A hydroelectric power station is supplied with water from a reservoir. A pipeline connects the reservoir to the turbine hall. The flow of water through the pipeline is controlled by a valve which is located 500 metres below the surface of the water in the reservoir. The lower end of the pipeline is 0.30m in diameter where it enters the valve. Q1 If the valve is opened to give a flow rate of 5m^3 per second, calculate the pressure at the valve. Answer. I have previously calculated the pressure at the valuve when it is closed by adding atmospheric pressure with pgh to give a static pressure of 5 x 10^6 Pa. I am given a flow rate of 5 m^3/s and i also know the diameter of the pipe, so to obtain velocity I have said v = flow rate / area = 5 / (pi x (0.15)^2) = 70.736 m/s (is this correct?) Now I can use Bernoullis equation which would be P (which is static pressure at valve) + 1/2pv^2 I am assuming that pgh is not needed here as we are no longer considering a change in height. Thus the pressure should be (5 x 10^6 Pa) + (1/2 * 1.00 x 10^3 * (70.736)^2) = 7.5 x 10^6 Pa Although the magnitude looks ok, I was expecting the pressure at the valve to drop here, but its gone up... Have I gone wrong somewhere?