A projectile travels in air of pressure 10.1043N/cm210.1043N/cm2 at 10∘C10∘C at a speed of 1500 km per hour. Find the Mach number and Mach angle. Take K=1.4,R=287J/kg−KK=1.4,R=287J/kg−K

 A projectile travels in air of pressure 10.1043N/cm2$10.1043N/c{m}^2$ at 10∘C${10}^{\circ }C$ at a speed of 1500 km per hour. Find the Mach number and Mach angle. Take 

K=1.4,R=287J/kg−K$K=1.4,R=287J/kg-K$

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Solution: 

Given:

Pressure,  P=10.1043N/cm2=10.1043×104N/m2$P=10.1043N/c{m}^2=10.1043\times {10}^{4}N/m^2$

R=287.14J/kg−K$R=287.14J/kg-K$

K=1.4

t=10∘C$t={10}^{\circ }C$

∴T=10+273=283K$\therefore T=10+273=283K$

Spped, V=1500km/hr$V=1500km/hr$

V=1500×100060×60=416.67m/s$V={\displaystyle \frac{1500\times 1000}{60\times 60}}=416.67m/s$

For adiabatic process, the velocity of sound is given by

c=KRT−−−−−√$c=\sqrt{KRT}$

c=1.4×287.14×283−−−−−−−−−−−−−−−√$c=\sqrt{1.4\times 287.14\times 283}$

c=337.20m/s

Mach Number,

M=Vc$M={\displaystyle \frac{V}{c}}$

M=416.67337.20$M={\displaystyle \frac{416.67}{337.20}}$

M=1.235$M=1.235$

Using the relation,

sina=cV=1M$\sin a={\displaystyle \frac{c}{V}}={\displaystyle \frac{1}{M}}$

∴sina=11.235$\therefore \sin a={\displaystyle \frac{1}{1.235}}$

∴sina=0.8097$\therefore \sin a=0.8097$

a=sin−1(0.8097)$a={\sin }^{-1}(0.8097)$

∴a=54.06∘$\therefore a={54.06}^{\circ }$ is the mach angle