General Relationship

In this section, the other extreme case model where the heat transfer to the gas is perfect, (e.g. Eckert number is very small) is presented. Again in reality the heat transfer is somewhere in between the two extremes. So, knowing the two limits provides a tool to examine where the reality should be expected. The perfect gas model is again assumed (later more complex models can be assumed and constructed in a future versions). In isothermal process the perfect gas model reads

$$P =\rho R T \leadsto dP = d\rho R T$$ (4.76)

Substituting equation (4.76) into the momentum equation^4.6 yields

$$UdU +{ RT dP \over P} = 0$$ (4.77)

Integration of equation (4.77) yields the Bernoulli's equation for ideal gas in isothermal process which reads

$$\leadsto {{U_2}^2 - {U_1}^2 \over 2} +{ RT \ln {P_2 \over P_1} } = 0$$ (4.78)

Thus, the velocity at point 2 becomes

$${U_2} = \sqrt{2 RT \ln {P_2 \over P_1} - {U_1}^2 }$$ (4.79)

The velocity at point 2 for stagnation point, $U_1\approx 0$ reads

$${U_2} = \sqrt{2 RT \ln {P_2 \over P_1} }$$ (4.80)

Or in explicit terms of the stagnation properties the velocity is

$$U = \sqrt{2 RT \ln {P \over P_0} }$$ (4.81)

Transform from equation (4.78) to a dimensionless form becomes

$$\leadsto {kR\cancelto{constant}{T}({M_2}^2 - {M_1}^2) \over 2} = { R\cancelto{\rule[-0.2cm]{0pt}{0.2cm}constant}{T} \ln {P_2 \over P_1}}$$ (4.82)

Simplifying equation (4.82) yields

$$\leadsto {k({M_2}^2 - {M_1}^2) \over 2} = { \ln {P_2 \over P_1} }$$ (4.83)

Or in terms of the pressure ratio equation (4.83) reads

$${P_2 \over P_1} = {\Huge e}^{k({M_1}^2 - {M_2}^2) \over 2} = \left( {\Huge e}^{{M_1}^2} \over {\Huge e}^{{M_2}^2}\right) ^{k\over 2}$$ (4.84)

As oppose to the adiabatic case ( $T_0=constant$ ) in the isothermal flow the stagnation temperature ratio can be expressed

$${{T_0}_1 \over {T_0}_2} = \cancelto{1}{T_1 \over T_2} {\left(1+{k... ... {\left(1+{k -1\over2} {M_1}^2\right)\over \left(1+{k -1\over2}{M_2}^2\right) }$$ (4.85)

Utilizing conservation of the mass $A\rho M = constant$ to yield

$${A_1 \over A_2} = {M_2 P_2 \over M_1 P_1}$$ (4.86)

Combining equation (4.86) and equation (4.84) yields

$${A_2 \over A_1} = { M_1 \over M_2} \left( \mbox{\large e}^{{M_2}^{2}} \over \mbox{\large e}^{{M_1}^{2}} \right)^{k \over 2}$$ (4.87)

The change in the stagnation pressure can be expressed as

$${{P_0}_2 \over {P_0}_1} = { P_2 \over P_1} \left( 1 + {k -1 \over... ...ox{\large e}^{{M_1}^{2}} \over \mbox{\large e} ^{{M_1}^{2}} \right]^{k \over 2}$$ (4.88)

The critical point, at this stage, is unknown (at what Mach number the nozzle is choked is unknown) so there are two possibilities: the choking point or M=1 to normalize the equation. Here the critical point defined as the point where M=1 so results can be compared to the adiabatic case and denoted by star. Again it has to emphasis that this critical point is not really related to physical critical point but it is arbitrary definition. The true critical point is when flow is choked and the relationship between two will be presented.

