TITLE.PM5

154 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-4.pm5

i.e., vdp

1

2

z =

C^2

2 in the case of a nozzle.

If ∆ PE = 0 and ∆ KE = 0, as in case of a compressor, – vdp

1

2

z = W

or W = vdp

1

2

z in the case of a compressor.

The integral pdv

1

2

z and 1 vdp

2

z are shown in Fig. 4.31 (a) and (b).

(a) Work done in non-flow process. (b) Work done in flow process.

Fig. 4.31. Representation of work on p-v diagram.

The work done during non-flow process is given by

1 pdv

2

z = Q – ∆u ...[4.50 (c)]

For isothermal process, we have

∆ u = 0 and ∆h = 0.

Substituting these values in (equations) 4.50 and [4.50 (c)]

• vdp
  1

2

z = Q and 1 pdv

2

z = Q

∴ pdv

1

2

z = – 1 vdp

2
z
The above equation indicates that the area under both curves is same for an isothermal
process.
Note. In all the above equations ‘v’ represents volume per unit mass as mass flow is considered unity.
Now let us find out expressions for work done for different flow processes as follows :
(i)Steady flow constant pressure process :

W = – 1 vdp.

2

z = 0 [Q dp = 0] ...(4.51)