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Studentka2010 [4]
3 years ago
13

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.
Engineering
1 answer:
natita [175]3 years ago
5 0

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

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