39A (Waves)

In the frame in which the shock is stationary, the upstream gas approaches it with speed

V

, while the downstream gas recedes from it with speed

V - u_{1}

. The Rankine-Hugoniot conditions for a perfect gas are therefore

ρ_{0} V = ρ_{1} (V - u_{1}),

(1)

p_{0} + ρ_{0} V^{2} = p_{1} + ρ_{1} (V - u_{1})^{2},

(2)

and

\frac{γ p _{0}}{( γ - 1 ) ρ _{0}} + \frac{1}{2} V^{2} = \frac{γ p _{1}}{( γ - 1 ) ρ _{1}} + \frac{1}{2} (V - u_{1})^{2} .

(3)

These equations express conservation of mass flux, momentum flux, and total specific enthalpy respectively.

b

 0  0

Solution

 0  0

When

u_{1} = V /3

, the downstream speed in the shock frame is

2 V /3

. Mass conservation gives

ρ_{0} V = ρ_{1} \frac{2 V}{3},

(1)

\frac{ρ _{1}}{ρ _{0}} = \frac{3}{2} .

(2)

c

 0  0

Solution

 0  0

Momentum conservation and the density ratio from part (b) give

p_{1} - p_{0} = ρ_{0} V^{2} - \frac{3}{2} ρ_{0} (\frac{2 V}{3})^{2} = \frac{1}{3} ρ_{0} V^{2} .

(1)

For

γ = 3

, energy conservation becomes

\frac{3 p _{0}}{2 ρ _{0}} + \frac{1}{2} V^{2} = \frac{p _{1}}{ρ _{0}} + \frac{2}{9} V^{2} .

(2)

Eliminating

ρ_{0} V^{2}

between these equations yields

p_{1} = 4 p_{0}

. Hence

\frac{p _{1}}{p _{0}} = 4 .

(3)

d

 0  0

Solution

 0  0

The momentum equation from part (c) is

p_{1} - p_{0} = \frac{1}{3} ρ_{0} V^{2} .

(1)

Since

p_{1} - p_{0} = 3 p_{0}

, it follows that

ρ_{0} V^{2} = 9 p_{0}, V = 3 \frac{p _{0}}{ρ _{0}} .

(2)

This completes the piston-driven shock with specific-heat ratio three calculation.

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 39A (Waves)

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