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The effects of Real Gases

To obtained expressions for non-ideal gas it is communally done by reusing the ideal gas model and introducing a new variable which is a function of the gas properties like the critical pressure and critical temperature. Thus, a real gas equation can be expressed in equation (3.19). Differentiating equation (3.19) and dividing by equation (3.19) yields

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

(4.116)

Again, Gibb's equation (4.27) is reused to related the entropy change to the change in thermodynamics properties and applied on non-ideal gas. Since

and utilizing the equation of the state $dh = dP/\rho$ . The enthalpy is a function of the temperature and pressure thus,

and full differential is

$\displaystyle dh = \left(\partial h \over \partial T \right)_P dT + \left(\partial h \over \partial P \right)_T dP$

(4.117)

The definition of pressure specific heat is $C_p \equiv {\partial h \over \partial T }$ and second derivative is Maxwell relation hence,

$\displaystyle \left(\partial h \over \partial P \right)_T = v - T \left(\partial s \over \partial T \right)_P$

(4.118)

First, the differential of enthalpy is calculated for real gas equation of state as

$\displaystyle dh = C_p dT -\left( T \over Z \right) \left( \partial z \over \partial T \right)_{P} {dP \over \rho}$

(4.119)

Equations (4.27) and (3.19) are combined to form

$\displaystyle {ds \over R} = {C_p \over R }{dT \over T} - z \left[ 1 + \left( T... ...er Z \right)\left( \partial z \over \partial T \right)_{P} \right] {dP \over P}$

(4.120)

The mechanical energy equation can be expressed as

$\displaystyle \int d \left( U^2 \over 2 \right) = -\int {dP \over \rho}$

(4.121)

At the stagnation the definition requires that the velocity is zero. To carry the integration of the right hand side the relationship between the pressure and the density has to be defined. The following power relationship is assumed

$\displaystyle {\rho \over \rho_0} = \left( P \over P_0 \right)^ {1 \over n}$

(4.122)

Notice, that for perfect gas the n is substituted by

. With integration of equation (4.121) when using relationship which is defined in equation (4.122) results

$\displaystyle {{U } ^ 2 \over 2} = \int _{P_0} ^{P_1} {dP \over \rho} = \int _{P_0} ^{P} { 1 \over \rho_0} \left( P_0 \over P \right)^ {1 \over n} dP$

(4.123)

Substituting relation for stagnation density (3.19) results

$\displaystyle {{U } ^ 2 \over 2} = \int _{P_0} ^{P} {z_0 R T_0 \over P_0} \left( P_0 \over P \right)^ {1 \over n} dP$

(4.124)

For

the integration results in

$\displaystyle U = \sqrt{z_0 R T_0 {2n \over n -1 } \left[1 - \left( P \over P_0 \right)^{\left( n-1 \over n \right)} \right] }$

(4.125)

For

the integration becomes

$\displaystyle U = \sqrt{2 z_0 R T_0 \ln \left( P_0 \over P \right) }$

(4.126)

It must be noted that n is a function of the critical temperature and critical pressure. The mass flow rate is regardless to equation of state as following

$\displaystyle \dot{m} = \rho^{*} A^{*}U^{*}$ (4.127)

Where $\rho^{*}$ is the density at the throat (assuming the chocking condition) and $A^{*}$ is the cross area of the throat. Thus, the mass flow rate in our properties

$\displaystyle \dot{m} = A^{*} \overbrace{{P_0 \over z_0 R T_0 } \left( P \over ... ...[1 - \left( P \over P_0 \right)^{\left( n-1 \over n \right)} \right] }}^{U^{*}}$ (4.128)

For the case of

$\displaystyle \dot{m} = A^{*} \overbrace{{P_0 \over z_0 R T_0 } \left( P \over ... ...rho^{*}} \overbrace{\sqrt{2 z_0 R T_0 \ln \left( P_0 \over P \right)} }^{U*{*}}$ (4.129)

The Mach number can be obtained by utilizing equation (3.34) to defined the Mach number as

$\displaystyle M = {U \over \sqrt{z n R T} }$ (4.130)

Integrating equation (4.120) when results

$\displaystyle \int _{T_1}^{T_2}{C_p \over R }{dT \over T} = \int _{P_1} ^{P_2} ... ...r Z \right)\left( \partial z \over \partial T \right)_{P} {dP \over P } \right)$ (4.131)

To carryout the integration of equation (4.131) looks at Bernnolli's equation which is

$\displaystyle \int {dU^2 \over 2} = - \int {dP \over \rho}$ (4.132)

After integration of the velocity

$\displaystyle {dU^2 \over 2} = - \int^{P/P_0}_{1} {\rho_0 \over \rho} d \left(P \over P_0 \right)$ (4.133)

It was shown in Chapter (3) that (3.33) is applicable for some ranges of relative temperature and pressure (relative to critical temperature and pressure and not the stagnation conditions).

