# 11.4.7: The Impulse Function

Fig. 11.9 Schematic to explain the significances of the Impulse function.

$F_{net}= \dot{m} (U_2 -U_1) + P_2 A_2 - P_1 A_1 \label{gd:iso:eq:monImp} \tag{64}$
The net force is denoted here as $$F_{net}$$. The mass conservation also can be applied to our control volume
$\dot{m} = \rho_1A_1U_1 = \rho_2A_2U_2 \label{gd:iso:eq:massImp} \tag{65}$
Combining equation (64) with equation (??) and by utilizing the identity in equation (50) results in
$F_{net}= kP_2A_2{M_2}^2 - kP_1A_1{M_1}^2 + P_2 A_2 - P_1 A_1 \label{gd:iso:eq:MomMassImp} \tag{66}$
Rearranging equation (66) and dividing it by $$P_0 A^{*}$$ results in
${F_{net} \over P_0 A^{*}} = \overbrace{P_2A_2 \over P_0 A^{*}}^{f(M_2)} \overbrace{\left( 1 + k{M_2}^2 \right)}^{f(M_2)} - \overbrace{P_1A_1 \over P_0 A^{*}}^{f(M_1)} \overbrace{\left( 1 + k{M_1}^2 \right)}^{f(M_1)} \label{gd:iso:eq:beforeDefa} \tag{67}$
Examining equation (67) 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
$F = PA\left( 1 + k{M_2}^2 \right) \label{gd:iso:eq:impulsDef} \tag{68}$
In the Impulse function when $$F$$ ($$M=1$$) is denoted as $$F^{*}$$
$F^{*} = P^{*}A^{*}\left( 1 + k \right) \label{gd:iso:eq:impulsDefStar} \tag{69}$
The ratio of the Impulse function is defined as
${F \over F^{*}} = {P_1A_1 \over P^{*}A^{*}} {\left( 1 + k{M_1}^2 \right) \over \left( 1 + k \right) } = {1 \over \underbrace{P^{*}\over P_{0} }_ {\left(2 \over k+1 \right)^{k \over k-1}}} \overbrace ParseError: EOF expected (click for details) Callstack: at (Under_Construction/Book:_Fluid_Mechanics_(Bar-Meir)__-_old_copy/11:_Compressible_Flow_One_Dimensional/11.4_Isentropic_Flow/11.4.7:_The_Impulse_Function), /content/body/p[3]/span, line 1, column 10  ^{\hbox{see function qref{gd:iso:eq:beforeDefa}}} {1 \over \left( 1 + k \right) } \label{gd:iso:eq:ImpulseRatio} \tag{70}$
This ratio is different only in a coefficient from the ratio defined in equation (67) which makes the ratio a function of $$k$$ and the Mach number. Hence, the net force is
$F_{net} = P_0 A^{*} (1+k) {\left( k+1 \over 2 \right)^{k \over k-1}} \left( {F_2 \over F^{*} } - { F_1 \over F^{*}}\right) \label{gd:iso:eq:NetForce} \tag{71}$
To demonstrate the usefulness of the this function consider a simple situation of the flow through a converging nozzle.

Example 11.11

Fig. 11.10 Schematic of a flow of a compressible substance (gas) through a converging nozzle for example (??)

Consider a flow of gas into a converging nozzle with a mass flow rate of $$1[kg/sec]$$ and the entrance area is $$0.009[m^2]$$ and the exit area is $$0.003[m^2]$$. The stagnation temperature is $$400K$$ and the pressure at point 2 was measured as $$5[Bar]$$. Calculate the net force acting on the nozzle and pressure at point 1.

Solution 11.11

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. The denominator is needed to be determined to obtain this ratio. Utilizing Fliegner's equation (59), provides the following

$A^{*} P_0 = \dfrac{\dot{m} \sqrt{R\,T} }{ 0.058} = \dfrac{1.0 \times \sqrt{400 \times 287} }{ 0.058} \sim 70061.76 [N]$
and
$\dfrac{A_2\, P_2 }{ A^{\star}\, P_0} = \dfrac{ 500000 \times 0.003 }{ 70061.76 } \sim 2.1$

 Isentropic Flow Input: \dfrac{A\, P }{ A^{\star} \, P_0} k = 1.4 $$M$$ $$\dfrac{T}{T_0}$$ $$\dfrac{\rho}{\rho_0}$$ $$\dfrac{A}{A^{\star} }$$ $$\dfrac{P}{P_0}$$ $$\dfrac{A\, P }{ A^{\star} \, P_0}$$ $$\dfrac{F }{ F^{\star}}$$ 0.27353 0.98526 0.96355 2.2121 0.94934 2.1000 0.96666

With the area ratio of $${A \over A^{\star}}= 2.2121$$ the area ratio of at point 1 can be calculated.
$\dfrac{ A_1 }{ A^{\star}} = \dfrac{A_2 }{ A^{\star}} \dfrac{A_1 }{ A_2} = 2.2121 \times \dfrac{0.009 }{ 0.003} = 5.2227$
And utilizing again Potto-GDC provides

 Isentropic Flow Input: \dfrac{A\, P }{ A^{\star} \, P_0} k = 1.4 $$M$$ $$\dfrac{T}{T_0}$$ $$\dfrac{\rho}{\rho_0}$$ $$\dfrac{A}{A^{\star} }$$ $$\dfrac{P}{P_0}$$ $$\dfrac{A\, P }{ A^{\star} \, P_0}$$ $$\dfrac{F }{ F^{\star}}$$ 0.11164 0.99751 0.99380 5.2227 0.99132 5.1774 2.1949

The pressure at point $$1$$ is
$P_1 = P_2 {P_0 \over P_2} { P_1 \over P_0} = 5.0 times 0.94934 / 0.99380 \sim 4.776[Bar]$
The net force is obtained by utilizing equation (71)
$F_{net} &= P_2 A_2 {P_0 A^{*} \over P_2 A_2} (1+k) {\left( k+1 \over 2 \right)^{k \over k-1}} \left( {F_2 \over F^{*} } - { F_1 \over F^{*}}\right) \ & = 500000 \times {1 \over 2.1}\times 2.4 \times 1.2^{3.5} \times \left( 2.1949 - 0.96666 \right) \sim 614[kN]$

### Contributors

• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.
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