The air in the reservoir has an absolute pressure of 500 kPakPa and a temperature of 30∘C∘C. The throat has a diameter of 50 mmmm. (Figure 1) Figure1 of 1 Part ADetermine the greatest possible mass flow through the nozzle. The air in the reservoir has an absolute pressure of500 kPa and a temperature of 30° C. The throat has adiameter of 50 mm. (Figure 1)Figure50 mm1 of 1Part ADetermine the greatest possible mass flow through the nozzle.Express your answer using three significant figures.IVE| ΑΣΦ | 11 | vecm =SubmitPrevious Answers Request AnswerX Incorrect; Try Again; 4 attempts remainingProvide Feedbackkg/s

Introduction The nozzle allows very powerful air to exit the engine. It is well-known for its function in generating thrust. The force created by the hot airflow energy and the cool airflow surrounding the engine core causes propulsion. In this problem, we will calculate the mass flow rate from the nozzle. Given data: absolute pressure of airP=500 kPatemperature of airT=30°Cdiameter of throatd=50 mm Find out: The greatest possible mass flow rate through the nozzle (m˙)Calculation of mass flow rate through the nozzle (m˙) For the calculation of the mass flow rate through the nozzle the following things we have to consider, The temperature of the air at the throat of the nozzle (T) To=T1+γ-12M2To=absolute air temperatureTo=30+273 KTo=303 Kγ=1.4 (adiabatic index of air)M=Mach numberM=1 (at the throat section of nozzle)To=T1+γ-12M2303 K=T1+1.4-1212T=252.5 K The pressure of air at the throat of the nozzle (P); Po=P×1+γ-12×M21.41.4-1Po=absolute pressure of airPo=500 kPa 500 kPa=P×1+1.4-12×121.41.4-1P=264.141 kPa The density of air (ρ); ρ=PRTρ=264.141×103 Pa287 J/kg K×252.5 K(ρ)=3.645 kg/m3The velocity of air (V); V=γ×R×TR=287 J/kg KV=1.4×287J/kg K×252.5KV=318.519 m/s The cross-sectional area of the nozzle (A)c; Ac=π4×d2d=50 mmd=0.05 mAc=π4×0.052Ac=0.00196 m2 The mass flow rate through the nozzle (m˙); (m˙)=ρ×A×V(m˙)=mass flow rateρ=3.645 kg/m3A=0.00196 m2V=318.519 m/s(m˙)=ρ×A×V(m˙)=3.645 kg/m3×0.00196 m2×318.519 m/s(m˙)=2.276 kg/s Hence the mass flow rate through the nozzle is 2.276 kg/s.Result: For the given problem the greatest mass flow rate through the nozzle (m˙)=2.276 kg/s

Answered: The air in the reservoir has an…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Part A

Determine the greatest possible mass flow through the nozzle.

The air in the reservoir has an absolute pressure of
500 kPa and a temperature of 30° C. The throat has a
diameter of 50 mm. (Figure 1)
Figure
50 mm
1 of 1
Part A
Determine the greatest possible mass flow through the nozzle.
Express your answer using three significant figures.
IVE| ΑΣΦ | 11 | vec
m =
Submit
Previous Answers Request Answer
X Incorrect; Try Again; 4 attempts remaining
Provide Feedback
kg/s

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