Compressible Flow Free Textbook

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Impulse in Isentropic Adiabatic Nozzle

One of the functions that is used in calculating the forces is the Impulse function. The Impulse function is denoted here as , but in the literature some denote this function as . To explain the motivation for using this definition consider the calculation of the net forces that acting on section shown in Figure (4.9). To calculate the net forces acting in the x-direction the momentum equation has to be applied

$\displaystyle F_{net}= \dot{m} (U_2 -U_1) + P_2 A_2 - P_1 A_1$

(4.104)

The net force is denoted here as $F_{net}$ . The mass conservation also can be applied to our control volume

$\displaystyle \dot{m} = \rho_1A_1U_1 = \rho_2A_2U_2$ (4.105)

Combining equation (4.104) with equation (4.105) and by utilizing the identity in equation (4.42) results in

$\displaystyle F_{net}= kP_2A_2{M_2}^2 - kP_1A_1{M_1}^2 - P_2 A_2 - P_1 A_1$ (4.106)

Rearranging equation (4.106) and dividing it by $P_0 A^{*}$ results in

$\displaystyle {F_{net} \over P_0 A^{*}} = \overbrace{P_2A_2 \over P_0 A^{*}}^{f... ...1A_1 \over P_0 A^{*}}^{f(M_1)} \overbrace{\left( 1 + k{M_1}^2 \right)}^{f(M_1)}$ (4.107)

Figure 4.9: Schematic to explain the significances of the Impulse function

Examining equation (4.107) shows that the right hand side is only a function of Mach number and specific heat ratio, . Hence, if the right hand side is only a function of the Mach number and than the left hand side must be function of only the same parameters, and . Defining a function that depends only on the Mach number creates the convenience for calculating the net forces acting on any device. Thus, defining the Impulse function as

$\displaystyle F = PA\left( 1 + k{M_2}^2 \right)$ (4.108)

In the Impulse function when ( M=1 ) is denoted as $F^{*}$

$\displaystyle F^{*} = P^{*}A^{*}\left( 1 + k \right)$ (4.109)

The ratio of the Impulse function is defined as

$\displaystyle {F \over F^{*}} = {P_1A_1 \over P^{*}A^{*}} {\left( 1 + k{M_1}^2 ... ...e function \eqref{variableArea:eq:beforeDefa}}} {1 \over \left( 1 + k \right) }$ (4.110)

This ratio is different only in a coefficient from the ratio defined in equation (4.107) which makes the ratio a function of and the Mach number. Hence, the net force is

$\displaystyle F_{net} = P_0 A^{*} (1+k) {\left( k+1 \over 2 \right)^{k \over k-1}} \left( {F_2 \over F^{*} } - { F_1 \over F^{*}}\right)$ (4.111)

To demonstrate the usefulness of the this function consider a simple situation of the flow through a converging nozzle

Figure: Schematic of a flow of a compressible substance (gas) thorough a converging nozzle for example (4.7)
$\begin{figure}\centerline{\includegraphics{cont/variableArea/exampleNozzle}} \end{figure}$

$\begin{examl} Consider a flow of gas into a converging nozzle with a mass flow r... ...alculate the net force acting on the nozzle and pressure at point 1. \end{examl}$
Solution

The solution is obtained by getting the data for the Mach number. To obtained the Mach number, the ratio of $P_1A_1/A^{*}P_0$ is needed to be calculated. To obtain this ratio the denominator is needed to be obtained. Utilizing Fliegner's equation (4.51), provides the following

$\displaystyle A^{*} P_0 = {\dot{m} \sqrt{RT} \over 0.058} = {1.0 \times \sqrt{400 \times 287} \over 0.058} \sim 70061.76 [N]$

and

$\displaystyle {A_2 P_2 \over A^{*} P_0} = { 500000 \times 0.003 \over 70061.76 } \sim 2.1$

Isentropic Flow Input: PAR k = 1.4

M T/T0 ρ/ρ0 A/A* P/P0 PAR F/F*

0.273534 0.985256 0.963548 2.21206 0.949342 2.1 0.966656

With the area ratio of ${A \over A^{\star}}= 2.2121$ the area ratio of at point 1 can be calculated.

$\displaystyle { A_1 \over A^{\star}} = {A_2 \over A^{\star}} {A_1 \over A_2} = 2.2121 \times {0.009 \over 0.003} = 5.2227$

And utilizing again Potto-GDC provides

Isentropic Flow Input: A/A* k = 1.4

M T/T0 ρ/ρ0 A/A* P/P0 PAR F/F*

0.111636 0.997514 0.993796 5.2227 0.991325 5.1774 2.19489

The pressure at point 1 is

$\displaystyle P_1 = P_2 {P_0 \over P_2} { P_1 \over P_0} = 5.0 times 0.94934 / 0.99380 \sim 4.776[Bar]$

The net force is obtained by utilizing equation (4.111)

$\displaystyle F_{net}$ $\displaystyle = P_2 A_2 {P_0 A^{*} \over P_2 A_2} (1+k) {\left( k+1 \over 2 \right)^{k \over k-1}} \left( {F_2 \over F^{*} } - { F_1 \over F^{*}}\right)$

$\displaystyle = 500000 \times {1 \over 2.1}\times 2.4 \times 1.2^{3.5} \times \left( 2.1949 - 0.96666 \right) \sim 614[kN]$

Next: The Impulse Function in Up: The Impulse Function Previous: The Impulse Function Index
Created by:Genick Bar-Meir, Ph.D.
On: 2007-11-21

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