Basic Mechanical Engineering (3110006)

Question-3c

BE | Semester-1 Summer-2020 | 11-06-2020

Q3) (c)

Given data:

m = 1.5 kg

p_{1} = 10 bar

T_{\sup, 1} = 300 ° C

p_{2} = 2.8 bar

x_{2} = 0.9

Calculate:

du = ?

Solution:

By using steam table for superheated steam (at state 1): at

p_{1} = 10 bar

and

T_{\sup, 1} = 300 ° C

h_{\sup, 1} = 2943 \frac{kJ}{kg}

v_{\sup, 1} = 0.2328 \frac{m^{3}}{kg}

By using steam table (for dry saturated steam): at

p_{2} = 2.8 bar

h_{f, 2} = 551.4 \frac{kJ}{kg}

h_{g, 2} = 2721.5 \frac{kJ}{kg}

h_{f g, 2} = 2170.1 \frac{kJ}{kg}

v_{g, 2} = 0.646 \frac{m^{3}}{kg}

The internal energy per kg of superheated steam,

u_{\sup, 1} = h_{\sup, 1} - p_{1} \times v_{\sup, 1}

u_{\sup, 1} = 2943 - 1000 \times 0.2328

u_{\sup, 1} = 2710.2 \frac{kJ}{kg}

But, the change of internal energy

= u_{2} - u_{\sup, 1} \frac{kJ}{kg}

Finding unknown

u_{2}

Internal energy of wet steam after expansion,

= u_{2} = h_{2} - p \cdot v_{2}

Finding unknown,

h_{2}

Specific enthalpy of wet steam,

h_{2} = h_{f, 2} + x_{2} \cdot h_{fg, 2}

h_{2} = 551.4 + 0.9 \cdot 2170.1

h_{2} = 2309.181 \frac{kJ}{kg}

Finding unknown,

v_{2}

Specific enthalpy of wet steam,

v_{2} = x_{2} \cdot v_{g, 2}

v_{2} = 0.9 \times 0.646

v_{2} = 0.5814 \frac{m^{3}}{kg}

Therefore,

u_{2} = h_{2} - p \cdot v_{2} = 2309.181 - 1000 \times 0.5814 = 1721.781 \frac{kJ}{kg}

Hence, change of internal energy,

u_{2} - u_{\sup, 1} = 1721.181 - 2710.2 = - 988.419 \frac{kJ}{kg}

But, change of internal energy for 1.5 kg steam is,

= 1.5 \times (- 988.419) = - 1482.628 \frac{kJ}{kg}

Negative sign indicates decrease in internal energy.