Question

6. Air is cooled in a process with
constant pressure of 150 kPa.
Before the process begins, air has
a specific volume of 0.062 m^3/kg
The final specific volume is 0.027
m^3/kg. Find the specific work in
the process.​

Answer 1

Given :-

Pressure = P = 150 kPa

Initial Specific Volume = 0.062 m³/Kg

Final Specific Volume = 0.027 m³/Kg

As given Pressure is constant.

W = ∫P•dv

W = $P({V}_{f} -{V}_{i})$

Now put the all given Values.

W = 150 × 10³(0.027 - 0.062)

W = -5.25 × 10³ Pa m³/Kg

W = -5250 J/Kg

Hence,

Specific Work = W = -5250 J/Kg.

Answer 2

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

$W=vi∫vfPdv=P(vf−vi)=150kPa∗(0.0027−0.062)m3/kg=−5.25kPa∗m3/kg.$

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.