Non-isentropic Diffuser Analysis Air, at 100 kPa, 300 K, enters a 90% efficient adiabatic diffuser with a velocity of 300 m/s. If the exit velocity is 50 m/s, determine (a) the exit pressure, (b) pressure recovery factor, (c) pressure rise coefficient, and (d) exit-to-inlet area ratio. What-if

Question

Non-isentropic Diffuser Analysis

Air, at 100 kPa, 300 K, enters a 90% efficient adiabatic diffuser with a velocity of 300 m/s. If the exit velocity is 50 m/s, determine (a) the exit pressure, (b) pressure recovery factor, (c) pressure rise coefficient, and (d) exit-to-inlet area ratio. What-if scenario: (e) What would the answers be if the exit velocity was 75 m/s?

Accepted Answer

Different principal states of the diffuser are shown in Figure 15.43. States 1 and 2 are the actual inlet and exit states, with t1 and t2 denoting the corresponding total states. State-3 is the ideal inlet state (with reduced kinetic energy) such that the corresponding total state has the same total pressure at the exit state, that is, [latex] p_{t 3}=p_{t 2} [/latex]. The diffuser efficiency is the ratio of [latex] ke _{3} ext { to } ke _{1} [/latex], which relates [latex] T_{t 3}, T_{t 1}, ext { and } T_{1} [/latex].

Assumptions
One-dimensional, adiabatic flow of a perfect gas.
Analysis
For the inlet state, State-1:

[latex] M_{1}=\frac{V_{1}}{\sqrt{(1000 N / kN ) k R T_{1}}}=\frac{300}{\sqrt{(1000)(1.4)(0.287)(300)}}=0.864 [/latex]

Using the isentropic table (or the gas dynamics TESTcalc):

[latex]\begin{gathered}T_{t 1}=T_{1}\left(\frac{T_{1}}{T_{t 1}} ight)^{-1}=T_{1}\left[\left(\frac{T}{T_{t}} ight)_{@ M_{1}} ight]^{-1}=(300)(0.870)^{-1}=344.8 K \p_{t 1}=p_{1}\left(\frac{p_{1}}{p_{t 1}} ight)^{-1}=p_{1}\left[\left(\frac{p}{p_{t}} ight)_{@ M_{1}} ight]^{-1}=(100)(0.6144)^{-1}=162.76 kPa\end{gathered}[/latex]

Irrespective of the presence of friction, the energy equation, Eq. (15.10), for the diffuser can be written as [latex] T_{t 1}=T_{t 2} [/latex], where State-2 is the actual exit state. With the exit velocity given, the static temperature at the exit can be obtained from Eq. (15.8) as:

[latex] h_{t}=h+\frac{V^{2}}{2000} ;\left[\frac{ kJ }{ kg } ight] \quad ext { and }, \quad T_{t}=T+\frac{V^{2}}{2000 c_{p}} \quad [K] [/latex] (15.8)

[latex] j_{i}=j_{e} ; \quad \Rightarrow \quad h_{t i}=h_{t e} ;\left[\frac{ kJ }{ kg } ight] [/latex] (15.10)

[latex]T_{2}=T_{i 2}- ke _{2}=T_{i 2}-\frac{V^{2}}{2(1000 J / kJ ) c_{p}}=344.8-\frac{50^{2}}{2000(1.005)}=343.56 K[/latex]

The Mach number and the static-to-total pressure ratio at the exit now can be calculated:

[latex] M_{2}=\frac{V_{2}}{\sqrt{(1000 N / kN ) k R T_{2}}}=0.135 ; \quad \Rightarrow \quad \frac{p_{2}}{p_{t 2}}=\left(\frac{p}{p_{t}} ight)_{@ M_{2}}=0.987 [/latex]

Because the flow is not isentropic in the diffuser, [latex] p_{t 2}

[latex]\begin{gathered}\eta_{ ext {diffuser }}=\frac{ ke _{3}}{ ke _{1}}=\frac{h_{t 3}-h_{3}}{h_{t 1}-h_{1}}=\frac{T_{t 3}-T_{1}}{T_{t 1}-T_{1}} \\Rightarrow \quad T_{t 3}=300+0.9(344.8-300)=340.3 K\end{gathered}[/latex]