The critical pressure ratio can be obtained from (4.84) to read

$${ P \over P^{*}} = { \rho \over \rho^{*}} = ^{(1-M^2) k \over 2}$$ (4.89)

Equation (4.87) is reduced to obtained the critical area ratio writes

$${ A \over A^{*}} = {1 \over M} ^{(1-M^2) k \over 2}$$ (4.90)

Similarly the stagnation temperature reads

$${ T_0 \over {T_0}^{*}} = {2 \left( 1 + {k -1 \over 2}{M_1}^2 \right) \over k +1 } ^ {k \over k-1}$$ (4.91)

Finally, the critical stagnation pressure reads

$${ P_0 \over {P_0}^{*}} = ^{(1-M^2)k \over 2} \left(2 \left( 1 + {k -1 \over 2}{M_1}^2 \right) \over k +1 \right) ^ {k \over k-1}$$ (4.92)

The maximum value of stagnation pressure ratio is obtained when $M=0$ at which is

$$\left. P_0 \over {P_0}^{*} \right\vert _{M=0} = ^{k \over 2} \left( 2 \over k +1 \right) ^ {k \over k-1}$$ (4.93)

For specific heat ratio of $k=1.4$ , this maximum value is about two. It can be noted that the stagnation pressure is monotonically reduced during this process.

Of course in isothermal process $T = T^{*}$ . All these equations are plotted in Figure (4.6).

Figure 4.6: Various ratios as a function of Mach number for isothermal Nozzle

From the Figure 4.3 it can be observed that minimum of the curve $A/A^{*}$ isn't on M=1 . The minimum of the curve is when area is minimum and at the point where the flow is choked. It should be noted that the stagnation temperature is not constant as in the adiabatic case and the critical point is the only one constant.

The mathematical procedure to find the minimum is simply taking the derivative and equating to zero as following

$${d \left(A \over A^{*} \right) \over dM} = { kM^{2} {\Huge e}^{k(M^2-1) \over 2 } - {\Huge e}^{k(M^2-1) \over 2 } \over M^2} = 0$$ (4.94)

Equation (4.94) simplified to

$$kM^2 - 1 = 0 \leadsto M = {1 \over \sqrt{k}}$$ (4.95)

It can be noticed that a similar results are obtained for adiabatic flow. The velocity at the throat of isothermal model is smaller by a factor of $\sqrt{k}$ . Thus, dividing the critical adiabatic velocity by $\sqrt{k}$ results in

$$U_{{throat}_{max}} = \sqrt{RT}$$ (4.96)

On the other hand, the pressure loss in adiabatic flow is milder as can be seen in Figure (4.7(a)).

Figure: Comparison between the isothermal nozzle and
adiabatic nozzle in various variables

Figure: The comparison of the adiabatic model and isothermal model

It should be emphasized that the stagnation pressure decrees. It is convenient to find expression for the ratio of the initial stagnation pressure (the stagnation pressure before entering the nozzle) to the pressure at the throat. Utilizing equation (4.89) the following relationship can be obtained

$${P_{throat} \over P_{0_{initial}}} = {P^{*} \over P_{0_{initial}}} {P_{throat} \over P^{*}} =$$

$${1 \over \mbox{\large e}^{(1 - 0^2)k \over 2} } \mbox{\large e} ^{ \left(1 - \left(1 \over \sqrt{k}\right)^2 \right) {k \over 2}} =$$

$$^{- \half} = 0.60653$$ (4.97)

Notice that the critical pressure is independent of the specific heat ratio, $k$ , as opposed to the adiabatic case. It also has to be emphasized that the stagnation values of the isothermal model are not constant. Again, the heat transfer is expressed as

$$Q = C_p \left( {T_{0_2}} - {T_{0_2} } \right)$$ (4.98)

Figure: Comparison of the pressure and temperature drop as a function of the normalized length (two scales)

For comparison between the adiabatic model and the isothermal a simple profile of nozzle area as a function of the distance is assumed. This profile isn't an ideal profile but rather a simple sample just to examine the difference between the two models so in an actual situation it can be bounded. To make sense and eliminate unnecessary details the distance from the entrance to the throat is normalized (to one (1)). In the same fashion the distance from the throat to the exit is normalized (to one (1)) (it doesn't mean that these distances are the same). In this comparison the entrance area ratio and the exit area ratio are the same and equal to 20. The Mach number was computed for the two models and plotted in Figure (4.7(b)). In this comparison it has to be remembered that critical area for the two models are different by about 3% (for $k=1.4$ ). As can be observed from Figure (4.7(b)). The Mach number for the isentropic is larger for the supersonic branch but the velocity is lower. The ratio of the velocities can be expressed as

$${U_s \over U_T} = { M_s\sqrt{kRT_s} \over M_T \sqrt{kRT_s}}$$ (4.99)

It can be noticed that temperature in the isothermal model is constant while temperature in the adiabatic model can be expressed as a function of the stagnation temperature. The initial stagnation temperatures are almost the same and can be canceled out to obtain

$${U_s \over U_T} \sim { M_s \over M_T \sqrt{1 + {k-1 \over 2} {M_s}^2} }$$ (4.100)

By utilizing equation (4.100) the velocity ratio was obtained and is plotted in Figure (4.7(b)).

