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
Sorry for reposting this, but i still don't understand it. Could someone please give a step for step equation or explaination on how to do this. Thank you.
Sure, I can give you a step-by-step explanation of how to approach this problem.
(a) To find the speed at which water leaves the showerhead, we can use Bernoulli's equation, which relates the pressure, height, and velocity of a fluid. The equation is:
P + ρgh + 1/2 ρv^2 = constant
Where:
P is the pressure
ρ is the density of the fluid (water in this case)
g is the acceleration due to gravity
h is the height from the reference point (in this case, the pump's output pipe)
v is the velocity of the fluid
First, let's find the pressure at the showerhead:
P = gauge pressure + atmospheric pressure
= 420 kPa + 101.3 kPa (assuming atmospheric pressure is 101.3 kPa)
= 521.3 kPa
Next, let's find the height difference between the pump's output pipe and the showerhead:
height difference = 7.5 m
Now we can solve for the velocity of water leaving the showerhead. Since the pressure at the showerhead is higher, the term ρgh will be positive. We can ignore the term 1/2 ρv^2 since the problem statement instructs us to ignore viscosity. Rearranging the equation, we have:
v = sqrt(2(P1 - P2)/ρ)
= sqrt(2(521.3 kPa - 0)/1000 kg/m^3) (assuming the density of water is 1000 kg/m^3)
≈ 32.26 m/s
So, the speed at which water leaves the showerhead is approximately 32.26 m/s.
(b) To find the speed of water through the output pipe of the pump, we can use the Bernoulli's equation again. Except this time, the height is the same (since the pipe and showerhead are on the same level). We also know the radius of the pipe and the number of holes in the showerhead.
Using the equation P + ρgh + 1/2 ρv^2 = constant, we can ignore the term ρgh since the height is the same. Rearranging the equation, we have:
v = sqrt(2(P1 - P2)/ρ)
= sqrt(2(420 kPa - 0)/1000 kg/m^3) (assuming the density of water is 1000 kg/m^3)
≈ 28.98 m/s
So, the speed at which water moves through the output pipe of the pump is approximately 28.98 m/s.
I hope this helps! Let me know if you have any more questions.
To solve this problem, we will use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid flowing through a pipe.
(a) To find the speed at which water leaves the showerhead, we need to calculate the velocity of water at the showerhead. We can use Bernoulli's equation to do this. Bernoulli's equation states that the sum of the pressure, kinetic energy, and potential energy per unit volume of a fluid remains constant along a streamline.
The equation for Bernoulli's principle is:
P + ½ρv² + ρgh = constant
Where:
- P is the pressure of the fluid
- ρ is the density of the fluid
- v is the velocity of the fluid
- g is the acceleration due to gravity
- h is the height of the fluid above a reference point
In this case, we can assume that the pressure at the showerhead is atmospheric pressure (P₀ = 1 atm) since the problem doesn't mention any specific values.
So, at the showerhead:
P + ½ρv² + ρgh = P₀ + 0 + 0
P + ½ρv² + ρgh = P₀
Since we are interested in the velocity (v), we need to solve the equation for v:
½ρv² = P₀ - P - ρgh
Multiply both sides of the equation by 2 to isolate v:
ρv² = 2P₀ - 2P - 2ρgh
Divide both sides by ρ to find v²:
v² = (2P₀ - 2P - 2ρgh) / ρ
Substitute the given values into the equation:
v² = (2 * 1 atm - P - ρgh) / ρ
The density of water (ρ) is 1000 kg/m³, and the acceleration due to gravity (g) is approximately 9.81 m/s².
v² = (2 * 1 atm - P - 1000 kg/m³ * 9.81 m/s² * 7.5 m) / (1000 kg/m³)
The pressure at the showerhead is atmospheric pressure, which is approximately 101325 Pa.
v² = (2 * 101325 Pa - P - 1000 kg/m³ * 9.81 m/s² * 7.5 m) / (1000 kg/m³)
Finally, take the square root of both sides to find the velocity (v):
v = √[(2 * 101325 Pa - P - 1000 kg/m³ * 9.81 m/s² * 7.5 m) / (1000 kg/m³)]
Now, substitute the given value of P (gauge pressure) into the equation and solve for v.
(b) To find the speed at which water moves through the output pipe of the pump, we can use the same formula as in part (a), but with different values.
The pressure at the pump's output pipe is given as a gauge pressure of 420 kPa. To convert this to absolute pressure, we add it to atmospheric pressure, which is approximately 101325 Pa:
P = 420 kPa + 101325 Pa
Substitute this value of P into the equation for v and solve for v.
Note: The final answers to both parts (a) and (b) will depend on the given values of P and the other variables.