TITLE.PM5

152 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-4.pm5

C = Velocity of fluid ,

Z = Height above datum,

p = Pressure of the fluid,

u = Internal energy per kg of fluid, and

pv = Energy required for 1 kg of fluid.

This equation is applicable to any medium in any steady flow. It is applicable not only to

rotary machines such as centrifugal fans, pumps and compressors but also to reciprocating machines

such as steam engines.

In a steady flow the rate of mass flow of fluid at any section is the same as at any other

section. Consider any section of cross-sectional area A, where the fluid velocity is C, the rate of

volume flow past the section is CA. Also, since mass flow is volume flow divided by specific volume,

∴ Mass flow rate, m& CA

v

= ...(4.46)

(where v = Specific volume at the section)
This equation is known as the continuity of mass equation.
With reference to Fig. 4.30.
∴ m&
CA
v

CA

v

==^11

1

22

2

...[4.46 (a)]

4.11. Energy Relations for Flow Process

The energy equation (m kg of fluid) for a steady flow system is given as follows :

m u

C Zg pv

1 1

2

+++ 2 111

F

HG

I

KJ

+ Q = m u

C Zg pv

2 2

2

+++ 2 222

F

HG

I

KJ

+ W

i.e., Q = m ()(uu ZgZg) ( )
CC pv pv
21 2 1^2

2

1

2

−+ − + − 22 22 11

F

HG

I

KJ

+−

L

N

M

M

O

Q

P

P

+ W

i.e., Q = m ()( )uu gZZ ( )

CC pv pv

21 2 1^2

2

1

2

−+ − + 2 22 11

F −

HG

I

KJ

+−

L

N

M

M

O

Q

P
P + W
= ∆U + ∆PE + ∆KE + ∆ (pv) + W
where ∆U = m (u 2 – u 1 )
∆PE = mg (Z 2 – Z 1 )

∆KE = m

CC 22 12

2

F −

HG

I

KJ

∆pv = m(p 2 v 2 – p 1 v 1 )

∴ Q – ∆U = [∆PE + ∆KE + ∆(pV) + W] ...(4.47)

For non-flow process,

Q = ∆U + W = ∆U + pdV

1

2

z

i.e., Q – ∆U = 1 pdV.

2

z ...(4.48)