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Page-1SOLUTIONLets evaluate areas at L//2 and L//3 Given linear variation.Length is not given.Assume l=1m. Please refer to this Solution as a methodology 10 solve such problem.By geometry{:[(0.045)/(1)=(x)/( 0.5)],[x=0.045 xx0.5=0.0225],[gamma1//2=0.075+0.0225],[gamma1//2=0.0975m^(2)],[AY//2=pigamma^(2)],[AY//2=0.03m^(2)]:}Similarly(0.045)/(1)=(x)/(1//3)Page - 2{:[x=0.015],[gamma//1//3=x+0.075],[gamma1//3=0.09m],[" A "1//3=0.0254m^(2)],[" A exit "=pi xx0.12^(2)],[=0.045m^(2)]:}consider the region 1known parameters areP_(0)=220kpa,T_(0)=300k. To get the properties at 1, The throat area is known to us. Since shock is expected un diverging section, Throat of the nozzle will be chocked. i-e. Mach =1 at throat. So, area of the Throat itself is the A^(**){:[A" th "=A^(**)=0.0177m^(2)],[A_(1)=0.03m^(2)" (calcalated in the above "]:}Page-3The area Ratio for the flow in this region is A_(1)//A^(**)=1.695The Supersonic solution for Mach and Pressure ratio at this area Ratio are{:[" Mach "M_(1)=2],[P_(1)//P_(0)=01267quad(:.P_(0)=220kPa:}],[{:P_(1)=27.9kpa)]:}we now known the mach and pressure values ahead of the shock. using normal Shock relations, we lan arrive at Mach and pressure values behind the shock.Normal shock parameters at H=2 are" mach "{:[M_(2),=0.577,<=>P_(1),=27.9kpa],[P_(2)//P_(1),=4.5,P_(2),=125.55kpa],[P_(O_(2))//P_(01),=0.72,<=>P_(01),=220kpa],[,P_(0_(2)),=158.4kPa]:}The flow conditions at region2 follow mach 0.577 properties SO for A^(**), is A_(2)//A_(1)=1.2165. A_(2) is the area at shock which is 0.03m^(2)page-4A_(2)=0.0247m^(2). The critical area is Consistent throughout the flow till esit consider region 3, where flow is change in flow properties are Isentropic, Which obeys the properties.{:[A_(2)^(**)=A_(3)^(**)],[P_(O_(2))=P_(03)]:}The area Ratio at exit in the area at exit divided by A_(3)^(**).A_(3)//A_(3)**=(A_("esit "))/(A_(2)^(**))=(0.045)/(0.0247)=1.822Since it is known t ... See the full answer