Question

Air enters a compressor operating at steady state with pressure of 90 kPa, at a temperature of 350 K, and a volumetric flow rate of 0.6 m3/s. The air exits the compressor at a pressure of 700 kPa. Heat transfer from the compressor to its surrounding occurs at a rate of 30 kJ/kg of air flowing. The compressor power input is 75 kW. Neglecting kinetic and potential energy effects and assuming air to be an ideal gas, find the exit temperature of air.

Answer 1

Answer:

$T_{out} = 457.921\,K$

Explanation:

Before determining the exit temperature of air, it is required to find the specific enthalpy at outlet by using the First Law of Thermodynamics:

$-\dot Q_{out} + \dot W_{un} + \dot m \cdot (h_{in}-h_{out})=0$

$h_{out} = \frac{\dot W_{in}}{\dot m}- q_{out} +h_{in}$

An ideal gas observes the following mathematical model:

$P\cdot V = n\cdot R_{u}\cdot T$

Where:

$P$ - Absolute pressure, in kilopascals.

$V$ - Volume, in cubic meters.

$n$ - Quantity of moles, in kilomole.

$R_{u}$ - Ideal gas universal constant, in $\frac{kPa\cdot m^{3}}{kmole\cdot K}$.

$T$ - Absolute temperature, in kelvin.

The previous equation is re-arranged in order to calculate specific volume at inlet:

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

$\nu = \frac{R_{u}\cdot T}{P\cdot M}$

$\nu_{in} = \frac{(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (350\,K)}{(90\,kPa)\cdot (28.97\,\frac{kg}{kmol} )}$

$\nu_{in} = 1.116\,\frac{m^{3}}{kg}$

The mass flow is:

$\dot m = \frac{\dot V}{\nu_{in}}$

$\dot m = \frac{0.6\,\frac{m^{3}}{s} }{1.116\,\frac{m^{3}}{kg} }$

$\dot m = 0.538\,\frac{kg}{s}$

The specific enthalpy in ideal gases depends on temperature exclusively. Then:

$h_{in} = 350.49\,\frac{kJ}{kg}$

The specific enthalpy at outlet is:

$h_{out} = \frac{75\,kW}{0.538\,\frac{kg}{s} }-30\,\frac{kJ}{kg} + 350.49\,\frac{kJ}{kg}$

$h_{out} = 459.895\,\frac{kJ}{kg}$

The exit temperature of air is:

$T_{out} = 457.921\,K$