Heat transfer is a path function. The amount of heat transfer not only depends on end states but also depends on the path followed by it. Thus we can write aQ for heat transfer.

It is an inexact differential. So

Whenever there is temperature difference, there will be heat flow. The temperature difference is the cause and transfer of heat is the effect.

The heat transfer is written mathematically by

where T is the intensive property and X is the extensive property of the substance. It is valid for a quasi-static process only. The amount of heat transfer can be calculated from the area under (T, X) coordinates diagram

dQ = T dX

dX = dQ / T

here 1 / T is called integrating factor used with inexact quantity to make it exact differentials.

**SOLVED EXAMPLES**

**Example 3.1.** A centrifugal pump forces 2 m3/min of water horizontally from an open well to a closed tank, where the pressure is 0.45 MPa. Compute the work the pump must do upon the water in an hour just to force the water into the tank against the pressure.

**Solution. **P = 0.45 x 10^{6} Pa = 450 kPa

dV = 2m^{3} , time , t = 1 min

= V_{2} – V_{1}

In one minute work done = ʃ = pd V = P ( V_{2}-V_{1})

= 450 (2-0) = 900 kNm.

Thus in one hour the work done= 900 x 60 KJ / hr.

= 54000 KJ/hr.

**Example 3.2.** A petrol engine cylinder has a piston of cross section 0.15 m^{2 }It contains gas at a pressure of 2MPa. The gas expands according to a straight line process on P – V diagram. The final pressure is 0.2 Mpa. Calculate the work done by the gas on the piston if the stroke is 0.15 m.

**Solution .**

W_{I-2} = area under the curve (Straight line) 1-2

= 1/2 [ 2 + 0.2] x 10^{6} x 0.15 x 0.15 Nm

= 24.75 KJ.Ans.

**Example 3.3.** A mass of 2 kg of air is compressed in a quasi-static process from 0.15 MPa to

0.75 MPa for which PV = Canst. The initial density of air is 1.5 kg/m^{3 }Calculate the work done by the piston to compress the air.

** **

**Solution.** We have W_{1 2} = P_{1} V_{1} In P_{1} / P_{2}

= 0.15

= – 0.15 x 106 x 2 / 1.5 In 0.15 / 0.75 J

= – 0.321888 x 10^{6} J

= – 321.8 KJ

The negative sign of work shows that the work is done on the system.

**Example 3.4.** A quantity of gas is compressed in a quasi-equilibrium process from 90 KPa,

0.15 m3 to 0.5 MPa, 0.04 m^{3 }Assuming that the pressure and volume are related by PV” =Canst.,

calculate the workdone on the gas system. (Take n = 1.3)

**solution. **W_{1} = P_{1} V_{1} – P_{2} –P_{2} / n-1 = 90 x 103 x 0.15 – 0.5 x 106 x 0.04 J / 1,3 – 1

= – 21.7 Kj.

Negative sign shows that the work is done on the system.

**Example 3.5**. Calculate the total work done by a gas system in a expansion process shown in

Fig. 3.12 (1 bar = Uf N/m^{2})

**Solution.** W_{1-2-3} = W_{1-2} + W_{2-3}

Now W_{1-2} = 50 x 105 x (0.4 -0.2 ) J = 10 x 10^{5 }J [ Note.P_{1}V_{2} = P_{3} V_{3}]

W_{2-3} = P_{2} V_{2}-P_{3} V3 / n-1 = P_{2} V_{2} / n-1 [ 1 – (P_{3} / P_{2}) n-1 / n

=P2 V_{2} / n-1 [ 1 – ( V_{2} / V_{3}) n-1]

= 50 x105 x 0.4 / 13.1 [ 1- (0.4 / 0.8)0.3]

= 12.516 x 10^{5} J

= 2.2156 x 10^{6} J = 2.2516 MJ. Ans.