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.
Bernoulli's equation.
Pressurebase+1/2 rho*V1^2 + rho*gh1 =Pressureabove+1/2*18 rho*V2^2 + rho*gh2
let the base be h1=0 ; pressure above=0, pressure base= 420kPa
h2=7.5m
rho is density water.
one equation, two unknowns (V2, V1)
Equation of Continunity:
Area1*V1=Area1*V2
now you can solve the equation for V1, V2, just a tad of algebra :)
I got this
(420+(1/2)1000(V1^2)+1000(0)= 0+(1/2)18(1000)V2^2+1000(9.8)(7.5)
so it's 920 V1^2=82500 V2^2
but I don't understand how to get part a and b n how to solve for V1^2 and V2^2 and what does Equation of Continunity:
Area1*V1=Area1*V2 mean I mean what is the area and how can I get V1 and V2. Sorry, but am confused! Thank you.
To solve this problem, we can use the principles of Bernoulli's equation, which relates the pressure, height, and velocity of a fluid in a horizontal pipe or tube.
Step 1: Determine the pressure at the showerhead
First, let's find the pressure at the showerhead using Bernoulli's equation. The equation is:
P + 1/2 * ρ * v^2 + ρ * g * h = constant
Where P is the pressure, ρ is the density of the fluid, v is the velocity of the fluid, g is the acceleration due to gravity, and h is the height.
We can assume the constant is the pressure at the pump output pipe, which is 420 kPa.
Now, let's substitute the values:
P_shower + 1/2 * ρ * v_shower^2 + ρ * g * h_shower = P_pump
(here "shower" refers to the showerhead and "pump" refers to the pump output pipe).
- The pressure at the showerhead (P_shower) is atmospheric pressure, which is approximately 101.3 kPa.
- The density of water (ρ) is about 1000 kg/m^3.
- Acceleration due to gravity (g) is approximately 9.81 m/s^2.
- The height (h_shower) is 7.5 m above the pump output pipe.
- The velocity (v_shower) is what we need to find.
Step 2: Find the velocity of water at the showerhead
Rearranging the equation from step 1, we get:
P_shower + 1/2 * ρ * v_shower^2 + ρ * g * h_shower = P_pump
v_shower^2 = (2 * (P_pump - P_shower) + 2 * ρ * g * h_shower) / ρ
Now, substitute the values we know:
v_shower^2 = (2 * (420,000 - 101,300) + 2 * 1000 * 9.81 * 7.5) / 1000
After evaluating the right side of the equation, take the square root of both sides to get the velocity (v_shower) in m/s. This will give you the answer to part (a).
Step 3: Find the velocity of water through the pump output pipe
Next, let's find the velocity of water through the pump output pipe. We will again use Bernoulli's equation. The equation at the pump output pipe is:
P_pump + 1/2 * ρ * v_pump^2 + ρ * g * h_pump = P_output
Where P_output is the atmospheric pressure.
- The pressure at the pump output pipe (P_pump) is given as 420 kPa.
- The height at the pump output pipe (h_pump) is 0, as it is at the same level as the pump.
- The pressure at the output (P_output) is atmospheric pressure, which is approximately 101.3 kPa.
- The density of water (ρ) is about 1000 kg/m^3.
- The velocity (v_pump) is what we need to find.
Using the equation above, solve for v_pump. This will give you the answer to part (b).
By following these steps, you can find the velocities at the showerhead and through the pump output pipe.