Basic Mechanical Engineering (3110006)

Question-2c

BE | Semester-1 Summer-2019 | 04-06-2023

Q2) (c)

Given Data:

At state-1 :

p_{1} = 7 bar

and

x_{1} = 0.9

At state-2 :

p_{2} = 3 bar

w_{t} = 20 kJ / kg

Calculate:

x_{2} = ?

Solution:

From Steam Table:

@ p_{1} = 7 bar \Rightarrow h_{f} = 697.1 \frac{kJ}{kg}, h_{fg} = 2064.9 \frac{kJ}{kg}

@ p_{2} = 3 bar \Rightarrow h_{f} = 561.5 \frac{kJ}{kg}, h_{fg} = 2163.2 \frac{kJ}{kg}

Now,

h_{1} = {(h_{f 1} + x_{1} h_{fg 1})}_{@ 7 bar}

h_{1} = 697.1 + 0.9 \times 2064.9

h_{1} = 2555.51 \frac{kJ}{kg}

After losing

125 kJ

from the steam at constant pressure, we get

h_{1} = 2555.51 - 125 = 2430.51 \frac{kJ}{kg}

Now,

w_{t} = (h_{1} - h_{2}) \frac{kJ}{kg}

20 = 2430.51 - h_{2}

h_{2} = 2410.51 \frac{kJ}{kg}

Now,

h_{2} = {(h_{f 2} + x_{2} h_{fg 2})}_{@ 3 bar}

2410.51 = 561.5 + x_{2} \times 2163.2

x_{2} = 0.85