Question

A gas mixture at 560.0°C and 129 kPa absolute enters a heat exchanger at a rate of 70.0 m/hr. The qas leaves the heat exchanger at 205.0°C. The change in enthalpy of the gas during the cooling process is -4.50 kJ/mol. What is the heat required in kW? Assume the gas behaves ideally and that the changes in kinetic and potential energy are negligible. Number kW

Answer 1

Solution-

Given from the question we have

P1 = 129 kPa

T1 = 560 C

T2 = 205.0 C

V1 = 70 m^3/hr

Δh = -4.5 kJ/mol

Using equation

PV = nRT
So we have the following equation

=>P1*V1 = n*R*T1

129*10^3*70/3600 = n*8.314*(560+273)

n = 0.362 mol/s

Now the heat lost by the gas mixture can be found as below

Q(lost) = n*delta(h)

Qlost = 0.362*-4.50 = 1.629 kW

Now the Heat required = the heat lost

Heat required = 1.629 kW