Question 2.6: Helium gas at 250 kPa and 100 °C flows through a rectangular...

Helium gas at 250 kPa and 100^{\circ} C flows through a rectangular duct of cross-section 40 cm × 50 cm, with a velocity of 500 m s ^{-1}. Determine the Mach number, stagnation temperature, stagnation pressure, and the mass flow rate. Assume helium to be an ideal gas.

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Given, p=250 kPa , T=100^{\circ} C =373.15 K , \text { Area }=0.40 \times 0.50=0.2 m ^{2}, V=500 m s ^{-1} The gas constant for helium (of molecular weight 4.003) is

R=\frac{8314}{4.003}=2077 J ( kg K )^{-1}

 

The speed of sound is

a=\sqrt{\gamma R T}

 

=\sqrt{1.66 \times 2077 \times 373.15}=1134.3 m s ^{-1}

 

since \gamma=1.66 for helium gas. Thus,

M=\frac{V}{a}=\frac{500}{1134.3}=0.44

 

By isentropic relation, we have the temperature ratio as

\frac{T_{0}}{T}=1+\frac{\gamma-1}{2} M^{2}

 

Therefore,

T_{0}=T\left(1+\frac{\gamma-1}{2} M^{2}\right)

 

=373.15\left(1+0.33 \times 0.44^{2}\right)=397 K

 

By isentropic relation, the pressure ratio is given by

\frac{p_{0}}{p}=\left(1+\frac{\gamma-1}{2} M^{2}\right)^{\frac{\gamma}{\gamma-1}}

 

Thus,

p_{0}=250\left(1+0.33 \times 0.44^{2}\right)^{2.515}

 

= 292.13 kPa

 

The mass flow rate through the duct is

\dot{m}=\rho A V=\frac{p}{R T} A V

 

=\frac{250 \times 10^{3}}{2077 \times 373.15} \times 0.2 \times 500=32.26 kg s ^{-1}

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