Compressible Flow Free Textbook

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``Naughty Professor'' Problems in Isentropic Flow

To explain the material better some instructors invented problems, which have mostly academic proposes, (see for example, Shapiro (problem 4.5)). While these problems have a limit applicability in reality, they have substantial academic value and therefore presented here. The situation where the mass flow rate per area given with one of the stagnation properties and one of the static properties, e.g. and or T₀ and present difficulty for the calculations. The use of the regular isentropic Table is not possible because there isn't variable represent this kind problems. For this kind of problems a new Table was constructed and present here^4.5.

The case of T₀ and

This case considered to be simplest case and will first presented here. Using energy equation (4.9) and substituting for Mach number $M = \dot{m} / A \rho c$ results in

$\displaystyle {T_0 \over T} = 1 + { k -1 \over 2} \left(\dot{m} \over A \rho c \right)^2$

(4.56)

Rearranging equation (4.56) result in

$\displaystyle T_0 \rho^2 = \overbrace{T \rho}^{p \over R} \rho + \overbrace{\left( T \over c^2 \right)}^{1/kR} {k -1 \over 2} \left( \dot{m} \over A \right) ^2$

(4.57)

And further Rearranging equation (4.57) transformed it into

$\displaystyle \rho^2 = {P \rho \over T_0 R} + {k -1 \over 2kRT_0} \left( \dot{m} \over A \right) ^2$

(4.58)

Equation (4.58) is quadratic equation for density, ρ when all other variables are known. It is convenient to change it into

$\displaystyle \rho^2 - {P \rho \over T_0 R} - {k -1 \over 2kRT_0} \left( \dot{m} \over A \right) ^2 = 0$

(4.59)

The only physical solution is when the density is positive and thus the only solution is

$\displaystyle \rho = \half \left[ { P \over R T_0} + \sqrt{ \left( P \over RT_0... ...} \over A \right)^2}_{\hookrightarrow (M\rightarrow 0 ) \rightarrow 0}} \right]$

(4.60)

For almost incompressible flow the density is reduced and the familiar form of perfect gas model is seen since stagnation temperature is approaching the static temperature for very small Mach number ( $\rho = {P \over R T_0}$ ). In other words, the terms for the group over the under-brace approaches zero when the flow rate (Mach number) is very small.

It is convenient to denote a new dimensionless density as

$\displaystyle \hat{\rho} = {\rho \over {p \over R T_0 }} = {\rho R T_0 \over P} = { 1 \over \bar{T}}$

(4.61)

With this new definition equation (4.60) is transformed into

$\displaystyle \hat{\rho} = {1 \over 2} \left( 1 + \sqrt{1 + 2 {(k -1) R T_0 \over k P^2} \left( \dot{m} \over A \right) ^2 } \right)$

(4.62)

The dimensionless density now is related to a dimensionless group that is a function of Fn number and Mach number only! Thus, this dimensionless group is function of Mach number only.

$\displaystyle { R T_0 \over P^2} \left( \dot{m} \over A \right) ^2 = { 1 \over ... ... \over A \right)^2 \left( P_0 \over P \right)^2}^{ {A^{*}P_0 \over AP} = f(M) }$

(4.63)

Thus,

$\displaystyle { R T_0 \over P^2} \left( \dot{m} \over A \right) ^2 = {Fn^2\over k} \left( A^{*} P_0 \over A P \right)^2$

(4.64)

Hence, the dimensionless density is

$\displaystyle \hat{\rho} = {1 \over 2} \left( 1 + \sqrt{1 + 2 {(k -1) Fn^2 \over k^2} \left( A^{*} P_0 \over A P \right) ^2 } \right)$

(4.65)

Again notice that the right hand side of equation (4.65) is only function of Mach number (well, also the specific heat,

). And the values of $A P \over A^{*} P_0$ were tabulated in Table (4.2) and Fn is tabulated in the next Table (4.1). Thus, the problems is reduced to finding tabulated values.

The case of P₀ and T

A similar problem can be described for the case of stagnation pressure, , and static temperature, .

First, it is shown that the dimensionless group is a function of Mach number only (well, again the specific heat ratio, also).

$\displaystyle { R T \over {P_0}^2} \left( \dot{m} \over A \right) ^2 = {Fn^2\ov... ...P_0 \over A P \right)^2 \left( T \over T_0 \right) \left( P_0 \over P \right)^2$

(4.66)

It can be noticed that

$\displaystyle {{Fn}^2 \over k} = \left( T \over T_0 \right) \left( P_0 \over P \right)^2$

(4.67)

Thus equation (4.66) became

$\displaystyle { R T \over {P_0}^2} \left( \dot{m} \over A \right) ^2 = \left( A^{*} P_0 \over A P \right)^2$

(4.68)

The right hand side is tabulated in the ``regular'' isentropic Table such (4.2). This example shows how a dimensional analysis is used to solve a problems without actually solving any equations. The actual solution of the equation is left as exercise (this example under construction). What is the legitimacy of this method? The explanation simply based the previous experience in which for a given ratio of area or pressure ratio (etcetera) determines the Mach number. Based on the same arguments, if it was shown that a group of parameters depends only Mach number than the Mach is determined by this group.

