11.7.2: Dimensionless Representation

In this section the equations are transformed into the dimensionless form and presented as such. First it must be recalled that the temperature is constant and therefore, equation of state reads

${dP \over P} = {d \rho \over \rho} \label{isothermal:eq:stateDa} \tag{6}$
It is convenient to define a hydraulic diameter
$D_{H} = {4 \times \hbox{Cross Section Area} \over \hbox{ wetted perimeter }} \label{isothermal:eq:HDdef} \tag{7}$

The Fanning friction factor is introduced, this factor is a dimensionless friction factor sometimes referred to as the friction coefficient as

$f = { \tau_w \over \dfrac{1}{2} \rho U^{2} } \label{isothermal:eq:f} \tag{8}$
Substituting equation (8) into momentum equation (2) yields
$- dP - \dfrac{4\, dx }{ D_H} f \left( \dfrac{1}{2}\, \rho\, U^{2} \right) = { \overbrace{\rho\, U}^{\dot {m} \over A } dU } \label{isothermal:eq:momentumD1} \tag{9}$
Rearranging equation (9) and using the identify for perfect gas $$M^{2} = \rho U^{2} / k P$$ yields:
$- \dfrac{dP }{ P} - \dfrac{4\,f\, dx }{ D_H} \left( \dfrac{ k\,P\, M^{2} }{ 2} \right) = \dfrac{k\,P\, M^{2}\, dU }{ U} \label{isothermal:eq:momentumD} \tag{10}$

The pressure, $$P$$ as a function of the Mach number has to substitute along with velocity, $$U$$ as

$U^{2} = {k\,R\,T} M^{2} \label{isothermal:eq:MtoUfun} \tag{11}$
Differentiation of equation (11) yields
$d(U^{2}) = k\,R\,\left(M^{2}\; dT + T \; d (M^{2}) \; \right) \label{isothermal:eq:anotherMtoU} \tag{12}$
$\dfrac{d(M^2) }{ M^{2}} = \dfrac{d(U^2) }{ U^{2}} - \dfrac{ dT }{ T} \label{isothermal:eq:MtoUfunD} \tag{13}$
It can be noticed that $$dT = 0$$ for isothermal process and therefore
${d(M^2) \over M^{2}} = {d(U^2) \over U^{2}} = {2U \; dU \over U^{2}} = {2 dU \over U} \label{isothermal:eq:yetMtoUfunD} \tag{14}$
The dimensionalization of the mass conservation equation yields
${d \rho \over \rho} + {dU \over U } = {d \rho \over \rho} + {2\, U\, dU \over 2 U^{2} } = {d \rho \over \rho} + {d (U^{2}) \over 2\; U^{2}} = 0 \label{isothermal:eq:massD} \tag{15}$

Differentiation of the isotropic (stagnation) relationship of the pressure (??) yields

$\dfrac{dP_0 }{ P_0} = \dfrac{dP }{ P } + \left( \dfrac{ \dfrac { k\, M^{2} }{2}} { 1 + \dfrac{ k-1 }{ 2 } M^{2} } \right) \; {d M^{2} \over M^{2} } \label{isothermal:eq:PzeroD} \tag{16}$
Differentiation of equation (??) yields:
$dT_{0} = dT \left( 1 + {k - 1 \over 2} M^{2} \right) + T \; {k - 1 \over 2} \; d M^{2} \label{isothermal:eq:dT0T} \tag{17}$

Notice that $$dT_{0} \neq 0$$ in an isothermal flow. There is no change in the actual temperature of the flow but the stagnation temperature increases or decreases depending on the Mach number (supersonic flow of subsonic flow). Substituting $$T$$ for equation (17) yields:

