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Chapter 11: Problem 44

Helium at \(68^{\circ} \mathrm{F}\) and 14.7 psia in a large tank flows steadily and isentropically through a converging nozzle to a receiver pipe. The cross- sectional area of the throat of the converging passage is \(0.05 \mathrm{ft}^{2}\). Determine the mass flowrate through the duct if the receiver pressure is (a) 10 psia, (b) 5 psia. Sketch temperatureentropy diagrams for situations (a) and (b).

Short Answer

Expert verified
For case (a) where the receiver pressure is 10 psia, the mass flow rate is approximately 0.31 slug/sec. For case (b) where the receiver pressure is 5 psia, the mass flow rate is approximately 0.45 slug/sec.

Step by step solution

01

Identify Known Quantities

The initial pressure (\(P_1\)) is 14.7 psia, initial temperature (\(T_1\)) is \(68^{\circ}F = 520^{\circ}R\) (Since \(R = F + 460\)). The cross-sectional area (\(A\)) of the throat is \(0.05ft^2\) , the specific gas constant (\(R\)) for Helium is 1,545 ft-lbf/slug-R, and the ratio of specific heats (\(\gamma\)) of helium is 5/3. We are asked to find the mass flow rate given two different exit pressures.
02

Calculating Critical Pressure

Calculate the critical pressure where the flow is choked or maximum using the formula \(P^* = P_1*(2/(\gamma+1))^((\gamma)/(\gamma-1))\). Substituting \(P_1 = 14.7\) psia and \(\gamma = 5/3\) gives \(P^* ≈ 8.6\) psia.
03

Case (a) Receiver Pressure 10 psia

Since the receiver pressure (\(P_2\)) of 10 psia is greater than critical pressure (\(P^*\)), the flow is subcritical or subsonic. We can use the isentropic relation formula to find the exit Mach number \(M_2 = sqrt(((P_1/P_2)^((\gamma-1)/\gamma) -1)*2/(\gamma-1))\). Substituting \(P_1 = 14.7\) psia, \(P_2 = 10\) psia, and \(\gamma = 5/3\) we find that \(M_2 ≈ 0.58\). Using this Mach number we can find the mass flow rate using the formula \(\dot{m} = A * P_1 *sqrt(\frac{\gamma}{R*T_1})*\frac{M_2}{(1 + ((\gamma -1)/2)*M_2^2))^((\gamma+1)/(2*(\gamma-1)))\). This gives us \(\dot{m} ≈ 0.31 slug/sec\) for case (a).
04

Case (b) Receiver Pressure 5 psia

Since the receiver pressure (\(P_2\)) of 5 psia is less than the critical pressure (\(P^*\)), the flow is supersonic and choked at the throat. This means our mass flow rate equation simplifies to \(\dot{m} = A * P_1 * sqrt(\frac{\gamma}{R * T_1}) * ((\gamma +1)/2)^((\gamma+1)/(2*(\gamma −1))\). This gives us \(\dot{m} ≈ 0.45 slug/sec\).

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Most popular questions from this chapter

At some point for air flow in a duct, \(p=20\) psia, \(T=500^{\circ} \mathrm{R}\) and \(V=500 \mathrm{ft} / \mathrm{s}\). Can a normal shock occur at this point?

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