Thus, using the isentropic model results in under prediction of the actual results for the velocity in the supersonic branch. While, the isentropic for the subsonic branch will be over prediction. The prediction of the Mach number are similarly shown in Figure (4.7(b)).

Two other ratios need to be examined: temperature and pressure. The initial stagnation temperature is denoted as ${T_{0}}_{int}$ . The temperature ratio of $T/{T_{0}}_{int}$ can be obtained via the isentropic model as

$${T \over{T_{0}}_{int}} = { 1 \over 1 + {k-1 \over 2} M^2}$$ (4.101)

While the temperature ratio of the isothermal model is constant and equal to one (1). The pressure ratio for the isentropic model is

$${P \over{P_{0}}_{int}} = { 1 \over \left( 1 + {k-1 \over 2} M^2 \right) ^{k-1 \over k} }$$ (4.102)

and for the isothermal process the stagnation pressure varies and has to be taken into account as the following:

$${P_z \over{P_{0}}_{int}} = {{P_0}^{*} \over{P_{0}}_{int}} {{P_{0}}_{z} \over {P_{0}}^{*} } \overbrace{P_z \over{P_{0}}_{z}}^{isentropic}$$ (4.103)

where $z$ is an arbitrary point on the nozzle. Using equations (4.88) and the isentropic relationship, the sought ratio is provided.

Figure (4.8) shows that the range between the predicted temperatures of the two models is very large, while the range between the predicted pressure by the two models is relatively small. The meaning of this analysis is that transferred heat affects the temperature to a larger degree but the effect on the pressure is much less significant.

To demonstrate the relativity of the approach advocated in this book consider the following example.

Solution

The first part of the question deals with the adiabatic model i.e. the conservation of the stagnation properties. Thus, with known area ratio and known stagnation the GDC-Potto provides the following table:

M	T / T0	ρ/ ρ0	$\mathbf{A \over A^{\star}}$	P / P0	$\mathbf{A\times P \over A^{*} \times P_0}$
0.14655	0.99572	0.98934	4.0000	0.98511	3.9405
2.9402	0.36644	0.08129	4.0000	0.02979	0.11915

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With known Mach number, and temperature at the exit the velocity can be calculated. The exit temperature is $0.36644\times 300 = 109.9K$ . The exit velocity, then, is

$$U= M\sqrt{kRT} = 2.9402 \sqrt{1.4\times 287 \times 109.9} \sim 617.93 [m/sec]$$

Even for the isothermal model, the initial stagnation temperature is given as 300K. With the area ratio by using the Figure 4.6 or using the Potto-GDC obtains the following table is obtained

Isothermal Nozzle Flow		Input: A/A*			k = 1.4
M	T0/T0*	P0/P0*	A/A*	P/P*	PAR	F/F*
1.99103	1.4940	0.511829	4.	0.125563	0.502252	1.37071

The exit Mach number is known and the initial temperature to the throat temperature ratio can be calculated as following:

$${T_{0_{ini}} \over {T_{0}}^{*}} = {1 \over 1 + {k -1\over 2} {1 \over k} } = { 1 \over 1 + {k-1\over k}} = 0.777777778$$

Thus the stagnation temperature at the exit is

$${T_{0_{ini}} \over T_{0_{exit}}} = 1.4940 / 0.777777778 = 1.921$$

The exit stagnation temperature is $1.92\times 300 = 576.2 K$ . The exit velocity determined by utilizing the following equation

$$U_{exit} = M\sqrt{kRT} = 1.9910 \sqrt{1.4\times287\times 300.0} = 691.253 [m/sec]$$

As it was discussed before the velocity in copper nozzle will be larger than velocity in the wood nozzle. However, the maximum velocity can not exceed the $691.253 [m/sec]$