TITLE.PM5

FIRST LAW OF THERMODYNAMICS 155

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/M-therm/Th4-4.pm5

(ii)Steady flow constant volume process :

W = – 1 Vdp

2

z = – V(p 2 – p 1 ) = V(p 1 – p 2 )

i.e., W = V(p 1 – p 2 ) ...(4.52)

(iii)Steady flow constant temperature process :

The constant temperature process is represented by

pV = p 1 V 1 = p 2 V 2 = C (constant)

∴ W = – Vdp

1

2

z

= –

C

1 pdp

2

z QV

C

=p

L

NM

O

QP

= – C

dp

p

Cpe

1

2

1

2

z =− log

= – C logeelog

p

p C

p

p

2

1

1

2

=

i.e., W = p 1 V 1 loge

p

p

1

2

F

HG

I

KJ ...(4.53)

Now substituting the values of W in the equation (4.49), considering unit mass flow :

(a) The energy equation for constant pressure flow process

dQ = ∆ PE + ∆ KE + ∆ h

= ∆ h (if ∆PE = 0 and ∆ KE = 0).

(b) The energy equation for constant volume flow process

dQ = – vdp

1

2

z + ∆^ PE + ∆^ KE + ∆^ u + pdv + vdp

= ∆ PE + ∆ KE + ∆ u Qpdv==v dp vdp

L

N

M

O

Q

(^0) z 1 P
2
and.
∴ dQ = ∆ u (if ∆ PE = 0 and ∆ KE = 0)

4.12. Engineering Applications of Steady Flow Energy Equation (S.F.E.E.)

4.12.1. Water turbine

Refer to Fig. 4.32. In a water turbine, water is supplied from a height. The potential energy
of water is converted into kinetic energy when it enters into the turbine and part of it is converted
into useful work which is used to generate electricity.
Considering centre of turbine shaft as datum, the energy equation can be written as follows :

upvZg 1111 C^1

2

+++ 2

F

HG

I

KJ + Q =

upvZg 2222 C^2

2

+++ 2

F

HG

I

KJ + W

In this case,

Q = 0

∆ u = u 2 – u 1 = 0

∴ v 1 = v 2 = v

Z 2 = 0