$dT_{0} = \dfrac{ T_{0} \; \dfrac{k - 1 }{ 2} \; d\,M^{2} }{ \left( 1 + \dfrac{k - 1 }{ 2} M^{2} \right) } \; {M^{2} \over M^{2} } \label{isothermal:eq:TOTtmp} \tag{18}$
Rearranging equation (18) yields
${ dT_{0} \over T_{0} } = \dfrac{ {(k - 1 )} \; M^{2} }{ 2 \left( 1 + \dfrac{k - 1 }{ 2} \right) } \; {dM^{2} \over M^{2} } \label{isothermal:eq:logT0T} \tag{19}$
By utilizing the momentum equation it is possible to obtain a relation between the pressure and density. Recalling that an isothermal flow ($$dT=0$$) and combining it with perfect gas model yields
${dP \over P } = { d\rho \over \rho} \label{isothermal:eq:stateD} \tag{20}$
From the continuity equation (see equation (14)) leads
${d M^{2} \over M^{2} } = { 2 dU \over U } \label{isothermal:eq:massAnother} \tag{21}$

The four equations momentum, continuity (mass), energy, state are described above. There are 4 unknowns ($$M, T, P, \rho$$) and with these four equations the solution is attainable. One can notice that there are two possible solutions (because of the square power). These different solutions are supersonic and subsonic solution. The distance friction, $$t\dfrac{4\,f\,L}{D}$$, is selected as the choice for the independent variable. Thus, the equations need to be obtained as a function of $$\dfrac{4f\,L}{D}$$. The density is eliminated from equation (15) when combined with equation (20) to become

$\dfrac{dP }{ P } = - \dfrac{dU }{ U} \label{isothermal:eq:stateMassD} \tag{22}$
After substituting the velocity (22) into equation (10), one can obtain
$- \dfrac{dP }{ P} - \dfrac{4\,f\, dx }{ D_H} \left( \dfrac{ k\,P\, M^{2} }{ 2} \right) = {k\,P\, M^{2} }\, \dfrac{dP }{ P } \label{isothermal:eq:mm} \tag{23}$
Equation (23) can be rearranged into
$\dfrac{dP }{ P } = \dfrac{d\rho }{ \rho} = - \dfrac{dU }{ U} = - \dfrac{1 }{ 2} \dfrac{dM^{2} }{ M^{2} } = - \dfrac{ k \,M^{2} }{ 2\left( 1 - k\,M^{2} \right) } 4\,f \,\dfrac{dx }{ D} \label{isothermal:eq:pressureGov} \tag{24}$

Similarly or by other paths, the stagnation pressure can be expressed as a function of $$\dfrac{4\,f\,L}{D}$$

$\dfrac{dP_{0} }{ P_{0} } = ParseError: EOF expected (click for details) Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.2:_Dimensionless_Representation), /content/body/p[14]/span[1], line 1, column 3  4f {dx \over D} \label{isothermal:eq:stangnationPressureGov} \tag{25}$
${dT_{0} \over T_{0} } = ParseError: EOF expected (click for details) Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.2:_Dimensionless_Representation), /content/body/p[14]/span[2], line 1, column 4  \, 4\,f {dx \over D} \label{isothermal:eq:stangnationTempGov} \tag{26}$
The variables in equation (24) can be separated to obtain integrable form as follows
$\int_{0}^{L} { 4\,f\, dx \over D} = \int_{M^{2}}^{1/k} { 1 - k\,M{2} \over k\,M{2}} dM^{2} \label{isothermal:eq:integralMach} \tag{27}$

It can be noticed that at the entrance $$(x = 0)$$ for which $$M = M_{x=0}$$ (the initial velocity in the tube isn't zero). The term $$\dfrac{4f\,L}{D}$$ is positive for any $$x$$, thus, the term on the other side has to be positive as well. To obtain this restriction $$1= k\,M^{2}$$. Thus, the value $$M= {1\over \sqrt{k}}$$ is the limiting case from a mathematical point of view. When Mach number larger than $$M > {1\over \sqrt{k}}$$ it makes the right hand side of the integrate negative. The physical meaning of this value is similar to $$M=1$$ choked flow which was discussed in a variable area flow in Section 11.4. Further it can be noticed from equation (26) that when $$M \rightarrow {1\over \sqrt{k}}$$ the value of right hand side approaches infinity ($$\infty$$). Since the stagnation temperature ($$T_{0}$$) has a finite value which means that $$dT_{0} \rightarrow \infty$$. Heat transfer has a limited value therefore the model of the flow must be changed. A more appropriate model is an adiabatic flow model yet this model can serve as a bounding boundary (or limit).  Integration of equation (27) requires information about the relationship between the length, $$x$$, and friction factor $$f$$. The friction is a function of the Reynolds number along the tube. Knowing the Reynolds number variations is important. The Reynolds number is defined as

