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sashaice [31]
4 years ago
7

A converging-diverging nozzle with an exit to throat area ratio of 4.0 is designed to expand air isentropically to atmospheric p

ressure. Determine the exit Mach number at design conditions and the required inlet stagnation pressure. Also, calculate the back pressures corresponding to the various flow regimes for the nozzle.

Engineering
1 answer:
34kurt4 years ago
4 0

Answer

0.9, 1172.35kPa

Explanation:

<em>Question (in proper order)  Attached below</em>

Air is flowing inside the throat has following inlet conditions

P_{0}=1000 kPa

T_{0}=500 K

M=1.8

M=\frac{u}{c}=1.8

'u' is the speed of sound in the air

\Rightarrow u=1.8\times c

=1.8\times 340.29

 =612.522\frac{m}{sec}

Therefore volumetric flow rate entering,

Q=612.522\times 0.0008

=0.4900176\frac{m^{3}}{sec}

Using ideal gas equation

PV=nRT

n=\frac{PV}{RT}

=\frac{1000\times 0.4900176}{8.314\times 500}

=0.117878 gmoles/sec

Therefore , mass flow rate

Mass = 0.117878\times 29

=3.4184 grams/sec

Given

\frac{A}{A_{0}}=2

\Rightarrow A=0.0016.m^{2}

Using continuity equation

A_{1}V_{1}=A_{2}V_{2}

\Rightarrow V_{2}=\frac{A_{1}V_{1}}{A_{2}}

=\frac{0.0008\times 612.522}{0.0016}

=306.261\frac{m}{sec}

Hence exit velocity = 306.261 m/sec

Exit Mach number

M=\frac{u}{c}=\frac{306.261}{340.29}=0.9

Temperature will remain same as 500 K

Now

Using Bernoulli's equation

\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\rho g}+\frac{v_{2}^{2}}{2g}+z_{2}

Here

z_{1} = z_{2}

\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}-\frac{v_{2}^{2}}{2g}=\frac{P_{2}}{\rho g}

\Rightarrow \frac{1000000}{\rho g}+\frac{612.522^{2}}{2g}-\frac{306.261^{2}}{2g}=\frac{P_{2}}{\rho g}

\Rightarrow \frac{1000000}{1.225}+\frac{612.522^{2}}{2}-\frac{306.261^{2}}{2}=\frac{P_{2}}{1.225}

\Rightarrow P_{2}=1172.35kPa

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