Question

A gas mixture at 490.0°C and 127 kPa absolute enters a heat exchanger at a rate of 60.0 m3/hr. The gas leaves the heat exchanger at 170.0°C. The change in enthalpy of the gas during the cooling process is -4.70 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.

Answer 1

Explanation:

According to the ideal gas equation, PV = nRT.

or,            n = $\frac{PV}{RT}$

As it is given that pressure is 127 kPa or 127000 Pa (as 1 kPa = 1000 Pa), volume is 60.0 $m^{3}/hr$, R is gas constant equals 8.314 J/K/mol, and temperature is (490 + 273) K = 763 K.

Hence, putting these values into the above equation as follows.

                    n = $\frac{PV}{RT}$

                        = $\frac{127000 Pa \times \frac{60sec}{3600}}{8.314 J/K/mol \times 763 K}$

                        = 0.320 mol/s

Therefore, heat required will be calculated as follows.

  change in enthalpy of the gas during the cooling process × mole flow

                      = 4.70 kJ/mol × 0.320 mol/s

                     = 1.50 kW

Thus, we can conclude that heat required in kW for the given situation will be 1.50 kW.