MB613d said:
Thank you for your answer.
I understand what you did there and I think it is correct.
So do you think the pressure loss of the pipe is the pressure that needs to be established before gas begins to flow?
If you understand what I am saying, then you know that this is a totally transient problem, and that there is no steady state. The initial pressure difference between the two vessels is zero, and thus the initial mass flow rate through the pipe is zero. As time progresses, the pressure difference between the vessels and the mass flow rate through the pipe both increase
gradually with time. So there is no magic pressure difference at which the gas starts flowing.
The mistake in your original analysis was assuming that the volume flow rate of gas through the pipe must match the volume flow rate of water between the two vessels. Because the air is compressible, this is not the case. The equations I gave in post #19 show how the pressures vary at short times when there is not enough of a pressure difference across the pipe to produce significant air flow.
So, initially, the flow Reynolds number is zero, and it builds up gradually with time, first passing through the laminar region, then the transition, and then to turbulent flow at later times.
The flow analysis would go something like the following if the zeta resistances were distributed uniformly along the length of the pipe: At any given time, the pressure gradient in the pipe would be $$\frac{dp}{dz}=-\left[\frac{\zeta}{L}+\frac{\lambda(Re)}{D}\right]\frac{1}{2}\rho w^2\tag{1}$$where ##p=p_2## at z = 0 and ##p=p_1## at z = L and Re is given by Eqn. 6 in post #17. The mass flow rate through the pipe is given by: $$\dot{m}=\rho wA\tag{2}$$where, from Eqn. 5 in post #17,
$$\dot{m}=\frac{M}{RT}\frac{d[p_1(V_1+\alpha t)]}{dt}\tag{3}$$
If we combine Eqns. 1 and 2, we obtain:$$\frac{dp}{dz}=-\left[\frac{\zeta}{L}+\frac{\lambda(Re)}{D}\right]\frac{1}{2}\frac{(\dot{m})^2}{\rho A^2}\tag{4}$$From the ideal gas law, the density is given by: $$\rho=\frac{pM}{RT}$$Substituting this into Eqns. 4 then yields:
$$p\frac{dp}{dz}=-\left[\frac{\zeta}{L}+\frac{\lambda(Re)}{D}\right]\frac{1}{2}\frac{RT(\dot{m})^2}{M A^2}\tag{5}$$Next, integrating this equation between z = 0 and z = L yields:$$p_2^2-p_1^2=\left[\zeta+\lambda(Re)\frac{L}{D}\right]\frac{RT(\dot{m})^2}{M A^2}\tag{6}$$It follows from an un-numbered equation in post # 17 that ##p_2## can be expressed in terms of ##p_1## as:
$$p_2=p_0+\frac{(V_{10}+\alpha t)}{(V_{20}-\alpha t)}(p_0-p_1)\tag{7}$$##p_2## can be eliminated from Eqn. 6 by substitution of Eqn. 7 to yield:
$$\left(p_0+\frac{(V_{10}+\alpha t)}{(V_{20}-\alpha t)}(p_0-p_1)\right)^2-p_1^2=\left[\zeta+\lambda(Re)\frac{L}{D}\right]\frac{RT(\dot{m})^2}{M A^2}\tag{8}$$Since the Reynolds number Re is a function of ##\dot{m}##, Eqn. 8 is of the mathematical form $$g(p_1,t)=f(\dot{m})\tag{9}$$If we formally invert this to solve for ##\dot{m}## and combine the result with Eqn. 3, we obtain: $$\frac{d[p_1(V_1+\alpha t)]}{dt}=\frac{RT}{M}h(p_1, t)\tag{9}$$
Integrating this ordinary differential equation with respect to time allows us to determine ##p_1## as a function of time.