Question

Air is compressed steadily from 150 kPa and 300 K to 600 kPa with a mass flow rate of 5 kg/s. Each unit of mass passing from the inlet to exit undergoes a process described by PVLS constant Heat transfer occurs at a rate of 50.35 kJ per kg of air flowing to cooling water. The cooling water is circulating in a water jacket enclosing the compressor. Assume the potential and kinetic energy changes of the air from the inlet to exit are negligible. Calculate the exit temperature of the air in K. [4 marks Derive an energy balance equation for the compressor and wont determine its power input in kW.

Answer 1

Step-by-step explanation:

As it is given that air is compressed steadily from 150 kPa and 300 K to 600 KPa with a mass flow rate of 5 kg/s.

`P_(1)` = 150 kPa,
`P_(1)` = 600 kPa

`T_(1)` = 300 K,
`T_(2)` = ?

Mass flow rate (m) = 5 kg/s.

`PV^(1.5)` = constant

Q = 50.35 kJ/kg, n = 1.5

(a) Calculate the exit temperature of air as follows.

`TP^{(1 - n)/(n)}` = constant

`T_(1)P^{(1 - n)/(n)}_(1)` =
`T_(2)P^{(1 - n)/(n)}_(2)`

`T_(2)` =
`T_(1) ((P_(1))/(P_(2)))^{(1 - 1.5)/(1.5)}`

=
`300 ((150)/(600))^{(- 0.5)/(1.5)}`

= 476.22 K

(b) An energy balance equation for the compressor and determine its power input in kW as follows.

Q - w = m (\Delta H + K.E + P.E)

As K.E and P.E will be negligible here. So, Q and w will be in kW.

Also,
`\Delta H = \Delta U + \Delta (PV)`

`\Delta H` =
`\Delta U + V \Delta P + P \Delta V`

As, Q - w =
`m \Delta H`

and
`\Delta H = H_(2) - H_(1)`

So, air enthalpy at 150 kPa and 300 K,
`h_(1)` = ?

Air enthalpy = 1.007 \times T(q) - 0.026

`h_(1)` =
`1.007 * (300 - 273) - 0.026`

= 27.163 kJ/kg

Also,
`h_(2)` =
`1.007 * (476.22 - 273) - 0.026`

= 204.62 kJ/kg

Q = mQ

=
`50 * 50.35`

= 251.75 kJ/s

Q - w =
`m * (204.62 - 27.163)`

- w = 887.285 - 251.75

= -635.535 kJ/s

or, = 635.535 kW

Whereas negative sign indicates that work is done on the system.