A house with its own well has a pump in the basement with an output pipe of inner radius 6.4 mm. Assume that the pump can maintain a gauge pressure of 420 kPa in the output pipe. A showerhead on the second floor (7.5 m above the pump's output pipe) has 38 holes, each of radius 0.34 mm. The shower is on "full blast" and no other faucet in the house is open.
(a) Ignoring viscosity, with what speed does water leave the showerhead?
m/s
(b) With what speed does water move through the output pipe of the pump?
m/s
I can not even guess how to start this :/. Please help. Thank you.
To solve this problem, we can begin by applying Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a vertical column.
(a) To find the speed at which water leaves the showerhead, we need to calculate the pressure difference between the showerhead and the pump's output pipe using Bernoulli's equation:
P1 + 1/2 ρV1^2 + ρgh1 = P2 + 1/2 ρV2^2 + ρgh2
Where:
P1 and P2 are the pressures at points 1 (showerhead) and 2 (pump's output).
V1 and V2 are the velocities at points 1 and 2.
ρ is the density of water.
g is the acceleration due to gravity.
h1 and h2 are the heights of points 1 and 2.
Since the system is at rest and no other faucet in the house is open, the velocity of the water at the output pipe of the pump (V2) will be zero. We can also assume the pipe is horizontal, so the height difference (h1 - h2) is the height between the pump's output and the showerhead. Therefore, the equation simplifies to:
P1 + 1/2 ρV1^2 = P2
Since the pressure at the showerhead (P1) is atmospheric pressure and the pump maintains a gauge pressure of 420 kPa in the output pipe, we have:
P1 = 0 kPa
P2 = 420 kPa
Substituting these values into the equation, we have:
0 + 1/2 ρV1^2 = 420,000 Pa
We can rearrange this equation to solve for V1:
V1 = √((2 * P2) / ρ)
To find the density (ρ) of water, we can use its known value of 1000 kg/m^3.
Substituting all the values into the equation, we get:
V1 = √((2 * 420,000) / 1000)
Calculating V1 gives us the speed at which water leaves the showerhead.
(b) To find the speed at which water moves through the output pipe of the pump, we can assume there is no change in pressure along a horizontal pipe. Therefore, the speed of water leaving the pump's output (V2) will be the same as the speed of water in the pipe.
V2 = V1
So, the speed at which water moves through the output pipe of the pump is the same as the speed at which water leaves the showerhead.
(a) Use the Bernoulli equation, with elevation change, with the assumption that the gauge pressure in the shower holes is zero.
(b) Knowing the flow velocity in the shower holes from (a), and the total area of all shower holes, use the staedy state continuity equation.
If you are not familiar with the Bernoulli and continuity equations, you need to review your text materials, or do some Googling.