Question

A pump is used to circulate hot water in a home heating system. Water enters the well-insulated pump operating at steady state at a rate of 0.42 gal/min. The inlet pressure and temperature are 14.7 lbf/in.², and 180°F, respectively; at the exit the pressure is 90 lbf/in.² The pump requires 1/35 hp of power input. Water can be modeled as an incompressible substance with constant density of 60.58 lb/ft³ and constant specific heat of 1 Btu/lb · °R.


Neglecting kinetic and potential energy effects, determine the temperature change, in °R, as the water flows through the pump.

Answer 1

Given flow rate = 0.42 gal/min

1 gal = 3.79 kg

.42 gal/min = (0.42*3.79)/60 =0.027kg/s

T₁ = 180^oF = 355.37K

T₂ = ?

Pump Power = (1/35)HP = 0.03kW

(We know 1HP = 0.75kW)

Specific heat = 1BTU/lb-^oR = 4.1868 kJ/kg-K

Applying steady flow energy equation:

Neglecting kinetic and potential energy terms we get:

Here there is no heat loss Q = 0

The pump requires work input W= -0.03kW

1 kelvin(K) = 1.8 ^oR

0.265K = 0.477 ^oR

The temperature change = 0.477 ^oR