The total pressure [latex] p_{t 2} [/latex] can be calculated from [latex] \eta_{ ext {diffuser }} ext { and } p_{t 1} [/latex] using Eq. (15.46). A more fundamental approach is to use the isentropic relation, Eq. (3.71), between State-3 and State-t3 to obtain:

[latex] \frac{p_{2}}{p_{1}}=\left(\frac{T_{2}}{T_{1}} ight)^{\frac{k}{k-1}}=\left(\frac{ ho_{2}}{ ho_{1}} ight)^{k}=\left(\frac{v_{1}}{v_{2}} ight)^{k} [/latex] (3.71)

[latex] \eta_{ ext {diffuser }}=\frac{\left[1+M_{i}^{2}(k-1) / 2 ight]\left(F_{p} ight)^{(k-1) / k}-1}{M_{i}^{2}(k-1) / 2} [/latex] (15.46)

[latex] p_{t 2}=p_{t 3}=\frac{p_{t 3}}{p_{3}} p_{3}=p_{3}\left(\frac{T_{t 3}}{T_{3}} ight)^{k /(k-1)}=p_{1}\left(\frac{T_{t 3}}{T_{1}} ight)^{k /(k-1)}=155.5 kPa [/latex]

Therefore,

[latex] p_{2}=\frac{p_{2}}{p_{t 2}} p_{t 2}=\left(\frac{p}{p_{t}} ight)_{@ M_{2}=0.135} p_{t 2}=(0.987)(155.5)=153.5 kPa [/latex]

The pressure recovery factor and pressure rise coefficient can now be calculated:

[latex] F_{p} \equiv \frac{p_{t 2}}{p_{t 1}}=\frac{155.5}{162.76}=0.955 \quad ext { and } \quad C_{p}=\frac{p_{2}-p_{1}}{p_{t 1}-p_{1}}=\frac{153.5-100}{162.76-100}=0.85 [/latex]

Due to frictional losses in the diffuser, the critical area between the inlet and exit must be different. The mass flow rate expression of Eq. (15.28), when applied to the inlet and exit, yields:

[latex] \dot{m}=A_{*} p_{t} \sqrt{\frac{(1000 J / kJ ) k}{R T_{t}}}\left(\frac{2}{k+1} ight)^{(k+1) /[2(k-1)]} ;\left[\frac{ kg }{ s } ight] [/latex] (15.28)

[latex] A_{* 1} p_{t 1}=A_{* 2} p_{t 2} ; \quad \Rightarrow \quad \frac{A_{* 1}}{A_{* 2}}=\frac{p_{t 2}}{p_{t 1}}=F_{p}=0.955 [/latex] (15.48)

Therefore, the area ratio between the exit and inlet can be calculated by using the isentropic table as:

[latex] \frac{A_{2}}{A_{1}}=\frac{A_{2}}{A_{* 2}} \frac{A_{*_{2}}}{A_{* 1}} \frac{A_{* 1}}{A_{1}}=\left(\frac{A}{A_{*}} ight)_{@ M_{2}}\left(\frac{A_{*}}{A} ight)_{@ M_{1}} \frac{A_{* 2}}{A_{* 1}}=(4.35)\left(\frac{1}{1.017} ight)\left(\frac{1}{0.955} ight)=4.48 [/latex] (15.49)

TEST Solution
Launch the gas dynamics TESTcalcs and select Air as the working fluid. Calculate States 1, 2, and 3 as described in the TEST-code (see TEST > TEST-codes). Although A1 is assumed to be 1 m^2, the answers can be verified to be independent of the value of A1.

What-if scenario
Change Vel2 to 75 m/s and click Super-Calculate. The new p2 is found to be 151.1 kPa. While the pressure recovery factor remains unchanged, the pressure rise factor is calculated in the I/O panel as 0.814 and the area ratio as A2/A1 = 3.02.

Discussion
The exit velocity can be seen to have a significant effect on the shape and pressure rise factor of a diffuser. Therefore, the exit velocity should not be neglected in a diffuser analysis, as in Example 4-6.

Question 15.11: Non-isentropic Diffuser Analysis Air, at 100 kPa, 300 K, ent...

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