Question 5.17: (a) Show that, when critical pressure ratio occurs, the velo...
(a) Show that, when critical pressure ratio occurs, the velocity of a compressible fluid at the exit of a convergent nozzle is given by
C_{2} = \sqrt{ \frac{2}{γ + 1} . a_{1}}where a_{1} is the sonic velocity corresponding to the intial conditions. Assume critical pressure
ratio = (\frac{2}{γ + 1}) ^{\frac{γ }{γ + 1}} , where γ is the adiabatic index.
(b) State the factors on which nozzle efficiency depends.
(c) Determine the throat and exit height of a Delaval nozzle to discharge 27 kg of a perfect gas per minute. The inlet and exit pressures are 480 kPa and 138 kPa respectively. Initial temperature of the gas is 535°C. Nozzle efficiency is 90% and frictional losses occur only after the throat. The molecular weight of the gas is 29 and its adiabatic index = 1.4. Assume square cross-section of the nozzle. (AMIE Summer, 1998)
(a) \frac{p_{2}}{p_{1}} = ( \frac{2}{γ + 1}) ^{\frac{γ }{γ – 1}} ….Given
The exit Mach. number is 1, ∴ T_{2} = T^{*}
\frac{ T^{*}}{T_{1}} = ( \frac{p^{*}}{p_{1}} )^{\frac{γ – 1}{γ }}= [ (\frac{2}{γ + 1}) ^{\frac{γ }{γ – 1}}]^{\frac{γ – 1}{γ } } = \frac{2}{γ + 1}
The exit velocity is sonic velocity,
= \sqrt{γ R T_{2}} = \sqrt{γ R T^{*}}
= \sqrt{γ R \frac{2 T_{1}}{γ + 1}} = \sqrt{\frac{2 }{γ + 1}} × \sqrt{γ R T_{1}}
= \sqrt{\frac{2 }{γ + 1}}a_{1} . Proved.
(b) The factors on which nozzle efficiency depends are :
1. Material of the nozzle.
2. Workmanship of the manufacture of nozzle.
3. Size and shape of the nozzle.
4. Reynolds number of flow.
5. Angle of divergence of divergent portion.
6. Nature of fluid flowing and its state.
7. Turbulence in fluid and its state.
(c) Given : m_{g} = 27 kg/min ; p_{1} = 480 kPa ; p_{3} = 138 kPa ; T_{1} = 535 + 273 = 808 K ; η_{nozzle} = 90%,
Molecular weight of the gas = 29 ; γ = 1.4.
\frac{p_{2}}{p_{1}} = ( \frac{2}{γ + 1}) ^{\frac{γ }{γ – 1}} = ( \frac{2}{1.4 + 1})^{ \frac{1.4}{1.4 – 1}} = 0.528
∴ p_{2} = 480 × 0.528 = 253.44 kPa
\frac{T_{2}}{T_{1}} = \frac{2}{γ + 1} = \frac{2}{1.4 + 1} = 0.8333
∴ T_{2} = 808 × 0.8333 = 673.3 K
R = \frac{8.314}{29} = 0.2867 kJ/kg°C
c_{p} = \frac{R_{γ }}{γ – 1} = \frac{0.2867 × 1.4}{(1.4 – 1)} = 1.0034 kJ/kg°C
Now, \frac{C_{2}²}{2} = c_{p} (T_{1} – T_{2})
or C_{2}² = 2 c_{p} (T_{1} – T_{2} ) = 2 × 1.0034 × 1000(808 – 673.3)
∴ C_{2} = 519.9 m/s
\dot{m}_{g} = ρ _{2} A _{2} C _{2} = \frac{p_{2} }{R T_{2} } A _{2} C _{2}∴ A _{2} = \frac{\dot{m}_{g} R T_{2}}{p_{2} C _{2} } = \frac{(27 / 60) (0.2867 × 1000) × 673.3}{(253.44 × 1000) × 519.9} = 6.59 × 10^{–4} m²
∴ Height of nozzle (square section) at throat
= \sqrt{ 6.59 × 10^{–4} } = 0.0257 m or 25.7 mm.
At the exit :
\frac{T_{3}}{T_{1}} =( \frac{p_{3}}{p_{1}}) ^{\frac{γ – 1}{γ }}∴ T_{3} = T_{1} × ( \frac{p_{3}}{p_{1}}) ^{\frac{γ – 1}{γ }} = 808 × ( \frac{138}{480})^{\frac{1.4 – 1}{1.4}} = 565.9 K
η_{nozzle} = \frac{T_{1} – T_{3}′ }{T_{1} – T_{3}}or 0.9 = \frac{808 – T_{3}′ }{808 – 565.9}
∴ T_{3}′ = 808 – 0.9 (808 – 565.9) = 590.1 K
Now, \frac{C_{3}′²}{2} = c_{p} (T_{1} – T_{3}′)
or C_{3}′² = 2 c_{p} (T_{1} – T_{3}′ ) = 2 × 1.0034 × 1000 (808 – 590.1)
or C_{3}′ = 661.27 m/s
∴ A_{3} = \frac{\dot{m} R T_{3}′}{ p_{3} C_{3}′ } = \frac{(27 / 60) × (0.2867 × 1000) × 590.1}{ (138 × 1000) × 661.27} = 8.343 × 10^{-4} m²
∴ Height of nozzle (square section) at the exit
= \sqrt{8.343 × 10^{-4}} = 0.02888 m or 28.88 mm.
