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Chapter 8: Problem 7

The flow just upstream of a normal shock wave is given by \(p_{1}=1\) atm, \(T_{1}=288 \mathrm{~K}\), and \(M_{1}=2.6\). Calculate the following properties just downstream of the shock: \(p_{2}, T_{2}, \rho_{2}, M_{2}, p_{0,2}, T_{0,2}\), and the change in entropy across the shock.

Short Answer

Expert verified
Downstream properties are \(p_2 = 4.929\) atm, \(T_2 = 471.48\) K, \(\rho_2 = 3.014 \rho_1\), \(M_2 = 0.482\), \(p_{0,2} = 0.907 p_{0,1}\), \(T_{0,2} = 540.96\) K, and entropy change is 134.6 J/kg K.

Step by step solution

01

Calculate Downstream Pressure

For a normal shock, the pressure ratio across the shock can be calculated using the formula:\[\frac{p_2}{p_1} = 1 + \frac{2\gamma}{\gamma + 1}(M_1^2 - 1)\]Substitute \(M_1 = 2.6\) and \(\gamma = 1.4\) (a typical value for air),\[\frac{p_2}{p_1} = 1 + \frac{2 \times 1.4}{1.4 + 1}(2.6^2 - 1)= 4.929\]Thus, \(p_2 = 4.929 \, \text{atm} \times p_1 = 4.929 \, \text{atm}\).
02

Calculate Downstream Temperature

The temperature ratio across the shock wave is given by:\[\frac{T_2}{T_1} = \frac{(1 + (\gamma - 1)/2 \times M_1^2)(\gamma + 1)}{(\gamma + 1)^2 M_1^2}\]Substitute known values,\[\frac{T_2}{T_1} = \frac{(1 + 0.2 \times 2.6^2)(2.4)}{(2.4)^2 \times 2.6^2} = 1.635 \]Thus, \(T_2 = 1.635 \times 288 \, \text{K} = 471.48 \, \text{K}\).
03

Calculate Downstream Density

The density ratio is related to pressure and temperature as follows:\[\frac{\rho_2}{\rho_1} = \frac{p_2/p_1}{T_2/T_1}\]Using the calculated ratios \(\frac{p_2}{p_1} = 4.929\) and \(\frac{T_2}{T_1} = 1.635\), we have:\[\frac{\rho_2}{\rho_1} = \frac{4.929}{1.635} = 3.014\]If \(\rho_1\) is known, multiply by this factor to find \(\rho_2\).
04

Calculate Downstream Mach Number

The Mach number downstream of a normal shock is given by:\[M_2 = \sqrt{\frac{1 + (\gamma - 1)/2 M_1^2}{\gamma M_1^2 - (\gamma - 1)/2}}\]Substitute known values,\[M_2 = \sqrt{\frac{1 + 0.2 \times 2.6^2}{1.4 \times 2.6^2 - 0.2}} = 0.482\]
05

Calculate Downstream Stagnation Pressure

Stagnation pressure relationships in shocks are given by:\[\frac{p_{0,2}}{p_{0,1}} = \frac{\left(\frac{p_2}{p_1}\right)^{\frac{\gamma}{\gamma - 1}}}{\left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma - 1}}}\]Considering \(\frac{p_{0,2}}{p_{0,1}}\), we have:\[\frac{p_{0,2}}{p_{0,1}} = \frac{4.929^{3.5}}{1.635^{3.5}} = 0.907 \]If \(p_{0,1}\) is known, multiply by this factor to get \(p_{0,2}\).
06

Calculate Downstream Stagnation Temperature

For normal shocks, the stagnation temperature remains constant across the shock wave:\[T_{0,2} = T_{0,1}\]Therefore, if \(T_{0,1}\) can be calculated from \(M_1\) using\[T_{0,1} = T_1 \left(1 + \frac{\gamma - 1}{2}M_1^2\right) \approx 540.96 \, \text{K}\]Thus, \(T_{0,2} = 540.96 \, \text{K}\).
07

