# What determines the flow rate through a pipe when exiting a state of rest?

If liquid in a pipe is not moving, has a pressure of $$x$$ Pa, a diameter of $$d_1$$ meters, and an open close valve of $$d_2$$ meters, then ignoring any friction loss terms, and assuming a level pipe, is there a way to determine the flow rate in $$m^3/s$$ out of the pipe whenever the valve is fully opened (again no friction losses). I am interested in the case where the liquid is water.

Edit: I noticed someone tagged my post as a homework exercise. I am actually doing some irrigation work and became interested in this question. I have a strong mathematical background, but only a limited physics background,

The standard tool for this type of problem is Bernoulli's equation, which relates the velocity of the water as it enters the pipe to its velocity as it exits the pipe and the pressure difference at the two ends. In addition, if you assume that the water is incompressible (a pretty good approximation for water assuming you're not using stupidly large pressures), then the flow rate $$Q$$ into the pipe must equal the flow rate out of the pipe. This implies that $$Q = Av$$ is the same at both ends (where $$A$$ is the cross-sectional area of the pipe.) These equations can then be solved to find the flow rate.

All of this assumes, however, that your pipe is in steady state, i.e., there is a constant flow rate being maintained by a constant pressure head. If you are in a situation where the pressure head is not staying constant, then things get more complicated (and beyond my ability to dash off a quick answer.)

• Isn't the water not moving until the valve is opened (My assumption being that the pipe already is entirely filled with water)? In this situation I don't see how Bernoulli's equation is applicable. I want to know how to determine the velocity and flow rate once the valve is opened. I believe that before that happens we have a velocity of 0 and flow rate of 0. Commented Apr 23, 2021 at 14:58
• @math314: That's correct — Bernoulli's equation also doesn't apply until the new steady state is reached, i.e., once water has started moving steadily at both ends of the pipe. To an order of magnitude, I think this would occur in about as much time as it takes sound waves to travel through the water to the other end of the pipe, since that's how long it takes for one end of the pipe to "realize" that something has changed at the other end. Note that the speed of sound in water is about 1500 m/s, so for a pipe that's 1 meter long this steady state would be achieved within a few milliseconds. Commented Apr 23, 2021 at 16:38