A converging$-$diverging nozzle is to be designed where ideal gas air enters at a velocity, temperature and pressure of $50\frac{m}{s},15C,$ and $1083$kPa, respectively. To achieve a certain thrust, it was determined that $2\mathrm{.}8k\frac{g}{s}$ of air must exit the nozzle at a speed of $490\frac{m}{s}.$ Typical nozzles of this design have an isentropic efficiency of $69\%.$ For the following analysis, assume the cold$-$air$-$standard assumption, where $k=1\mathrm{.}4$ and $c_p=1\mathrm{.}005k\frac{J}{k}g*K.$ The enthalpy at the reference temperature of $25C$ can be taken to be $298\mathrm{.}33k\frac{J}{k}g.$ We define the isentropic efficiency of a nozzle as the kinetic energy exiting the real nozzle compared to the kinetic energy of an isentropic nozzle having the same exit pressure. The requested answers for your analysis should agree with the units indicated $($e$.$g$.$ for Part $($a$),$ since the unit is $k\frac{J}{k}g,$ it should be understood that the requested information is specific enthalpy, even though the word "specific" is not used.$).$

a$.$ The static and stagnation enthalpies of the incoming air are $k\frac{J}{k}g$ and

$k\frac{J}{k}g,$ respectively.

b$.$ The stagnation temperature of the incoming air is

C$.$

c$.$ The stagnation pressure of the incoming air is kPa.

d$.$ The kinetic energy of the exiting air is $k\frac{J}{k}g.$

If the nozzle was isentropic$,$ e$.$ The air's exit state would be $$C and kPa f$.$ The mach number of the air exiting the nozzle would be$.$ g$.$ The nozzles throat area would be $c{m}^2.$ For the actual nozzle having the efficiency shown, h$.$ The kinetic energy of the corresponding isentropic nozzle $($for the same exit pressure is$)$ kJ$/$kg$.$ i$.$ The air's exit state would be $$C and kPa. j$.$ The mach number of the air exiting the nozzle would be$.$