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For inviscid, compressible, but isentropic ($s=$ constant) flow of a diatomic gas, the relationship between the pressure and density is $$p = A(s)\rho^{\gamma}\quad\text{where}\quad \gamma = \frac{c_p}{c_v} = \frac{7}{5}$$ Give Bernoulli's equation for such a gas flow. Explain why the pressure term is more important here than in a constant-density liquid flow.

Trying to search through the online notes there is section called Bernoulli's Equation which says: $$\frac{\partial\underline{u}}{\partial t} - \underline{u}\times\underline{\omega} = -\underline{\nabla}\mathcal{H}\quad\text{where}\quad \mathcal{H}(\underline{r},t) = \underline{\nabla}\bigg[\int\frac{1}{\rho}\mathrm{d}p + \frac{1}{2}||\underline{u}||^2 +\chi\bigg]$$ is the Bernoulli function, where assuming $p = p(\rho)$ we have $$\frac{1}{\rho}\underline{\nabla}p = \underline{\nabla}\bigg[\int\frac{1}{\rho}\mathrm{d}p\bigg]$$ Do I just rearrange so that: $$\rho = \exp\Bigg({\frac{5\ln\bigg(\frac{p}{A(s)}\bigg)}{7}}\Bigg) = \bigg(\frac{p}{A(s)}\bigg)^{5/7}$$ And then try to just sub in $\rho$ into the LHS or RHS?

MRT
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  • You need to find either $$ \underline{\nabla}\bigg[\int\frac{1}{\rho}\mathrm{d}p\bigg]$$ or $$ \frac{1}{\rho}\underline{\nabla}p $$ in order to substitute it in the Bernoulli's equation. I suggest you take the gradient of pressure per viscosity (the second one) and substitute it in Bernoulli's equation. Note that this is pressure head. You also need to find velocity head and height head to fully determine the Bernoulli's equation. – Mehrdad Zandigohar Mar 12 '18 at 19:51
  • Hint: $\displaystyle \frac{1}{\rho}\underline{\nabla}p = \dfrac{\underline{\nabla} (A(s)\rho^{\gamma})}{\rho}=?$ – Mehrdad Zandigohar Mar 12 '18 at 19:55
  • I can't find anywhere how to find $\underline{\nabla} \rho$? It isn't a vector so I don't understand how to do it... @MehrdadZandigohar – MRT Mar 12 '18 at 20:03
  • It depends on the gas flow direction. For instance, if the flow is in a pipe, then it only has one direction and the gradient will convert to a derivative. Let's say $\underline{\nabla} \rho$ will be $\dfrac{d\rho}{dx}$ if the flow is in $x$ direction (i.e. density only changes with respect to $x$) – Mehrdad Zandigohar Mar 12 '18 at 21:41
  • But, from what I am given in the question, how is it possible to know? @MehrdadZandigohar – MRT Mar 12 '18 at 23:02
  • Then consider the general form. Your fluid is compressible and as a result of that it is expected for $\rho$ to be: $\rho=\rho(x,y,z)$ so the gradient will be like: $\nabla \rho=\dfrac{\partial \rho}{\partial x}i + \dfrac{\partial \rho}{\partial y}j + \dfrac{\partial \rho}{\partial z}k $ – Mehrdad Zandigohar Mar 13 '18 at 00:33

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