Answer to Question #77650 in Physics for Emy

Question

Water flows through a horizontal pipe at a speed of 14 m/s and pressure 8.4x10^5 Pa. The pipe narrows and the water speed goes up to a 42m/s. What is the pressure in the narrow section of the pipe?

Answer 1

We can find the pressure in the narrow section of the pipe from the Bernoulli’s equation:
P_1+1/2 ρv_1^2=P_2+1/2 ρv_2^2,
here, P_1, P_2 are the pressures in the wide and narrow section of the pipe, respectively; v_1, v_2 are the speeds of the water flowing through the wide and the narrow section of the pipe, respectively; ρ=1000 kg⁄m^3 is the density of the water.
From this equation we can find the pressure in the narrow section of the pipe:
P_2=P_1+1/2 ρ(v_1^2-v_2^2 )==8.4∙〖10〗^5 Pa+1/2∙1000 kg/m^3 ∙((14 m/s)^2-(42 m/s)^2 )==0.56∙〖10〗^5 Pa.