_3-H_1}$, we get

$$\begin{gathered} e=1-\frac{\left(2020.2\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)-\left(84\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)}{\left(3625\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)-\left(84\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)} \\ =0.453 \end{gathered}$$

Part(c): Step 1: Given Information

minimum temp is reduced to 10^oC

Part(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

$$\begin{gathered} x=\frac{\left(6.233\mathrm{kJ}·{\mathrm{K}}^{-1}·{\mathrm{kg}}^{-1}\right)-\left(8.901\mathrm{kJ}·{\mathrm{K}}^{-1}·{\mathrm{kg}}^{-1}\right)}{\left(0.151\mathrm{kJ}·{\mathrm{K}}^{-1}·{\mathrm{kg}}^{-1}\right)-\left(8.901\mathrm{kJ}·{\mathrm{K}}^{-1}·{\mathrm{kg}}^{-1}\right)} \\ =0.305 \end{gathered}$$

Substitute in equation (3)

x=0.305

H_Water = 42 kJ. kg^-1

H_Steam =2520 kJ. kg^-1

$$\begin{gathered} H_4=(0.305)\left(42\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)+(1-0.305)\left(2520\mathrm{kJ}·{\mathrm{kg}}^{-1}\right) \\ =1776.7\mathrm{kJ}·{\mathrm{kg}}^{-1} \end{gathered}$$

Now substitute these values in the efficience formula $e=1-\frac{H_4-H_1}{H_3-H_1}$

$$\begin{gathered} e=1-\frac{\left(1776.7\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)-\left(42\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)}{\left(34444\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)-\left(42\mathrm{kJ}·{\mathrm{kg}}^{-1}\right)} \\ =0.490 \end{gathered}$$