$Re = \dfrac{D\, U\, \rho}{\mu} \label{isothermal:eq:Re} \tag{28}$

The quantity $$U\,\rho$$ is constant along the tube (mass conservation) under constant area. Thus, only viscosity is varied along the tube. However under the assumption of ideal gas, viscosity is only a function of the temperature. The temperature in isothermal process (the definition) is constant and thus the viscosity is constant. In real gas, the pressure effects are very minimal as described in "Basic of fluid mechanics'' by this author. Thus, the friction factor can be integrated to yield

Friction Mach Isothermal Flow

$\label{isothermal:eq:workingEq} {\left.\dfrac{4\,f\,L}{D}\right|_{max}} = { 1- k\, M^{2} \over k\, M^{2} } + \ln \left(k\,M^{2} \right) \tag{29}$

The definition for perfect gas yields $$M^{2} = { U^{2} / k\,R\,T}$$ and noticing that $$T=constant$$ is used to describe the relation of the properties at $$M = 1 / \sqrt{k}$$. By denoting the superscript symbol $$*$$ for the choking condition, one can obtain that

${ M^{2} \over U^{2} } = ParseError: EOF expected (click for details) Callstack: at (Textbook_Maps/Map:_Fluid_Mechanics_(Bar-Meir)/11:_Compressible_Flow_One_Dimensional/11.70_Isothermal_Flow/11.7.2:_Dimensionless_Representation), /content/body/p[19]/span, line 1, column 5  \label{isothermal:eq:MachDef} \tag{30}$
Rearranging equation (30) is transformed into
${U \over U^{*}} = \sqrt{k} M \label{isothermal:eq:Ubar} \tag{31}$
Utilizing the continuity equation provides
$\rho U = \rho^{*} U^{*} \quad \Longrightarrow \quad \dfrac{\rho }{ \rho^{\star}} = \dfrac{ 1 }{ \sqrt{k} \, M} \tag{32}$
Reusing the perfect–gas relationship

Pressure Ratio

$\label{isothermal:eq:Pbar} \dfrac{P }{ P^{*}} = {\rho \over \rho^{*}} = \dfrac{ 1 }{ \sqrt{k} \, M} \tag{33}$

Utilizing the relation for stagnated isotropic pressure one can obtain
${P_{0} \over P_{0}^{*}} = {P \over P^{*}} \left[ {1 + { k -1 \over 2 } M ^ {2} \over { 1 + {k -1 \over 2k} } } \right] ^ { k \over k -1 } \tag{34}$
Substituting for $${P \over P^{*}}$$ equation (33) and rearranging yields

Stagnation Pressure Ratio

$\label{isothermal:eq:P0bar} \dfrac{P_{0} }{ P_{0}^{*}} = {1 \over \sqrt{k}} \left( {2\,k \over 3\,k- 1} \right)^ {\dfrac{k }{ k -1} } \left( 1 + {k -1 \over 2} M ^{2}\right)^{\dfrac{k }{ k-1} } \dfrac{ 1 }{ M} \tag{35}$

And the stagnation temperature at the critical point can be expressed as

Stagnation Pressure Ratio

$\label{isothermal:eq:T0bar} {T_{0} \over T_{0}^{*}} = { T \over T^{*}} \dfrac{ 1 + \dfrac{k -1 }{ 2} M ^{2} }{ 1 + \dfrac{k -1 }{ 2\,k} } = \dfrac{2\,k }{ 3\,k -1 } \left( 1 + {k -1 \over 2} \right) M ^{2} \tag{26}$

These equations (31), (??) are presented in Figure (??).

Fig. 11.18 Description of the pressure, temperature relationships as a function of the Mach number for isothermal flow.

Contributors

• 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.