$\displaystyle U = \sqrt{ z_0 R T_0 \left( 2 n \over n -1 \right) \left[ 1 - \left( P \over P_0 \right)^{n-1 \over n} \right] }$ (4.134)

When or when $n \rightarrow 1$

$\displaystyle U = \sqrt{2 z_0 R T_0 \ln \left( P_0 \over P \right) }$ (4.135)

The mass flow rate for the real gas $\dot{m} = \rho^{*} U^{*} A^{*}$

$\displaystyle \dot{m} = {A^{*}P_0 \over \sqrt{z_0 R T_0}} \sqrt{ 2 n \over n -1... ...left( P^{*} \over P_0\right) ^ {1 \over n} \left[ 1 - {P^{*} \over P_0} \right]$ (4.136)

And for

$\displaystyle \dot{m} = {A^{*}P_0 \over \sqrt{z_0 R T_0}} \sqrt{ 2 n \over n -1 } \sqrt{ 2z_0 R T_0 \ln \left( P_0 \over P \right) }$ (4.137)

Fliegner's number in this case is

$\displaystyle Fn = {\dot{m} c_0 \over A^{*} P_0} \sqrt{ 2 n \over n -1 } \left( P^{*} \over P_0\right) ^ {1 \over n} \left[ 1 - {P^{*} \over P_0} \right]$ (4.138)

Fliegner's number for is

$\displaystyle Fn = {\dot{m} c_0 \over A^{*} P_0} = 2 \left( P^{*} \over P_0\right) ^2 - \ln \left( P^{*} \over P_0\right)$ (4.139)

The critical ratio of the pressure is

$\displaystyle { P^{*} \over P_0} = \left(2 \over n+1 \right)^{n \over n-1}$ (4.140)

When or more generally when $n \rightarrow 1$ this is a ratio approach

$\displaystyle { P^{*} \over P_0} = \sqrt{\hbox{e}}$ (4.141)

To obtain the relationship between the temperature and pressure, equation (4.131) can be integrated

$\displaystyle {T_0 \over T} = \left( P_0 \over P \right) ^ { {R \over C_p}\left[ z + T \left( \partial z \over \partial T \right)_{P} \right] }$ (4.142)

The power of the pressure ratio is approaching $k-1 \over k$ when z approaches 1. Note that

$\displaystyle {T_0 \over T} = \left( z_0 \over z \right) \left( P_0 \over P \right) ^ {1-n \over n}$ (4.143)

The Mach number at every point at the nozzle can be expressed as

$\displaystyle M = \sqrt{\left( 2 \over n- 1\right) {z_0 \over z} {T_0 \over T} \left[ 1 - \left( P-0 \over P\right)^{1-n \over n} \right]}$ (4.144)

For the Mach number is

$\displaystyle M= \sqrt{2 {z_0 \over z} {T_0 \over T}\ln {P_0 \over P}}$ (4.145)

The pressure ratio at any point can be expressed as a function of the Mach number as

$\displaystyle {T_0 \over T} = \left[ 1 + { n -1 \over 2} M^2 \right] ^{\left(n-... ... n \right) \left[ z + T \left( \partial z \over \partial T \right)_{P} \right]}$ (4.146)

for

$\displaystyle {T_0 \over T} = \hbox{\LARGE e}^{M^2 \left[ z + T \left( \partial z \over \partial T \right)_{P} \right] }$ (4.147)

The critical temperature is given by

$\displaystyle {T^{*} \over T_0} = \left( 1+ n \over 2 \right)^ {\left( n \over 1 -n \right) \left[ z + T \left( \partial z \over \partial T \right)_{P} \right]}$ (4.148)

and for

$\displaystyle {T^{*} \over T_0} = \sqrt{\hbox{\LARGE e}^ {- \left[ z + T \left( \partial z \over \partial T \right)_{P} \right]}}$ (4.149)

The mass flow rate as a function of the Mach number is

$\displaystyle \dot{m} = {P_0 n \over c_0} M \sqrt{\left(1 + {n - 1 \over 2} M^2\right)^{n+1 \over n-1} }$ (4.150)

For the case of the mass flow rate is

$\displaystyle \dot{m} = {P_0 A^{*} n \over c_0} \sqrt{\hbox{\LARGE e}^{M^2}} \sqrt{\left(1 + {n - 1 \over 2} M^2\right)^{n+1 \over n-1} }$ (4.151)

$\begin{examl} A design is required that at a specific point the Mach number shou... ...ferent from the point? You can assume that $k=1.405$. \end{enumerate}\end{examl}$
Solution

The solution is simplified by using Potto-GDC for the results are

Isentropic Flow Input: M k = 1.4

M T/T0 ρ/ρ0 A/A* P/P0 PAR F/F*

2.61 0.423295 0.116575 2.92339 0.0493458 0.144257 0.633345

The stagnation pressure is obtained from

$\displaystyle P_0 = {P_0 \over P} P = {2.61 \over 0.04943} \sim 52.802 [Bar]$

The stagnation temperature is

$\displaystyle T_0 = {T_0 \over T} T = {300 \over 0.42027} \sim 713.82K$

Of course, the stagnation pressure is constant for isentropic flow.

Next: Normal Shock Up: Isentropic Flow Previous: Isothermal Table Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21