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Fluid Power Assignment-1

The document outlines the calculations for determining the throat and exit areas of a convergent-divergent nozzle through which steam expands isentropically from 20 bar and 300°C to 3 bar. It includes the necessary equations and data for calculating the velocities and areas, considering both ideal and real conditions with discharge and velocity coefficients. The final results indicate throat and exit areas of approximately 0.85 cm² and 1.85 cm², respectively, under ideal conditions, and 0.91 cm² and 1.97 cm² with real conditions.

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7 views4 pages

Fluid Power Assignment-1

The document outlines the calculations for determining the throat and exit areas of a convergent-divergent nozzle through which steam expands isentropically from 20 bar and 300°C to 3 bar. It includes the necessary equations and data for calculating the velocities and areas, considering both ideal and real conditions with discharge and velocity coefficients. The final results indicate throat and exit areas of approximately 0.85 cm² and 1.85 cm², respectively, under ideal conditions, and 0.91 cm² and 1.97 cm² with real conditions.

Uploaded by

shittum681
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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EXERCISE 1.

2
Steam at 20 bar and 3000C enters a convergent-divergent nozzle at the rate of 0.3kg/s with negligible
inlet velocity and expands into a space at 3 bar.
(a) Assuming that the steam expands isentropically according to a law 3 .1PV = const, estimate the
throat and exit areas of the nozzle.
(b) Recalculate the throat and exit areas of the nozzle taking a coefficient of discharge of 0.98 and a
coefficient of velocity of 0.92.

Given data
P0=20 ¯¿
0
t 0=300 c
P1=3 ¯¿

C d=0.98

∅ =0.92
For isentropic work
1.3
P V =Canstant
Unknown/find
¿
a . A =?
A1=¿?
Solution

The nozzle inlet is at superheated condition.

h 0=3023.5 KJ / Kg From table of superheated steam at p0=20 ¯¿, t 0=300 ℃

s0 =6.766 KJ / Kg
2
v 0=0.125 m / Kg
¿
P
But =0.546 for superheated condition.
P0
¿
P =10.92 ¯¿
From adiabatic work.
1.3 ¿ ¿ 1.3 1.3
P0 v 0 =P v =P1 v 1 ………….. ( 1 )

From ( 1 )

v ¿=
√
1.3 P 0 V 1.3
P ¿
0

√
1.3
¿ 1.3 20× ( 0.125 )
v=
10.92
¿ 3
v =0.1991 m / Kg
Similarly,

√
¿
1.3 P0 v 0
v 1=
P1

√
1.3
1.3 20 × ( 0.125 )
v 1=
3
3
v 1=0.54 m / Kg

1
V =44.72 ( h 0−h )
¿ ¿ 2

Along isentropic path


¿
S0 =S s=6.766 KJ / Kg
¿ ¿
S s=Sf + x s S fg
¿
¿S −S f
x= f
s
S sg
¿
At P =10.92 KJ / Kg
¿
S f =2.1786
S fg=4.3711

¿ 6.766−2.1786
x s= =1
4.3711
¿ ¿ ¿
Hence h f =hf + x s h fg at p =11 ¯
¿
¿
h s=751.1+(1× 1998.5)=2779.6 KJ / Kg
1
¿ 2
V =44.72 ( 3023.5−2779.6 ) =698.41 m/s
Also
1

{
V =44.72 ( h0 −h ) + ( h −h 1) }
¿ ¿
1
¿ 2

¿
S −S f
At exit P1=3 ¯¿
¿
x= f
s
S sg

¿ 6.766−1.6716
x s= ¿ 0.96
5.3193
Therefore,
h s=h f + x 1 s hfg

h s=561.4+ 0.96 ( 2163.2 )=2638.08 KJ / Kg

V 1=44.72¿ ¿

V 1=877.9 m/ s

¿
¿ m v 0.3 ×0.1991 2
a) A = ¿ = =0.85 c m
v 698.4

m v 1 0.3× 0.54 2
A 1= = =1.85 c m
v1 877.9

b) C d=0.98

ma
=0.98
ms

ma=0.98× 0.3=0.294

V1
∅=
V 1S
V 1=∅ V 1 S=0.92× 877.9=0.294 m/s
¿
V
∅= ¿
V 1s
¿ ¿
V =∅ V s=0.92 ×698.4=642.5 m/ s
Therefore

¿
¿ mv
A= ¿
v
¿ 0.294 × 0.1991
A=
642.5
¿ 2
A =0.91 c m

ma v 1
A 1=
v1
0.294 × 0.1991
A 1=
642.5
A1=¿1.97c m2

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