Question

 A steam turbine operates under steady flow conditions receiving steam at following state:-Pressure =10 bar, specific internal energy = 2800kJ/kg, specific volume = 0.2 m3/kg Velocity =100m/s. The exhaust of steam from the turbine is at 0.2 bar, with specific internal energy = 2200 kJ/kg, specific volume =15 cu.m/kg and velocity = 300 m/s. The intake is 3m above the exhaust. The turbine develops 25 kW and heat loss over the surface of turbine is 30 kJ/kg. Determine the steam flow rate (kg/s) through the turbine.

Answer 1

Given as,

Inlet condition of Turbine:

Pressure(P₁) =10 bar=10 x 10⁵ Pa,

Specific internal energy(u₁) = 2800x10³J/kg,

Specific volume (v₁) = 0.2 m3/kg

Velocity(C₁) =100m/s.

Height of Inlet=Z₁

Exhaust condition of turbine:

Pressure (P₂)= 0.2 bar=0.2 x 10⁵ Pa,

Specific internal energy(u²) = 2200 x 10³J/kg,

Specific volume (v₂)=15 m³/kg

velocity (C₂)= 300 m/s.

Height of exhaust(Z₂) =Z₁-3

Power Developed by Turbine

W=25x10³ watt

Heat loss over the surface

q=30 x10³ J/kg

To Find:

Steam flow rate (kg/s) through the turbine.

Solution:

By steady flow energy Equation

$\text{Let mass flow rate be}=\dot{m}\\ \dot{m}[h_1+\frac{C^2_1}{2}+gZ_1]-\dot{m}\times q=\dot{m}[h_2+\frac{C^2_2}{2}+gZ_2]+W$

$\dot{m}[(u_1+p_1\times v_1)+\frac{C^2_1}{2}+gZ_1]-\dot{m}\times q=\dot{m}[u_2+p_2\times v_2+\frac{C^2_2}{2}+gZ_2]+W$

$\dot{m}[(u_1-u_2) +(p_1\times v_1-p_2\times v_2)+(\frac{C^2_1}{2}-\frac{C^2_2}{2})+g(Z_1-Z_2)-q]=W$

Substituting the corresponding values,

$\dot{m}\times (430029.4)=25000\\\dot{m}=\frac{25000}{430029.4}\\\dot{m}=0.0581 kg/s$

Answer: Steam flow rate (kg/s) through the turbine=0.0581 kg/s