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

User TokyoToo
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1 Answer

6 votes

Answer

0.9, 1172.35kPa

Step-by-step explanation:

Question (in proper order) Attached below

Air is flowing inside the throat has following inlet conditions


P_(0)=1000 kPa


T_(0)=500 K


M=1.8


M=(u)/(c)=1.8

'u' is the speed of sound in the air


\Rightarrow u=1.8* c


=1.8* 340.29


=612.522(m)/(sec)

Therefore volumetric flow rate entering,


Q=612.522* 0.0008


=0.4900176(m^(3))/(sec)

Using ideal gas equation

PV=nRT


n=(PV)/(RT)


=(1000* 0.4900176)/(8.314* 500)

=0.117878 gmoles/sec

Therefore , mass flow rate


Mass = 0.117878* 29


=3.4184 grams/sec

Given


(A)/(A_(0))=2


\Rightarrow A=0.0016.m^(2)

Using continuity equation


A_(1)V_(1)=A_(2)V_(2)


\Rightarrow V_(2)=(A_(1)V_(1))/(A_(2))


=(0.0008* 612.522)/(0.0016)


=306.261(m)/(sec)

Hence exit velocity = 306.261 m/sec

Exit Mach number


M=(u)/(c)=(306.261)/(340.29)=0.9

Temperature will remain same as 500 K

Now

Using Bernoulli's equation


(P_(1))/(\rho g)+(v_(1)^(2))/(2g)+z_(1)=(P_(2))/(\rho g)+(v_(2)^(2))/(2g)+z_(2)

Here


z_(1) = z_(2)


(P_(1))/(\rho g)+(v_(1)^(2))/(2g)-(v_(2)^(2))/(2g)=(P_(2))/(\rho g)


\Rightarrow (1000000)/(\rho g)+(612.522^(2))/(2g)-(306.261^(2))/(2g)=(P_(2))/(\rho g)


\Rightarrow (1000000)/(1.225)+(612.522^(2))/(2)-(306.261^(2))/(2)=(P_(2))/(1.225)


\Rightarrow P_(2)=1172.35kPa

A converging-diverging nozzle with an exit to throat area ratio of 4.0 is designed-example-1
User Root
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