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Chapter 11: Problem 54

An ideal gas flows isentropically through a convergingdiverging nozzle. At a section in the converging portion of the nozzle. \(A_{1}=0.1 \mathrm{m}^{2}, p_{1}=600 \mathrm{kPa}(\mathrm{abs}), T_{1}=20^{\circ} \mathrm{C},\) and \(\mathrm{M} \varepsilon_{1}=0.6 .\) For section (2) in the diverging part of the nozzle, determine \(A_{2}, p_{2},\) and \(T_{2}\) if \(\mathrm{Ma}_{2}=3.0\) and the gas is air.

Short Answer

Expert verified
The cross-sectional area \(A_{2}\) at the diverging part of the nozzle is \(0.138 \, m^2\), the pressure \(p_{2}\) is \(71075 \, Pa\), and the temperature \(T_{2}\) is \(196.44 \, K\).

Step by step solution

01

Calculate the Initial Conditions

First, convert the given initial temperature from Celsius to Kelvin: \(T1 = 20 + 273.15 = 293.15 \, K\). Now, calculate the initial condition for the specific volume using the ideal gas law: \(v1 = RT1/p1\), where R is the gas constant for air: \(R = 287 \, J/(kg \times K)\). Then calculate the initial condition for the specific volume: \(v1 = (287 \times 293.15 )/600000 =0.1388 \, m^3/kg\).
02

Apply the Continuity Equation

Using the continuity equation, which states that the mass flow rate is constant throughout the nozzle, calculate the specific volume at Section 2. We know from the equation \(M2/M1 = sqrt[(1+ ((γ-1)/2)M1^2)/ (1+ ((γ-1)/2)M2^2)]*(v2/v1)^(γ-1)\), where γ is the heat capacity ratio (for air, γ = 1.4), that \(v2/v1 = (M2/M1) * sqrt[(1+ ((γ-1)/2)M2^2)/ (1+ ((γ-1)/2)M1^2)]^(1/(γ-1)). Substituting all known values, we have \(v2 = v1 * (M2/M1) * sqrt[(1+ ((γ-1)/2)M2^2)/ (1+ ((γ-1)/2)M1^2)] =12.53 \, m^3/kg \)
03

Use Isentropic Relations to Calculate Final Conditions

Apply the isentropic relations to calculate the final conditions. These relations state that the pressure and temperature at any point in an isentropic flow can be obtained by knowing the Mach number and the conditions at any other point. For pressure: \(p2/p1 = (1+ ((γ-1)/2)M1^2)^(γ/(γ-1)) / (1+ ((γ-1)/2)M2^2)^(γ/(γ-1))\). Substituting the known values gives: \(p2 = p1 * (1+ ((γ-1)/2)M1^2)^(γ/(γ-1)) / (1+ ((γ-1)/2)M2^2)^(γ/(γ-1)) = 71075 \, Pa \). For temperature: \(T2/T1 = (1+ ((γ-1)/2)M1^2) / (1+ ((γ-1)/2)M2^2)\). Substituting the known values gives: \(T2 = T1 * (1+ ((γ-1)/2)M1^2) / (1+ ((γ-1)/2)M2^2) = 196.44 \, K \).
04

Calculate the Final Area

The final area \(A2\) can be obtained by rearranging the definition of Mach number: \(M = V/a\), where \(V\) is velocity and \(a\) is speed of sound. The speed of sound in a gas is given by \(a = sqrt[γRT]\) and velocity by \(V = 1/v\), where \(v\) is the specific volume. Thus, the area can be calculated as \(A = m / (ρV)\), where \(ρ = 1/v\) is the density and \(m\) is the mass flow rate. The mass flow rate is equal in sections 1 and 2, thus \(A2/A1 = v2/v1\). So, \(A2 = A1 * (v2/v1) = 0.138 \, m^2.\)

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Most popular questions from this chapter

A jet engine is to be designed for an altitude of \(12,000 \mathrm{m}\) where the atmospheric pressure is \(19.3 \mathrm{kPa}\). The jet nozzle has a supersonic exit Mach number and is perfectly expanded. The stagnation pressure and temperature of the gas are 100 kpa and \(600^{\circ} \mathrm{C}\) The flow rate of gas is \(45 \mathrm{kg} / \mathrm{s}\). Calculate the throat area, exit area, and exit velocity. Use \(k=1.4\) and \(R=260 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) for the gas.

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