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Old   June 4, 2015, 05:32
Default Anderson's "Numerical solutions of quasi-one-dimensional nozzle flows"
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Leo Liljendal
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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
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Old   June 4, 2015, 08:44
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Troy Snyder
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Quote:
Originally Posted by liljendal View Post
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
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}
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