Question

In a steady flow apparatus, 135 kJ of work is done by each kg of fluid. The specific volume of the fluid, pressure and speed at the inlet are 0.37 m3 /kg, 600 kPa, and 16 m/s. The inlet is 32 m above the floor, and the discharge pipe is at floor level. The discharge conditions are 0.62 m3 /kg, 100 kPa, and 270 m/s. The total heat loss between the inlet and discharge is 9 kJ/kg of fluid. In flowing through this apparatus, does the specific internal energy increase or decrease, and by how much?

Answer 1

Solution;

Given;

Inlet conditions;

$v_1=0.37m^3/kg$

$p_1=600kPa$

$V_1=16m/s$

$z_1=32m$

Discharge conditions are;

$v_2=0.62m^3/kg$

$p_2=100kPa$

$V_2=270m/s$ ni

$z_2=0$

Heat loss;

$Q=-9kJ/kg$

Work done;

$W=135kJ$

From steady flow energy equation;

$u_1$+$p_1v_1$+$\frac{V_1^2}{2}$+$z_1g$ +$\frac{dQ}{dm}$ =$u_2$ +$p_2v_2$ +$\frac{V_2^2}{2}$ +$z_2g$ +$\frac{dW}{dm}$

Therefore;

$u_1$ -$u_2$ =($p_2v_2$ -$p_1v_1$)+$\frac{V_2^2-V_1^2}{2}$ + $(z_2-z_1)g$ +$\frac{dW}{dm}-\frac{dQ}{dm}$

By substitution;

$u_1-u_2$ =[($100×0.62$) -($600×0.37$ )]+$\frac{(270^2-16^2)×10^{-3}}{2}+(-32×9.81×10^{-3})+135-(-9)$

Hence;

$u_1-u_2=-160+36.322-0.31392+135+9$

$u_1-u_2=20.01kJ/kg$

Answer;

Specific internal energy decreases by $20.01kJ/kg$