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

$${dP \over P } = { d\rho \over \rho}$$ (8.6)

It is convenient to define a hydraulic diameter

$$D_{H} = {4 \times \hbox{Cross Section Area} \over \hbox{ wetted perimeter }}$$ (8.7)

Now, the Fanning friction factor^8.2 is introduced, this factor is a dimensionless friction factor sometimes referred to as the friction coefficient as

$$f = { \tau_w \over \half \rho U^{2} }$$ (8.8)

Substituting equation (8.8) into momentum equation (8.2) yields

$$- dP - {4 dx \over D_H} f \left( \half \rho U^{2} \right) = { \overbrace{\rho U}^{\dot {m} \over A } dU }$$ (8.9)

Rearranging equation (8.9) and using the identify for perfect gas $M^{2} = \rho U^{2} / k P$ yields:

$$- {dP \over P} - {4f dx \over D_H} \left( { kP M^{2} \over 2} \right) = {kP M^{2} dU \over U}$$ (8.10)

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

$$U^{2} = {kRT} M^{2}$$ (8.11)

Differentiation of equation (8.11) yields

$$d(U^{2}) = kR\left(M^{2}\; dT + T \; d (M^{2}) \; \right)$$ (8.12)

$${d(M^2) \over M^{2}} = {d(U^2) \over U^{2}} - { dT \over T}$$ (8.13)

Now it can be noticed that $dT = 0$ for isothermal process and therefore

$${d(M^2) \over M^{2}} = {d(U^2) \over U^{2}} = {2U \; dU \over U^{2}} = {2 dU \over U}$$ (8.14)

The dimensionalization of the mass conservation equation yields

$${d \rho \over \rho} + {dU \over U } = {d \rho \over \rho} + {2 U dU \over 2 U^{2} } = {d \rho \over \rho} + {d (U^{2}) \over 2\; U^{2}} = 0$$ (8.15)

Differentiation of the isotropic (stagnation) relationship of the pressure (4.11) yields

$${dP_0 \over P_0} = {dP \over P } + \left( { \half k M^{2} \over 1 + { k-1 \over 2 } M^{2} } \right) \; {d M^{2} \over M^{2} }$$ (8.16)

Differentiation of equation (4.9) yields:

$$dT_{0} = dT \left( 1 + {k - 1 \over 2} M^{2} \right) + T \; {k - 1 \over 2} \; d M^{2}$$ (8.17)

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 (8.17) yields:

$$dT_{0} = { T_{0} \; {k - 1 \over 2} \; d\;M^{2} \over \left( 1 + {k - 1 \over 2} M^{2} \right) } \; {M^{2} \over M^{2} }$$ (8.18)

Rearranging equation (8.18) yields

$${ dT_{0} \over T_{0} } = { {(k - 1 )} \; M^{2} \over 2 \left( 1 + {k - 1 \over 2} \right) } \; {dM^{2} \over M^{2} }$$ (8.19)

By utilizing the momentum equation it is possible to obtain a relation between the pressure and density. Recalling that an isothermal flow ($T=0$ ) and combining it with perfect gas model yields

$${dP \over P } = { d\rho \over \rho}$$ (8.20)

From the continuity equation (see equation (8.14)) leads

$${d M^{2} \over M^{2} } = { 2 dU \over U }$$ (8.21)

The four equations momentum, continuity (mass), energy, state are described above. There are 4 unknowns ( $M, T, P, \rho$ )^8.3 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, $\frac{4fL}{D}$ , is selected as the choice for the independent variable. Thus, the equations need to be obtained as a function of $\frac{4fL}{D}$ . The density is eliminated from equation (8.15) when combined with equation (8.20) to become

$${dP \over P } = - {dU \over U}$$ (8.22)

After substituting the velocity (8.22) into equation (8.10), one can obtain

$$- {dP \over P} - {4f dx \over D_H} \left( { kP M^{2} \over 2} \right) = {kP M^{2} {dP \over P }}$$ (8.23)

Equation (8.23) can be rearranged into

$${dP \over P } = {d\rho \over \rho} = - {dU \over U} = - {1 \over ... ... \over M^{2} } = - { k M^{2} \over 2\left( 1 - kM^{2} \right) } 4f {dx \over D}$$ (8.24)

Similarly or by other path the stagnation pressure can be expressed as a function of $\frac{4fL}{D}$

$${dP_{0} \over P_{0} } = {{k M^ {2} \left( 1 - {k + 1 \over 2}M^{2... ...( kM^{2} - 1 \right) \left( 1 + {k -1 \over 2} M^{2} \right) }} 4f {dx \over D}$$ (8.25)

$${dT_{0} \over T_{0} } = {{k \left( 1 - k \right) M^{2} } \over {2... ...t(1 - kM^{2} \right) \left( 1 + {k -1 \over 2} M^{2} \right) }} 4f {dx \over D}$$ (8.26)

The variables in equation (8.24) can be separated to obtain integrable form as follows

$$\int_{0}^{L} { 4 f dx \over D} = \int_{M^{2}}^{1/k} { 1 - kM{2} \over kM{2}} dM^{2}$$ (8.27)

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 $\frac{4fL}{D}$ is positive for any $x$ , thus, the term on the other side has to be positive as well. To obtain this restriction $1= kM^{2}$ . 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 Chapter (4).

Further it can be noticed from equation (8.26) 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 it can serve as a bounding boundary (or limit).

Integration of equation (8.27) yields

$${\left.\frac{4fL}{D}\right\vert _{max}} = { 1- k M^{2} \over k M^{2} } + \ln kM^{2}$$ (8.28)

The definition for perfect gas yields $M^{2} = { U^{2} / kRT}$ 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

$${ M^{2} \over U^{2} } = {{ 1/k} \over {U^{*}}^ {2}}$$ (8.29)

Rearranging equation (8.29) is transfered into

$${U \over U^{*}} = \sqrt{k} M$$ (8.30)

Utilizing the continuity equation provides

$$\rho U = \rho^{*} U^{*} ; \Longrightarrow {\rho \over \rho^{*}} = { 1 \over \sqrt{k} M} %\label{eq:}$$ (8.31)

Reusing the perfect-gas relationship

$${P \over P^{*}} = {\rho \over \rho^{*}} = { 1 \over \sqrt{k} M}$$ (8.32)

Now utilizing the relation for stagnated isotropic pressure one can obtain

$${P_{0} \over P_{0}^{*}} = {P \over P^{*}} \left[ {1 + { k -1 \ove... ... ^ {2} \over { 1 + {k -1 \over 2k} } } \right] ^ { k \over k -1 } %\label{eq:}$$ (8.33)

Substituting for ${P \over P^{*}}$ equation (8.32) and rearranging yields

$${P_{0} \over P_{0}^{*}} = {1 \over \sqrt{k}} \left( {2k \over 3k-... ...left( 1 + {k -1 \over 2} M ^{2}\right)^{k \over k-1} { 1 \over M} %\label{eq:}$$ (8.34)

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

$${T_{0} \over T_{0}^{*}} = { T \over T^{*}} { 1 + {k -1 \over 2} M... ... {k -1 \over 2k} } = {2k \over 3k -1 } \left( 1 + {k -1 \over 2} \right) M ^{2}$$ (8.35)

These equations (8.30)-(8.35) are presented on in Figure (8.2)

Figure: Description of the pressure, temperature relationships as a function of the Mach number for isothermal flow