Question

Assume 5 lb/sec of fluid enter a steady state, steady flow system with p1 = 100 psia, 1 =

0.2 lb/ft3

, v1 = 100 fps, u1 = 800 Btu/lb and leave with p2 = 20 psia, 2 = 0.05 lb/ft3

, v2 = 500

fps, u2 = 780 Btu/lb. During the passage through the open system, 10 Btu/lb of heat is

rejected. Find the work in hp.

Answer 1

Solution;

From the steady flow equation;

$u_1+\frac{p_1}{\rho_1}+\frac{v_1^2}{2000}+Q=u_2+\frac{p_1}{\rho_1}+\frac{v_2^2}{2000}+W$

Given;

$u_1=800Btu/lb=1860kJ/kg$

$P_1=100psia=689.5kPa$

$\rho_1=0.2lb/ft^3=3.2kg/m^3$

$V_1=100fps=30.48m/s$

$\dot{m}=5lb/s=2.268kg/s$

$u_2=780btu/lb=1814.18kJ/kg$

$p_2=20psia=137.895kPa$

$\rho_2=0.05lb/ft^3=0.8kg/m^3$

$V_2=500fps=152.4m/s$

$Q=10Btu/kg=23.26kJ/kg$

Substitute the values into the given equation;

$1860.7+\frac{689.7}{3.2×9.81}+\frac{30.48^2}{2000×9.81}-23.26=W+1814+\frac{137.895}{0.8×9.81}+\frac{152.4^2}{2000×9.81}$

$W=\dot{m}(55.23kJ/kg)$

$W=2.268×55.23=125.27kW$

Convert to horsepower;

168hp