# Frictionless flow in pipes

I am trying to understand the flow of an ideal, frictionless incompressible fluid through a pipe for these two cases. Note: Please consider Flow velocity in the tank is zero, and the fluid is incompressible.

Case 1: This is simple to understand. In this, I can calculate the flow velocity by applying Bernoulli's equation as the pressure difference is completely converted to kinetic energy.

i)So am I correct in understanding that the presence or absence of the pipe has no effect on the flow rate?

ii) Is the pressure at every point of the pipe the same as atmospheric pressure?

Case 2:

iii) Will the flow rate be the same as in case 1 or will the addition of the diverging section cause the flow rate to increase.

iv)Does the inclusion of the diverging section cause the flow to expand and so increase the static pressure (to atmospheric pressure) and if this is true will the static pressure before the expansion be less than atmospheric pressure?

Thank you

• For instantaneous flow, only the pipe diameter(s) and length(s), and fluid properties matter. If being very detailed, the fluid must be accelerated also. Otherwise use basic pipe flow formulas. However if the fluid is truly incompressible and the tank is truly rigid, the pressure difference will rapidly drop, unless P is maintained in the tank by something (which can be wall elasticity or something compressible like air in the tank, or regulator or just gravity) Aug 28, 2021 at 16:43
• @PeteW Assume the pressure and level remain constant what ever the flow in the pipe (infinite tank) and there is no frictional loss.. Aug 28, 2021 at 16:52
• by frictionless, do you mean the usual no-slip condition is abandoned completely? Aug 28, 2021 at 22:01
• @PeteW By friction less I mean there are no major or minor losses. Aug 29, 2021 at 3:06
• everything about fluid flow ceases to make sense if no friction exists. In your example, the only thing allowing the container pressure to be greater than atmospheric would be the inertia of the fluid. Aug 29, 2021 at 16:12

I suppose that theoretically you could get liquid to move fast enough to flash into vapor, but I've never heard of this being a practical design consideration. Maybe if the entire thing operated in a partial vacuum?

Case 1

(i) - Yes, correct.

(ii) - $$P_2 = P_1$$ Case 2

(i) - Yes, the flow rate is the same for both cases. No, the flow rate is constant but the velocity varies due to change in cross section ($$Q = v_1A_1 = v_2A_2$$).

(ii) - Again, let's write Bernoulli's equation:

$$P_2 = \dfrac{1}{2}\rho v_1^2 + P1 - \dfrac{1}{2}\rho v_2^2$$, since $$v_2 < v_1$$, so $$P_2 > P_1$$. The result is expected based on the Bernoulli's principle, that states, with constant datum, slower moving fluids create greater pressure (force) than faster moving fluids.

Due to the presence of head, both $$P_1$$ & $$P_2$$ are not likely less than $$P_{atm}$$.

• Thank you @r13 Case 1: ii) What about when 2 is at the end of the pipe? Will there be a sudden increase in pressure from P1 to Pout (atmospheric pressure)? Sep 6, 2021 at 4:52
• Following the release of the water, the pressure in the tank will continue to drop until there is no flow, then P1 = P2 = Patm.
– r13
Sep 6, 2021 at 15:45
• Please assume the tank level is always constant what ever the flow in the pipe(infinite tank). Sep 6, 2021 at 16:34
• The pressure in the tank will change, and it is always greater than the pressure at the outlet, otherwise, backflow will occur when Patm > P2. Sorry, I don't know the case you want to build.
– r13
Sep 6, 2021 at 16:47
• Yes pressure in the tank is grater than Pout and we are assuming the tank level won't vary much, say some one is putting the fluid back into the tank as flow takes place in the pipe. In your answer for case 1: ii) you mentioned P2=P1. What about when the point 2 near to the pipe outlet? Sep 6, 2021 at 16:54