The method of solution for given these parameters is by calculating the $P A \over P_0 A^{*}$ and then using the table to find the corresponding Mach number.

The case of &rho₀ and T or P

The last case sometimes referred to as the ``naughty professor's question'' case dealt here is when the stagnation density given with the static temperature/pressure. First, the dimensionless approach is used later analytical method is discussed (under construction).

$\displaystyle { 1 \over R {\rho_0}P } \left( \dot{m} \over A \right) ^2 = {\ove... ...} \left( \dot{m} \over A \right) ^2 = {Fn^2 \over k} \left( P_0 \over P \right)$

(4.69)

The last case dealt here is of the stagnation density with static pressure and the following is dimensionless group

$\displaystyle { 1 \over R {\rho_0}^2T } \left( \dot{m} \over A \right) ^2 = {\o... ...} \left( \dot{m} \over A \right) ^2 = {Fn^2 \over k} \left( T_0 \over T \right)$ (4.70)

It was hidden in the derivations/explanations of the above analysis didn't explicitly state under what conditions these analysis is correct. Unfortunately, not all the analysis valid for the same conditions and is as the regular ``isentropic'' Table, (4.2). The heat/temperature part is valid for enough adiabatic condition while the pressure condition requires also isentropic process. All the above conditions/situations require to have the perfect gas model as the equation of state. For example the first ``naughty professor'' question is sufficient that process is adiabatic only ( T₀, , mass flow rate per area.).

Fliegner's number a function of Mach number

M $\mathbf{Fn}$ $\mathbf{\hat{\rho} }$ $\mathbf{\left(P_0 A^{*} \over A P \right)^2}$ $\mathbf{{RT_0 \over P^2} \left( \dot{m}\over A \right)^2}$ $\mathbf{{1 \over R \rho_0 P} \left( \dot{m}\over A \right)^2}$ $\mathbf{{1 \over R {\rho_0}^2 T} \left( \dot{m}\over A \right)^2}$

