Compressible Flow credits Logo credits
Potto Home Contact Us

Potto Home

About Potto


  Hard copy
  Gas Dynamics Tables

Other things:
Other resources
Download Area

Other Resources

  Compare Other Books

Potto Statistics



next up previous index
Next: The Impulse Function in Up: The Impulse Function Previous: The Impulse Function   Index

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 $ F$ , but in the literature some denote this function as $ I$ . 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

The net force is denoted here as $ F_{net}$ . The mass conservation also can be applied to our control volume
Combining equation (4.104) with equation (4.105) and by utilizing the identity in equation (4.42) results in
Rearranging equation (4.106) and dividing it by $ P_0 A^{*}$ results in

Figure 4.9: Schematic to explain the significances of the Impulse function
Image impulse
Examining equation (4.107) shows that the right hand side is only a function of Mach number and specific heat ratio, $ k$ . Hence, if the right hand side is only a function of the Mach number and $ k$ than the left hand side must be function of only the same parameters, $ M$ and $ k$ . 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
In the Impulse function when $ F$ ( M=1 ) is denoted as $ F^{*}$
The ratio of the Impulse function is defined as
This ratio is different only in a coefficient from the ratio defined in equation (4.107) which makes the ratio a function of $ k$ and the Mach number. Hence, the net force is

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)

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.

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]


$\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 up previous index
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