Answer:
[tex]T_{out} = 457.921\,K[/tex]
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:
[tex]-\dot Q_{out} + \dot W_{un} + \dot m \cdot (h_{in}-h_{out})=0[/tex]
[tex]h_{out} = \frac{\dot W_{in}}{\dot m}- q_{out} +h_{in}[/tex]
An ideal gas observes the following mathematical model:
[tex]P\cdot V = n\cdot R_{u}\cdot T[/tex]
Where:
[tex]P[/tex] - Absolute pressure, in kilopascals.
[tex]V[/tex] - Volume, in cubic meters.
[tex]n[/tex] - Quantity of moles, in kilomole.
[tex]R_{u}[/tex] - Ideal gas universal constant, in [tex]\frac{kPa\cdot m^{3}}{kmole\cdot K}[/tex].
[tex]T[/tex] - Absolute temperature, in kelvin.
The previous equation is re-arranged in order to calculate specific volume at inlet:
[tex]P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T[/tex]
[tex]\nu = \frac{R_{u}\cdot T}{P\cdot M}[/tex]
[tex]\nu_{in} = \frac{(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (350\,K)}{(90\,kPa)\cdot (28.97\,\frac{kg}{kmol} )}[/tex]
[tex]\nu_{in} = 1.116\,\frac{m^{3}}{kg}[/tex]
The mass flow is:
[tex]\dot m = \frac{\dot V}{\nu_{in}}[/tex]
[tex]\dot m = \frac{0.6\,\frac{m^{3}}{s} }{1.116\,\frac{m^{3}}{kg} }[/tex]
[tex]\dot m = 0.538\,\frac{kg}{s}[/tex]
The specific enthalpy in ideal gases depends on temperature exclusively. Then:
[tex]h_{in} = 350.49\,\frac{kJ}{kg}[/tex]
The specific enthalpy at outlet is:
[tex]h_{out} = \frac{75\,kW}{0.538\,\frac{kg}{s} }-30\,\frac{kJ}{kg} + 350.49\,\frac{kJ}{kg}[/tex]
[tex]h_{out} = 459.895\,\frac{kJ}{kg}[/tex]
The exit temperature of air is:
[tex]T_{out} = 457.921\,K[/tex]
