Discussion
Consider steady, one-dimensional compressible flow of a gas in a pipe of diameter 1 m. At one location in the pipe, the density and velocity are 1 kg/m3 and 100 m/s, respectively. At a downstream location in the pipe, the velocity is 170 m/s. If the pressure drop between these two locations is 10 kPa, the force exerted by the gas on the pipe between these two locations is ____________ N.
Answer: B
The gas exerts a force of \(750\pi\) N on the pipe.
Apply the momentum equation to the gas between the two sections. The pipe area is \(A = \pi/4\) m2, and the mass flow is \(\dot{m} = \rho V A = 1 \times 100 \times \pi/4 = 25\pi\) kg/s, the same at both sections. With \(F\) the force of the pipe on the gas in the flow direction,
\[ \begin{aligned} (p_{1} - p_{2})A + F &= \dot{m}(V_{2} - V_{1}) \\ 10^{4} \times \frac{\pi}{4} + F &= 25\pi \times 70 \\ F &= 1750\pi - 2500\pi = -750\pi\ \text{N} \end{aligned} \]
The pipe pushes the gas backwards with \(750\pi\) N, so the gas pushes the pipe forwards with \(750\pi\) N.
The other values do not satisfy this balance; leaving out the momentum change would give \(2500\pi\), and leaving out the pressure force \(1750\pi\), neither of which is offered.
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