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Non-Dimensionalization of the Equations

Before solving the above equation a dimensionless process is applied. By utilizing the definition of the sound speed to produce the following identities for perfect gas

$\displaystyle M^{2} = \left( U \over c \right)^{2}$

$\displaystyle = {U^{2} \over k \underbrace{RT}_{P \over \rho}}$

(9.13)

Utilizing the definition of the perfect gas results in

$\displaystyle M^{2} = {\rho U^{2} \over kP}$

(9.14)

Using the identity in equation (9.14) and substituting it into equation (9.11) and after some rearrangement yields

$\displaystyle - dP + { 4 f dx \over D_H} \left( \half kP M^{2} \right) = {\rho U^{2} \over U} dU = \overbrace{kPM^2}^{\rho U^{2}} {dU \over U}$

(9.15)

By further rearranging equation (9.16) results in

$\displaystyle -{ dP \over P} - { 4 f dx \over D} \left( kM^2 \over 2\right) = kM^2 {dU \over U}$

(9.16)

It is convenient to relate expressions of (

) and

in terms of the Mach number and substituting it into equation (9.17). Derivative of mass conservation ((9.2)) results in

$\displaystyle {d\rho \over \rho} + \overbrace{\half {dU^2 \over U^2}}^{dU \over U} = 0$

(9.17)

The derivation of the equation of state (9.5) and dividing the results by equation of state (9.5) results

$\displaystyle {dP \over P} = { d \rho \over \rho} + {dT \over dT}$

(9.18)

Derivation of the Mach identity equation (9.14) and dividing by equation (9.14) yields

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

(9.19)

Dividing the energy equation (9.4) by

and by utilizing the definition Mach number yields

$\displaystyle {dT \over T} + {1 \over \underbrace{\left( kR \over (k -1) \right)}_{C_p}} { 1 \over T} {U^2 \over U ^2} d \left( U^2 \over 2 \right) =$
$\displaystyle \hookrightarrow {dT \over T} + {( k -1 ) \over \underbrace{kRT}_{c^2}} {U^2 \over U ^2} d \left( U^2 \over 2 \right) =$
$\displaystyle \hookrightarrow {dT \over T} + {k -1 \over 2} M^2 {dU^2 \over U^2} = 0$	(9.20)

Equations (9.17), (9.18), (9.19), (9.20), and (9.21) need to be solved. These equations are separable so one variable is a function of only single variable (the chosen as the independent variable). Explicit explanation is provided for only two variables, the rest variables can be done in a similar fashion. The dimensionless friction, $\frac{4fL}{D}$ , is chosen as the independent variable since the change in the dimensionless resistance, $\frac{4fL}{D}$ , causes the change in the other variables.

Combining equations (9.19) and (9.21) when eliminating results

$\displaystyle {dP \over P} = {d\rho \over \rho} - {(k -1 ) M^{2} \over 2} {dU^{2} \over U^{2}}$

(9.21)

The term ${d\rho \over \rho}$ can be eliminated by utilizing equation (9.18) and substituting it into equation (9.22) and rearrangement yields

$\displaystyle {dP \over P} = -{1 + (k -1 ) M^{2} \over 2} {dU^{2} \over U^{2}}$

(9.22)

The term

can be eliminated by using (9.23)

$\displaystyle {dP \over P} = - {kM^{2} \left( 1 + (k-1) M^{2} \right) \over 2 (1 - M^{2} ) } {4fdx \over D}$

(9.23)

The second equation for Mach number,

variable is obtained by combining equation (9.20) and (9.21) by eliminating

. Then $d\rho / \rho$ and

are eliminated by utilizing equation (9.18) and equation (9.22). The only variable that is left is

(or

) which can be eliminated by utilizing equation (9.24) and results in

$\displaystyle { 4 f dx \over D} = {{\left( 1 - M^2 \right) dM^2} \over {kM^4 ( 1 + {k-1 \over 2}M^2} ) }$

(9.24)

Rearranging equation (9.25) results in

$\displaystyle {dM^{2} \over M^2} = {kM^{2} \left( 1 + {k -1 \over 2}M^{2} \right) \over 1 -M^{2} } {4fdx \over D}$

(9.25)

After similar mathematical manipulation one can get the relationship for the velocity to read

$\displaystyle {dU \over U } = { kM^{2} \over 2 \left( 1 - M^{2} \right) } {4 f dx \over D}$

(9.26)

and the relationship for the temperature is

$\displaystyle {dT \over T} = \half { dc \over c} = - {k (k -1) M^{4} \over 2 (1 - M^{2} ) } {4 f dx \over D}$

(9.27)

density is obtained by utilizing equations (9.27) and (9.18) to obtain

$\displaystyle {d\rho \over \rho} = - { kM^{2} \over 2 \left(1 -M^{2} \right) } {4 f dx \over D}$

(9.28)

The stagnation pressure is similarly obtained as

$\displaystyle {dP_{0} \over P_{0}} = - { kM^{2} \over 2 } {4 f dx \over D}$

(9.29)

The second law reads

$\displaystyle ds = C_p \ln {dT \over T} - R \ln {dP \over P}$

(9.30)

The stagnation temperature expresses as

. Taking derivative of this expression when

remains constant yields

and thus when these equations are divided they yield

$\displaystyle dT /T = dT_0 / T_0$

(9.31)

In similar fashion the relationship between the stagnation pressure and the pressure can be substituted into the entropy equation and result in

$\displaystyle ds = C_p \ln { dT_0 \over T_0} - R \ln {dP_0 \over P_0}$

(9.32)

The first law requires that the stagnation temperature remains constant,

. Therefore the entropy change is

$\displaystyle {ds \over C_p} = - { (k - 1) \over k} { dP_0 \over P_0}$

(9.33)

Using the equation for stagnation pressure the entropy equation yields

$\displaystyle {ds \over C_p} = {(k-1) M^2 \over 2} {4fdx \over D}$

(9.34)

Next: The Mechanics and Why Up: Fanno Flow Previous: Model Index

Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21