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11.4.4: Isentropic Flow Examples

Example 11.3

Air is allowed to flow from a reservoir with temperature of \(21^{\circ}C\) and with pressure of 5[MPa] through a tube. It was measured that air mass flow rate is 1[kg/sec]. At some point on the tube static pressure was measured to be 3[MPa]. Assume that process is isentropic and neglect the velocity at the reservoir, calculate the Mach number, velocity, and the cross section area at that point where the static pressure was measured. Assume that the ratio of specific heat is \( k=C_p / C_v = 1.4\).

Solution 11.3

The stagnation conditions at the reservoir will be maintained throughout the tube because the process is isentropic. Hence the stagnation temperature can be written \(T_{0} = constant\) and \(P_{0} = constant\) and both of them are known (the condition at the reservoir). For the point where the static pressure is known, the Mach number can be calculated by utilizing the pressure ratio. With the known Mach number, the temperature, and velocity can be calculated. Finally, the cross section can be calculated with all these information. In the point where the static pressure known
    \bar{P} = \dfrac{P }{ P_0} = \dfrac{3 [MPa] }{ 5 [MPa] } = 0.6  
From Table (??) or from Figure (??) or utilizing the enclosed program, Potto-GDC, or simply using the equations shows that

\(\mathbf{M}\) \(\mathbf{\dfrac{T }{ T_0}}\) \(\mathbf{\dfrac{\rho }{ \rho_0}}\) \(\mathbf{\dfrac{A }{ A^{\star} }}\) \(\mathbf{\dfrac{P }{ P_0}}\) \(\mathbf{\dfrac{A \,P }{ A^{\star} \,P_0 }}\) \(\mathbf{\dfrac{F }{ F^{\star} }}\)
0.88639 0.86420 0.69428 1.0115 0.60000 0.60693 0.53105

With these values the static temperature and the density can be calculated.

    T  =  0.86420338 \times (273+ 21) = 254.076 K     \\
    \rho  =  \dfrac{\rho }{ \rho_0} \overbrace{\dfrac{P_{0}}{ R\, T_0}}^{\rho_0}=
        0.69428839 \times {5 \times 10^6 [Pa] \over 287.0  
        \left[\dfrac{J }{ kg\, K}\right]    \times 294 [K] } \\  
     =     41.1416 \left[\dfrac{kg }{ m^3 }\right]      
The velocity at that point is  
    U =  M \,\overbrace{\sqrt{k\,R\,T}}^{c} = 0.88638317 \times  
            \sqrt { 1.4 \times 287 \times 254.076} \sim 283 [m /sec]     
The tube area can be obtained from the mass conservation as
    A = \dfrac{\dot {m} }{ \rho\, U} = 8.26 \times 10^{-5} [m^{3}]    
For a circular tube the diameter is about 1[cm].

Example 11.4

The Mach number at point \(\mathbf{A}\) on tube is measured to be \(M=2\) and the static pressure is \(2 [Bar]\). Downstream at point B the pressure was measured to be 1.5[Bar]. Calculate the Mach number at point B under the isentropic flow assumption. Also, estimate the temperature at point B. Assume that the specific heat ratio \(k=1.4\) and assume a perfect gas model.

Well, this question is for academic purposes, there is no known way for the author to directly measure the Mach number. The best approximation is by using inserted cone for supersonic flow and measure the oblique shock. Here it is subsonic and this technique is not suitable.

Solution 11.4

With the known Mach number at point \(\mathbf{A}\) all the ratios of the static properties to total (stagnation) properties can be calculated. Therefore, the stagnation pressure at point \(\mathbf{A}\) is known and stagnation temperature can be calculated. At \(M=2\) (supersonic flow) the ratios are

Isentropic Flow Input: M k = 1.4  


\(\dfrac{\rho}{\rho_0}\) \(\dfrac{A}{A^{\star} }\) \(\dfrac{P}{P_0}\) \(\dfrac{A\, P }{ A^{\star} \, P_0}\) \(\dfrac{F }{ F^{\star}}\)
2.0 0.555556 0.230048 1.6875 0.127805 0.21567 0.593093

With this information the pressure at point B can be expressed as

    \dfrac{P_{A} }{ P_{0}} =  \overbrace{\dfrac{P_{B} }{ P_{0}} }^ {\text{from
    the table isentropic  @  M = 2}} \times
         \dfrac{P_{A} }{ P_{B}} = 0.12780453 \times \dfrac{2.0 }{ 1.5} = 0.17040604
The corresponding Mach number for this pressure ratio is 1.8137788 and \(T_{B} = 0.60315132\) \({P_{B} \over P_{0} }= 0.17040879\). The stagnation temperature can be "bypassed'' to calculate the temperature at point \(\mathbf{B}\)
    T_{B} = T_{A}\times \overbrace{T_{0} \over T_{A} }^{M=2} \times  
                 \overbrace{T_{B} \over T_{0}} ^{M=1.81..}
            = 250 [K] \times {1 \over 0.55555556} \times {0.60315132}  
            \simeq   271.42 [K] 

Example 11.5

Gas flows through a converging–diverging duct. At point "A'' the cross section area is \(50\) [\(cm^{2}\)] and the Mach number was measured to be 0.4. At point B in the duct the cross section area is 40 [\(cm^{2}\)]. Find the Mach number at point B. Assume that the flow is isentropic and the gas specific heat ratio is 1.4.

Solution 11.5

To obtain the Mach number at point B by finding the ratio of the area to the critical area. This relationship can be obtained by

    \dfrac{A_{B} }{ A{*} } = \dfrac{A _{B} }{ A_{A} } \times \dfrac{A_{A} }{ A^{\star}}
        = \dfrac{40 }{ 50} \times \overbrace{ 1.59014 }^{\text{from the 11.2}}
            = 1.272112  
With the value of \(\dfrac{A_{B} }{ A{\star} } \) from the Table or from Potto-GDC two solutions can be obtained. The two possible solutions: the first supersonic M = 1.6265306 and second subsonic M = 0.53884934. Both solution are possible and acceptable. The supersonic branch solution is possible only if there where a transition at throat where M=1.

Isentropic Flow input: \(\dfrac{A}{A^{\star}}\) k = 1.4  
\(M\) \(\dfrac{A}{A^{\star}}\) \(\dfrac{\rho}{\rho_0}\) \(\dfrac{A}{A^{\star} }\) \(\dfrac{P}{P_0}\) \(\dfrac{A\, P }{ A^{\star} \, P_0}\) \(\dfrac{F }{ F^{\star}}\)
0.538865 0.945112 0.868378 1.27211 0.820715 1.04404 0.611863
1.62655 0.653965 0.345848 1.27211 0.226172 0.287717 0.563918

Example 11.6

Engineer needs to redesign a syringe for medical applications. They complained that the syringe is "hard to push.'' The engineer analyzes the flow and conclude that the flow is choke. Upon this fact, what engineer should do with the syringe; increase the pushing diameter or decrease the diameter? Explain

Solution 11.6

This problem is a typical to compressible flow in the sense the solution is opposite the regular intuition. The diameter should be decreased. The pressure in the choke flow in the syringe is past the critical pressure ratio. Hence, the force is a function of the cross area of the syringe. So, to decrease the force one should decrease the area.


  • Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.