0.00E+00 1.400E-06 1.000 0.0 0.0 0.0 0.0

0.050001 0.070106 1.000 0.00747 2.62E-05 0.00352 0.00351

0.10000 0.14084 1.000 0.029920 0.000424 0.014268 0.014197

0.20000 0.28677 1.001 0.12039 0.00707 0.060404 0.059212

0.21000 0.30185 1.001 0.13284 0.00865 0.067111 0.065654

0.22000 0.31703 1.001 0.14592 0.010476 0.074254 0.072487

0.23000 0.33233 1.002 0.15963 0.012593 0.081847 0.079722

0.24000 0.34775 1.002 0.17397 0.015027 0.089910 0.087372

0.25000 0.36329 1.003 0.18896 0.017813 0.098460 0.095449

0.26000 0.37896 1.003 0.20458 0.020986 0.10752 0.10397

0.27000 0.39478 1.003 0.22085 0.024585 0.11710 0.11294

0.28000 0.41073 1.004 0.23777 0.028651 0.12724 0.12239

0.29000 0.42683 1.005 0.25535 0.033229 0.13796 0.13232

0.30000 0.44309 1.005 0.27358 0.038365 0.14927 0.14276

0.31000 0.45951 1.006 0.29247 0.044110 0.16121 0.15372

0.32000 0.47609 1.007 0.31203 0.050518 0.17381 0.16522

0.33000 0.49285 1.008 0.33226 0.057647 0.18709 0.17728

0.34000 0.50978 1.009 0.35316 0.065557 0.20109 0.18992

0.35000 0.52690 1.011 0.37474 0.074314 0.21584 0.20316

0.36000 0.54422 1.012 0.39701 0.083989 0.23137 0.21703

0.37000 0.56172 1.013 0.41997 0.094654 0.24773 0.23155

0.38000 0.57944 1.015 0.44363 0.10639 0.26495 0.24674

0.39000 0.59736 1.017 0.46798 0.11928 0.28307 0.26264

0.40000 0.61550 1.019 0.49305 0.13342 0.30214 0.27926

0.41000 0.63386 1.021 0.51882 0.14889 0.32220 0.29663

0.42000 0.65246 1.023 0.54531 0.16581 0.34330 0.31480

0.43000 0.67129 1.026 0.57253 0.18428 0.36550 0.33378

0.44000 0.69036 1.028 0.60047 0.20442 0.38884 0.35361

0.45000 0.70969 1.031 0.62915 0.22634 0.41338 0.37432

0.46000 0.72927 1.035 0.65857 0.25018 0.43919 0.39596

0.47000 0.74912 1.038 0.68875 0.27608 0.46633 0.41855

0.48000 0.76924 1.042 0.71967 0.30418 0.49485 0.44215

0.49000 0.78965 1.046 0.75136 0.33465 0.52485 0.46677

0.50000 0.81034 1.050 0.78382 0.36764 0.55637 0.49249

0.51000 0.83132 1.055 0.81706 0.40333 0.58952 0.51932

0.52000 0.85261 1.060 0.85107 0.44192 0.62436 0.54733

0.53000 0.87421 1.065 0.88588 0.48360 0.66098 0.57656

0.54000 0.89613 1.071 0.92149 0.52858 0.69948 0.60706

0.55000 0.91838 1.077 0.95791 0.57709 0.73995 0.63889

0.56000 0.94096 1.083 0.99514 0.62936 0.78250 0.67210

0.57000 0.96389 1.090 1.033 0.68565 0.82722 0.70675

0.58000 0.98717 1.097 1.072 0.74624 0.87424 0.74290

0.59000 1.011 1.105 1.112 0.81139 0.92366 0.78062

0.60000 1.035 1.113 1.152 0.88142 0.97562 0.81996

0.61000 1.059 1.122 1.194 0.95665 1.030 0.86101

0.62000 1.084 1.131 1.236 1.037 1.088 0.90382

0.63000 1.109 1.141 1.279 1.124 1.148 0.94848

0.64000 1.135 1.151 1.323 1.217 1.212 0.99507

0.65000 1.161 1.162 1.368 1.317 1.278 1.044

0.66000 1.187 1.173 1.414 1.423 1.349 1.094

0.67000 1.214 1.185 1.461 1.538 1.422 1.147

0.68000 1.241 1.198 1.508 1.660 1.500 1.202

0.69000 1.269 1.211 1.557 1.791 1.582 1.260

0.70000 1.297 1.225 1.607 1.931 1.667 1.320

0.71000 1.326 1.240 1.657 2.081 1.758 1.382

0.72000 1.355 1.255 1.708 2.241 1.853 1.448

0.73000 1.385 1.271 1.761 2.412 1.953 1.516

0.74000 1.415 1.288 1.814 2.595 2.058 1.587

0.75000 1.446 1.305 1.869 2.790 2.168 1.661

0.76000 1.477 1.324 1.924 2.998 2.284 1.738

0.77000 1.509 1.343 1.980 3.220 2.407 1.819

0.78000 1.541 1.362 2.038 3.457 2.536 1.903

0.79000 1.574 1.383 2.096 3.709 2.671 1.991

0.80000 1.607 1.405 2.156 3.979 2.813 2.082

0.81000 1.642 1.427 2.216 4.266 2.963 2.177

0.82000 1.676 1.450 2.278 4.571 3.121 2.277

0.83000 1.712 1.474 2.340 4.897 3.287 2.381

0.84000 1.747 1.500 2.404 5.244 3.462 2.489

0.85000 1.784 1.526 2.469 5.613 3.646 2.602

0.86000 1.821 1.553 2.535 6.006 3.840 2.720

0.87000 1.859 1.581 2.602 6.424 4.043 2.842

0.88000 1.898 1.610 2.670 6.869 4.258 2.971

0.89000 1.937 1.640 2.740 7.342 4.484 3.104

0.90000 1.977 1.671 2.810 7.846 4.721 3.244

0.91000 2.018 1.703 2.882 8.381 4.972 3.389

0.92000 2.059 1.736 2.955 8.949 5.235 3.541

0.93000 2.101 1.771 3.029 9.554 5.513 3.699

0.94000 2.144 1.806 3.105 10.20 5.805 3.865

0.95000 2.188 1.843 3.181 10.88 6.112 4.037

0.96000 2.233 1.881 3.259 11.60 6.436 4.217

0.97000 2.278 1.920 3.338 12.37 6.777 4.404

0.98000 2.324 1.961 3.419 13.19 7.136 4.600

0.99000 2.371 2.003 3.500 14.06 7.515 4.804

1.000 2.419 2.046 3.583 14.98 7.913 5.016

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$\begin{examl} A gas flows in the tube with mass flow rate of 0.1 [kg/sec] and tu... ...i>\end{rawhtml}, \begin{rawhtml} <i>R=287[j/kg K]</i>\end{rawhtml}. \end{examl}$
Solution

The first thing that need to be done is to find the mass flow per area and it is

$\displaystyle {\dot{m} \over A } = {0.1 / 0.001} = 100.0 [kg/sec/m^2]$

It can be noticed that the total temperature is and the static pressure is 1.5[Bar]. The solution is based on section equations (4.60) through (4.65). It is fortunate that Potto-GDC exist and it can be just plug into it and it provide that

Isentropic Flow Input: mDot P T0 k = 1.3

M T/T0 ρ/ρ0 A/A* P/P0 PAR F/F*

0.171236 0.995621 0.985478 3.47565 0.981162 3.41018 1.53921

The velocity can be calculated as

$\displaystyle U = Mc = \sqrt{kRT} M = 0.17\times \sqrt{1.3\times 287 \times 300\times } \sim 56.87 [m/sec]$

The stagnation pressure is

$\displaystyle P_0 = {P \over {P / P_0}} = 1.5 / 0.98116 = 1.5288 [Bar]$

Next: Flow with pressure losses Up: Mass Flow Rate (Number) Previous: Mass Flow Rate (Number) Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21

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