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Steam Nozzle Flow Calculations

1) The steam nozzle is supplied with steam at 15 bar and 350°C and discharges at 1 bar. The divergent portion is 80 mm long with a throat diameter of 6 mm. 2) The velocity of steam at the throat is 562.12 m/s and the temperature is 270°C. 3) The cone angle of the divergent portion is calculated to be 4° 22'.

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68 views2 pages

Steam Nozzle Flow Calculations

1) The steam nozzle is supplied with steam at 15 bar and 350°C and discharges at 1 bar. The divergent portion is 80 mm long with a throat diameter of 6 mm. 2) The velocity of steam at the throat is 562.12 m/s and the temperature is 270°C. 3) The cone angle of the divergent portion is calculated to be 4° 22'.

Uploaded by

Tharun Malar
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Example 18.14.

A steam nozzle is
15 bar 350°C and discharges steam at 1 suppliedster2am at
bar. If the
portion of the nozzle is 80 mm long and
wergi ng the
throatdio
divergentdiameter
is 6 mm, determine the cone
/kg K) angle of the ner
Assume 12% of the total available
friction in the divergent enthalpy drop isortion
lost in
portion.Also determine the
and temperature of the steam at throat. velocity
Solution. p = l5 bar, 350°C, pg = 1 bar,
k =
1-0.12 =
0.88
When steam supplied to the nozzle is
the pressure at throat,
superheated.
ond P2 0.546 p, =
0.546 x 15 =
8.19 bar
Ah (kJ/kg)
h,= 3150--.
/ 8.19
bar
8.19
b a r

350 C

/270 C
1 b a r

h 2992 -
1 b a

--

Saturation
h= 2580D line
g/s 3

s (kJkgK
Fig. 18.17
From Mollier chart
h3150 kJ/kg, h = 2992 kJ/kg
m"kg, h, 2580 kJ/kg
= 0.24 =

v'= 1.75 m°/kg, t2 270°C =

Temperature of the steam at throat

270°C. (Ans.)
Velocity of steam at throat,

Cy= 44.72 ha =44.72 h-ha)

= 44.72 (3150-2992)
= 562.12 m/s. (Ans.)
From the conditions at nozzle throat, mass flow rate,

m
AC2
U2
T/4 (6/ 1000) x 562.12
= 0.0662 kg/s
0.24
At exit:
C = 44.72 kha' = 44.72 0.88x (h -hg)
= 44.72 0.88x (3150 2580) 1001.5 m/s

Exit area of the


nozzle,
A C m xug 0.0662x 1.75 = 0.0001156 m2
1001.5

D=0.0001156
2
4x0.0001156
D,= = 0.012 m or 12.1 mm

If 6 be the cone angle of nozzle,


(12.1-6)
tan 6 = 2R
x 0 = 0.03812 or 0 = 2° 11
2x 80
Thus, cone angle = 2 x 2° 11' =4° 22. (Ans.)
Throat

.-
---------
Cone angle
Fig 18.17

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