Question

Circle your answer. Bare answers receive no credit. The horizontal pipe shown in the figure has a cross sectional area of 40 cm^2 at point A and C and 10 cm^2 at point B. Water (rho_w = 10^3 kg/m^3) is flowing in the pipe. The discharge is 0.006 m^3/s a) What is the velocity of the water at A? b) What is the velocity of the water at B) c) What is the pressure difference between A and B? d) What is the height difference between the mercury columns in the U -shaped tube? (rho_Hg - 13.6 times 10^6 kg/m^2)

Answer 1

The 6 liters/second gives the speed because 6 liters/second is the volume of water that flows across the cross-sectional area A with velocity V:

6 liters/second * (0.001 m^3 / 1 liter) = A * V
so vWide = 6 * 0.001 / (40 cm^2 * (1 meter/100cm)^2) = 1.5 m/s

The cross-sectional area A and speed V at 2 points are related by:
A1 * V1 = A2 * V2. You could use this to solve for the flow speed in the narrow section, or just use the previous equation with a different cross-sectional area as follows:
vNarrow = 6 * 0.001 / (10 cm^2 * (1 meter/100cm)^2) = 6 m/s

Bernoulli's Principle says
P1 + 1/2 * rho * V1^2 = P2 + 1/2 * rho * V2^2
so PWide - PNarrow = 1/2 * rho * (VNarrow^2 - VWide^2)
= 1/2 * 1000 kg/m^3 * (6^2 - 1.5^)
= 16,875 pascal

Converting the units for the density of mercury
13.6 g/cm^3 * (1kg / 1000 g) * (100cm/1m)^3 = 13600 kg/m^3

Bernoulli's Principle says
P1 + rho * g * H1 = P2 + rho * g * H2
so deltaH = deltaP / (rho * g)
= 16875Pa / (13600 kg/m^3 * 9.8m/s^2)
= 0.127 meters