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My questions are on the definitions of each of the terms in Bernoulli's equation for unsteady flows - which is the following equation (and which holds throughout the entire fluid):

$p+\frac{1}{2}\rho|\underline{u}|^2+\rho gz+\rho\frac{\partial \phi}{\partial t} = f(t)$.

I am really struggling to understand what some of these terms represent and hence this is affecting my ability to compute each of them. I essentially want to see if/where my understanding is failing. These are my questions:

i) Pressure, $p$ - Firstly am I right in saying that this is the pressure on the fluid at a given point? Intuitively I know that Pressure=Force/Area, and more formally: $d\boldsymbol{F}_n=-p\boldsymbol{n}dS$. So does this mean that is we want to calculate the pressure on a stationary fluid contained in a cylindrical cup of cross sectional area $A$ which is filled to height $h$, this pressure would be $p=\rho A(h-z)g+p_{atm}$ where $z$ is the height at which we are calculating the pressure?

ii) $|\underline{u}|$ - is this the speed of the fluid at the point at which we wish to apply Bernoulli's equation? For example, if we slowly drain (from the top) water out of a cylindrical cup then then is the speed of the fluid at the bottom of the cup $0$ (since only the top is moving)? If also we had fluid oscillating in a U tube, would the speed at the bottom of this tube also be $0$?

iii) $\rho gz$ - am I right in saying if we want to apply Bernoulli's equation at height $z=0$, then this term is just $0$?

iv) $\frac{\partial \phi}{\partial t}$ - this is the velocity potential. Let's say we have a fluid that is moving vertically only and we have $\phi=f(z)g(t)$, where $0\leq z\leq h$. Is it true $\frac{\partial \phi}{\partial t}=f(z)g'(t)$ - or is $z$ (the height at which we apply Bernoulli's equation?) a function of $t$?

I would be very grateful is someone could clear up my understanding?

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  • $\begingroup$ Another way of deriving the equation you are seeking is to write down the equation for the total kinetic plus potential energy of the fluid in the U tube. The rate of change of this total energy must be zero, since the fluid is inviscid. So set the derivative with respect to time equal to zero. This will give you what you want. Then you can compare the result with the derivation above term for term. $\endgroup$ – Chet Miller Dec 1 '20 at 12:24
  • $\begingroup$ @ChetMiller ok thank you, so would I be right in saying that the total kinetic energy of the fluid in the U tube is $1/2$ density $\times$ vol of fluid $\times$ rate of change of height^2 and the potential energy is density $\times$ vol of fluid $\times$ $g$ $\times$ (displaced height - equilibrium height)? $\endgroup$ – maths54321 Dec 1 '20 at 14:17
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The constraint on the U tube problem is that the sum of the kinetic energy and potential energy is constant.

The kinetic energy of the left leg is $$\frac{(\rho A_Lh_L)}{2}\left(\frac{dh_L}{dt}\right)^2$$and the potential energy of the left leg is $$\frac{\rho g A_Lh_L^2}{2}$$ Similarly for the right leg.

The other constraint is $$A_Lh_L+A_Rh_R=A_Lh_{L0}+A_Rh_{R0}$$ where the 0 subscripts denote the initial heights.

CONTINUATION OF SOLUTION

Based on the expressions developed above, the total of kinetic and potential energy of the fluid in the U tube is given by $$E=\frac{(\rho A_Lh_L)}{2}\left(\frac{dh_L}{dt}\right)^2+\frac{\rho g A_Lh_L^2}{2}+\frac{(\rho A_Lh_R)}{2}\left(\frac{dh_R}{dt}\right)^2+\frac{\rho g A_Lh_R^2}{2}$$If we take the time derivative of E and set it equal to zero, and make use of the constraint that $A_L\frac{dh_L}{dt}=-A_R\frac{dh_R}{dt}$, we obtain: $$\frac{\rho}{2}\left(\frac{dh_L}{dt}\right)^2+\rho h_L\frac{d^2h_L}{dt^2}+\rho g h_L=\frac{\rho}{2}\left(\frac{dh_R}{dt}\right)^2+\rho h_R\frac{d^2h_R}{dt^2}+\rho g h_R\tag{1}$$The Bernoulli form of this equation is apparent. The middle term on each side of the equation represents the time-dependent acceleration effect.

The height at equilibrium of the fluid is determined from the constraint condition with $h_L=h_R=H$, where $$H=\frac{(A_Lh_{L0}+A_Rh_{R0})}{A_L+A_R}$$

The constraint condition will always be satisfied exactly if we now express $h_L$ and $h_R$ in terms of H and in terms of a single other parameter $\delta$ as follows: $$h_L=H+\frac{A_R}{(A_L+A_R)}\delta$$ and $$h_R=H-\frac{A_L}{(A_L+A_R)}\delta$$If we substitute these relationships into our key differential Eqn. 1, we obtain a differential equation in term so the single time-varying parameter $\delta$:$$\frac{1}{2}\frac{(A_R-A_L)}{(A_R+A_L)}\left(\frac{d\delta}{dt}\right)^2+H\frac{d^2\delta}{dt^2}+\frac{(A_R-A_L)}{(A_R+A_L)}\delta \frac{d^2\delta}{dt^2}+g\delta=0$$ If the two cross sectional areas are equal, this reduces to $$H\frac{d^2\delta}{dt^2}+g\delta=0$$which has an obvious analytical solution.

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  • $\begingroup$ Thank you again. I just want to check whether $h_L$ is equal to $H+s_L(t)$ where $H$ is the height at equilibrium of the fluid and $s_L(t)$ is the displacement of the fluid on the left side? And hence is it true that pressure on the left side at arbitrary height $z$ in the fluid is equal to $\rho g(H+s_L-z)$ (+ atmospheric pressure) and the velocity potential at arbitrary height $z$ on the left side is equal to $z\frac{ds_L}{dt}$? $\endgroup$ – maths54321 Dec 1 '20 at 16:20
  • $\begingroup$ I took as a datum for potential energy the base of the U tube. I'm going to take the derivative of the potential energy and set it equal to zero, and you can see what I end up with so that we can better compare. Be back later. $\endgroup$ – Chet Miller Dec 1 '20 at 16:39
  • $\begingroup$ Your conclusion about the pressure is incorrect. That is only true if the fluid acceleration is zero. I've got the rest of the problem worked out, but I don't have time to write it out now. Be back later. $\endgroup$ – Chet Miller Dec 1 '20 at 18:56
  • $\begingroup$ See my continuation of the solution. $\endgroup$ – Chet Miller Dec 1 '20 at 22:06
  • $\begingroup$ I'm surprised you have not commented on my analysis. Do you have any questions about it? $\endgroup$ – Chet Miller Dec 2 '20 at 11:45

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