Question

While taking a shower, you notice that the shower head is made up of 44 small round openings, each with a radius of 2.00 mm. You also determine that it takes 3.00 s for the shower to completely fill a 1.00-liter container you hold in the water stream. The water for the shower is pumped by a pump that is 5.70 m below the level of the shower head. The pump maintains an absolute pressure of 1.50 atm. Use g = 10 m/s2, and assume that 1 atmosphere is 1.0 105 Pa. (a) At what speed does the water emerge from the shower head? (b) What is the speed of the water in the pipe connected to the pump? (c) What is the cross-sectional area of the pipe connected to the pump?

Answer 1

To solve the problem it is necessary to apply the concepts related to the calculation of discharge flow, Bernoulli equations and energy conservation in incompressible fluids.

PART A) For the calculation of the velocity we define the area and the flow, thus

$A = \pi r^2$

$A = pi (2*10^{-3})^2$

$A = 12.56*10^{-6}m^2$

At the same time the rate of flow would be

$Q = \frac{1L}{2s}$

$Q = 0.5L/s = 0.5*10^{-3}m^3/s$

By definition the discharge is expressed as

$Q = NAv$

Where,

A= Area

v = velocity

N = Number of exits

Q = NAv

Re-arrange to find v,

$v = \frac{Q}{NA}$

$v = \frac{0.5*10^{-3}}{44*12.56*10^{-6}}$

$v = 0.9047m/s$

PART B) From the continuity equations formulated by Bernoulli we can calculate the speed of water in the pipe

$P_1 + \frac{1}{2}\rho v_1^2+\rho gh_1 = P_2 +\frac{1}{2}\rho v^2_2 +\rho g h_2$

Replacing with our values we have that

$1.5*10^5 + \frac{1}{2}(1000) v_1^2+(1000)(9.8)(0) = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)$

$v_1^2 = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)$

$v_1 = \sqrt{10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)}$

$v_1 = 3.54097m/s$

PART C) Assuming that water is an incomprehensible fluid we have to,

$Q_{pipe} = Q_{shower}$

$v_{pipe}A_{pipe}=v_{shower}A_{shower}$

$3.54097*A_{pipe}=0.9047*12.56*10^{-6}$

$A_{pipe} = \frac{0.9047*12.56*10^{-6}}{3.54097}$

$A_{pipe = 3.209*10^{-6}m^2$