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

`T_(out) = 457.921\,K`

Step-by-step 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) = (\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
`(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 = (m)/(M)\cdot R_(u)\cdot T`

`\\u = (R_(u)\cdot T)/(P\cdot M)`

`\\u_(in) = ((8.314\,(kPa\cdot m^(3))/(kmol\cdot K) )\cdot (350\,K))/((90\,kPa)\cdot (28.97\,(kg)/(kmol) ))`

`\\u_(in) = 1.116\,(m^(3))/(kg)`

The mass flow is:

`\dot m = (\dot V)/(\\u_(in))`

`\dot m = (0.6\,(m^(3))/(s) )/(1.116\,(m^(3))/(kg) )`

`\dot m = 0.538\,(kg)/(s)`

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

`h_(in) = 350.49\,(kJ)/(kg)`

The specific enthalpy at outlet is:

`h_(out) = (75\,kW)/(0.538\,(kg)/(s) )-30\,(kJ)/(kg) + 350.49\,(kJ)/(kg)`

`h_(out) = 459.895\,(kJ)/(kg)`

The exit temperature of air is:

`T_(out) = 457.921\,K`