Question 2.5: An aluminum pitot probe is placed in a supersonic air stream...

An aluminum pitot probe is placed in a supersonic air stream of temperature 30^{\circ} C. If the melting temperature of aluminum is about 660^{\circ} C, determine the flow velocity at which the probe will begin to melt. Assume air to be an ideal gas.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

Given that aluminum will melt at 660^{\circ} C, the probe will start melting at a stream stagnation temperature T_{0} \text { of } 660^{\circ} C.

T_{0}=660^{\circ} C =660+273.15=933.15  K

 

The static temperature of the stream is T=30^{\circ} C =303.15 K. From isentropic relation, we have

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

 

where M is the stream Mach number. For the present problem, we have

\frac{933.15}{303.15}=1+0.2 M^{2}

 

since \gamma=1.4. Thus, we get

M = 3.22

 

 

The speed of sound in the air stream is

a=\sqrt{\gamma R T}

 

=\sqrt{1.4 \times 287 \times 303.15}=349 m s ^{-1}

 

Thus, the velocity at which the probe will just begin to melt is

V=M a=3.22 \times 349

 

=1123.78 m s ^{-1}

 

Alternatively

This problem can also be solved using the energy equation, as follows.

h_{0}=h+\frac{V_{2}}{2}

 

where h_{0} and h are the stagnation and static enthalpies of the air stream. For a perfect gas,

h_{0}=c_{p} T_{0}, \quad h=c_{p} T

 

where c_{p}=1004.5 J ( kg K )^{-1}, for air. Therefore,

T_{0}-T=\frac{V^{2}}{2 c_{p}}

 

V=\sqrt{2 c_{p}\left(T_{0}-T\right)}

 

=\sqrt{2 \times 1004.5(933.15-303.15)}

 

1125.02 m s ^{-1}

 

This is the stream velocity at which the probe will start melting.

Note: The difference between these two answers is due to the fact that the value of Mach number M was terminated at the second decimal place. For this problem, the accurate answer is

V=1125.02 m s ^{-1}.

Related Answered Questions