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Anderson's "Numerical solutions of quasi-one-dimensional nozzle flows"

Hi,

I am currently reading through chapter 7 "Numerical solutions of
quasi-one-dimensional nozzle flows" of the book "Computational Fluid
Dynamics, the basics with applications" by John D. Anderson - and got
stucked at the following step:

Let us obtain from Eq. (7.33) the nonconservation form expressed in
terms of internal energy by itself. The latter can be achieved by
multiplying Eq. (7.23)

$\hspace*{3cm} \frac{\partial (\rho VA)}{\partial t} + \frac{\partial (\rho V^2A)}{\partial x} = - A \frac{\partial p}{\partial x} \hspace*{3.8cm}(7.23)$

by V, obtaining

$\hspace*{3cm} \frac{\partial [\rho (V^2/2)A]}{\partial t} + \frac{\partial [\rho (V^3/2)A]}{\partial x} = - AV \frac{\partial p}{\partial x} \hspace*{2cm}(7.34)$

Could somebody please explain why Eq. (7.23) multiplied with V results
in Eq. (7.34)?

Thanks a lot,

Leo

Quote:

Originally Posted by liljendal (Post 548839)

Let us obtain from Eq. (7.33) the nonconservation form expressed in
terms of internal energy by itself. The latter can be achieved by
multiplying Eq. (7.23)

$\hspace*{3cm} \frac{\partial (\rho VA)}{\partial t} + \frac{\partial (\rho V^2A)}{\partial x} = - A \frac{\partial p}{\partial x} \hspace*{3.8cm}(7.23)$

by V, obtaining

$\hspace*{3cm} \frac{\partial [\rho (V^2/2)A]}{\partial t} + \frac{\partial [\rho (V^3/2)A]}{\partial x} = - AV \frac{\partial p}{\partial x} \hspace*{2cm}(7.34)$

Could somebody please explain why Eq. (7.23) multiplied with V results
in Eq. (7.34)?

Thanks a lot,

Leo

I think there is a typo in the second equation. It should be

$\hspace*{3cm} \frac{\partial [\rho (V^2/2)A]}{\partial t} + \frac{\partial [\rho (V^3/3)A]}{\partial x} = - AV \frac{\partial p}{\partial x}$

The arguments are a composite functions of V(x,t). Executing the chain rule
on the functions V^2/2 and V^3/3 allows you to pull out a single V yielding

$\hspace*{3cm} \frac{2V}{2} \frac{\partial [\rho V A]}{\partial t} +$ $\frac{3V}{3} \frac{\partial [\rho V^2 A]}{\partial x} = - AV \frac{\partial p}{\partial x}$

$\hspace*{3cm} V \frac{\partial [\rho V A]}{\partial t} +$ $V \frac{\partial [\rho V^2 A]}{\partial x} = - AV \frac{\partial p}{\partial x}$