Calculate Change in Entropy

The change in entropy across a shock is given by:\[\Delta S = R \ln\left(\frac{p_2/p_1}{\rho_2/\rho_1}\right)\]Where \(R\) is the specific gas constant (approximately 287 J/(kg K) for air),\[\Delta S = 287 \ln\left(\frac{4.929}{3.014}\right) = 287 \times 0.4689 = 134.6 \, \text{J/(kg K)}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Downstream Pressure
When a normal shock wave occurs, the pressure in the flow changes significantly. The downstream pressure, denoted as \(p_2\), can be found using the pressure ratio formula:
  • First, identify the upstream Mach number \(M_1\), which is the speed of the flow relative to the speed of sound.
  • Use the relation: \(\frac{p_2}{p_1} = 1 + \frac{2\gamma}{\gamma + 1}(M_1^2 - 1)\).
  • For standard air conditions, set \(\gamma = 1.4\) and substitute the known \(M_1\) to find the pressure ratio.
By solving the equation for \(M_1 = 2.6\), the pressure ratio gives us a downstream pressure \(p_2\) of approximately 4.929 atm. This increase shows how a normal shock wave compresses the flow.
Mach Number
The Mach number is a critical parameter in supersonic flows. It represents the speed of an object or flow relative to the speed of sound.
  • The upstream Mach number \(M_1\) indicates how fast the flow approaches the shock wave.
  • Post shock, the Mach number significantly drops, indicating a slower flow.
  • Using the formula for downstream Mach number, \(M_2 = \sqrt{\frac{1 + (\gamma - 1)/2 M_1^2}{\gamma M_1^2 - (\gamma - 1)/2}}\), helps in calculating this change.
For \(M_1 = 2.6\), the downstream Mach number \(M_2\) comes out to be about 0.482, illustrating the subsonic nature of the flow after the shock.
Entropy Change
Entropy is a measure of disorder in a system, and changes in entropy reflect energy dispersion and the second law of thermodynamics.
  • Across a normal shock wave, entropy increases due to irreversible processes.
  • The change in entropy \(\Delta S\) is calculated using: \(\Delta S = R \ln\left(\frac{p_2/p_1}{\rho_2/\rho_1}\right)\), where \(R\) is the specific gas constant.
In this particular case, \(\Delta S\) works out to be approximately 134.6 J/(kg K), indicating the energy transformation that occurs across the shock.
Stagnation Properties
Stagnation properties refer to conditions a fluid would reach if brought to rest isentropically.
  • Stagnation Pressure \(p_{0,2}\): Deals with losses due to shock waves. It's found using \(\frac{p_{0,2}}{p_{0,1}} = \frac{(p_2/p_1)^{\gamma/(\gamma-1)}}{(T_2/T_1)^{\gamma/(\gamma-1)}}\).
  • Despite shock, there is often a loss in stagnation pressure, reflecting the non-isentropic nature of shocks.
  • Stagnation Temperature \(T_{0,2}\): Remains constant across a normal shock, as no work or heat is exchanged with the surroundings in the system.
For the provided example, stagnation properties verify the entropic nature, with \(T_{0,2} = T_{0,1} \approx 540.96 \, \text{K}\), showing consistency in temperature across the shock.

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Most popular questions from this chapter

The temperature in the reservoir of a supersonic wind tunnel is \(519^{\circ} \mathrm{R}\). In the test section, the flow velocity is \(1385 \mathrm{ft} / \mathrm{s}\). Calculate the test-section Mach number. Assume the tunnel flow is adiabatic.

When the Apollo command module returned to earth from the moon, it entered the earth's atmosphere at a Mach number of 36 . Using the results from the present chapter for a calorically perfect gas with the ratio of specific heats equal to \(1.4\), predict the gas temperature at the stagnation point of the Apollo at Mach 36 at an altitude where the freestream temperature is \(300 \mathrm{~K}\). Comment on the validity of your answer.

On March 16, 1990, an Air Force SR-71 set a new continental speed record, averaging a velocity of \(2112 \mathrm{mi} / \mathrm{h}\) at an altitude of \(80,000 \mathrm{ft}\). Calculate the temperature (in degrees Fahrenheit) at a stagnation point on the vehicle.

Consider a flow with a pressure and temperature of \(1 \mathrm{~atm}\) and \(288 \mathrm{~K}\). A Pitot tube is inserted into this flow and measures a pressure of \(1.555 \mathrm{~atm}\). What is the velocity of the flow?

The flow just upstream of a normal shock wave is given by \(p_{1}=1800 \mathrm{lb} / \mathrm{ft}^{2}, T_{1}=480^{\circ} \mathrm{R}\), and \(M_{1}=3.1 .\) Calculate the velocity and \(M^{*}\) behind the shock.
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