Question

A steady flow, steady state thermodynamic system receives 45 kg/min of a fluid at 207kPa

and discharges it from point 24 m. above the entrance section at 1034 kPa. The fluid

enters with a velocity of 36.6 m/s and leaves with a velocity of 12 m/s. During this process,

there are supplied 7 kW of heat from an external source, and the increase in enthalpy is

4.6 kJ/kg. Determine the work in kW and in horsepower.

Answer 1

Solution;

Given;

$\dot{m}=45kg/min=0.75kg/s$

$P_1=207kPa$

$z_2=24m$

$P_2=1034kPa$

$v_1=36.6m/s$

$v_2=12m/s$

$Q=7kW$

$\Delta h=4.6kJ/kg$

From steady flow energy equation;

$W=Q+\dot{m}[(h_1-h_2)+\frac{v_1^2-v_2^2}{2000}+\frac{g(z_1-z_2}{1000}]$

$W=7kW+0.75[-4.6+\frac{36.6^2-12^2}{2000}+\frac{9.81(0-24)}{1000}]$

$W=7kW-3.178kW=3.822kW$

Convert to hp;

5.125hp