Basic Mechanical Engineering (3110006)

Question-2c-OR

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

Q2) (c)

Given Data:

m = 3 kg

p = 10 bar = 1200 kPa

C_{ps} = 2.1 \frac{kJ}{kg K}

Case 1: Wet steam with

x = 0.91

Case 2: Superheated steam at,

T_{\sup} = 200 ° C

Calculate:

h_{wet} and u_{wet}, when x = 0.91

h_{\sup} and u_{\sup}, when T_{\sup} = 200 ° C = 473 K

Solution:

From Steam Table at

10 bar

pressure

t_{sat} = 179.9 ° C = 452.9 K

h_{f} = 762.6 \frac{kJ}{kg}

h_{fg} = 2013.6 \frac{kJ}{kg}

h_{g} = 2776.2 \frac{kJ}{kg}

v_{f} = 0.001127 \frac{m^{3}}{kg}

v_{g} = 0.18548 \frac{m^{3}}{kg}

Case 1:

h_{wet} and u_{wet}, when x = 0.91

Enthalpy of wet steam is,

h_{wet} = h_{f} + x h_{fg} = 762.6 + 0.91 \times 2013.6 = 2594.976 \frac{kJ}{kg}

For 3 kg of steam,

h_{wet} = 2594.976 \frac{kJ}{kg} \times 3 kg = 7784.928 kJ

Case 2: $h_{\sup} and u_{\sup}, when T_{\sup} = 200 ° C = 473 K$

Enthaply of superheated steam is,

h_{\sup} = h_{g} + C_{ps} (T_{\sup} - T_{sat}) = 2776.2 + 2.1 \times (200 - 179.9) = 2818.41 \frac{kJ}{kg}

For 3 kg of steam,

h_{\sup} = 2818.41 \frac{kJ}{kg} \times 3 kg = 8455.23 kJ

Specific volume of superheated steam is,

v_{\sup} = v_{g} \frac{T_{\sup}}{T_{sat}} = 0.18548 \times \frac{553}{452.9} \frac{= 0.2264 m^{3}}{kg}

Internal energy of superheated steam is,

u_{\sup} = h_{\sup} - p v_{\sup} = 2818.41 - 1000 \times 0.2264 = 2592.01 \frac{kJ}{kg}

For 3 kg of steam,

u_{\sup} = 2592.01 \frac{kJ}{kg} \times 3 kg = 7776.03 kJ