A large water tank has an inlet pipe and an outlet pipe. The inlet pipe has a diameter of 2.19 cm and is 1.25 m above the bottom of the tank. The outlet pipe has a diameter of 5.93 cm and is 4.85 m above the bottom of the tank. A volume of 0.411 m^3 of water enters the tank every minute at a gauge pressure of 1 atm.
a) What is the velocity of the water in the outlet pipe?
b) What is the gauge pressure in the outlet pipe?
It seems that water is being pumped into an opening near the bottom of the tank and being removed from an outlet higher above. Is this supposed to be a steady state situation?
If so,
(a) V*(outlet pipe area)= (volume flow rate at inlet)
(b) The guage pressure difference is (rho)*g*h
where h is the difference of heights of inlet and outlet
A bat flying toward a wall at a speed of 6.47 m/s emits an ultrasound wave with a frequency of 27.5 kHz. What frequency does the reflected wave have when it reaches the flying bat? (Assume that the speed of sound is v = 341 m/s.)
To find the velocity of the water in the outlet pipe, we can use the Bernoulli's equation, which states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume of a fluid remains constant along a streamline.
The Bernoulli's equation can be written as:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
Where:
P1 is the pressure at point 1 (inlet pipe)
v1 is the velocity of the water at point 1 (inlet pipe)
h1 is the height of the water at point 1 (inlet pipe)
P2 is the pressure at point 2 (outlet pipe)
v2 is the velocity of the water at point 2 (outlet pipe)
h2 is the height of the water at point 2 (outlet pipe)
ρ is the density of the water
g is the acceleration due to gravity
Let's calculate the velocity of the water in the outlet pipe (v2) using the given information:
a) Velocity of the water in the outlet pipe (v2):
- P1 = 1 atm = 1.013 * 10^5 Pa (atmospheric pressure converted to Pascal)
- P2 = ?
- ρ = density of water = 1000 kg/m^3 (assuming at room temperature)
- g = acceleration due to gravity = 9.8 m/s^2
- h1 = 1.25 m
- h2 = 4.85 m
From the given information, we are missing the pressure at point 2 (P2). To find P2, we can use the hydrostatic pressure formula (pressure due to the height of the water column):
P2 = P1 + ρ * g * (h1 - h2)
Substituting the values:
P2 = 1.013 * 10^5 + 1000 * 9.8 * (1.25 - 4.85)
P2 = 1.013 * 10^5 - 39215
P2 = 1.013 * 10^5 - 39215 = 62085 Pa
Now that we have the pressure at point 2 (P2), we can substitute the known values into the Bernoulli's equation to find v2:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
1.013 * 10^5 + 0.5 * 1000 * v1^2 + 1000 * 9.8 * 1.25 = 62085 + 0.5 * 1000 * v2^2 + 1000 * 9.8 * 4.85
v2 = sqrt((1.013 * 10^5 + 0.5 * 1000 * v1^2 + 1000 * 9.8 * 1.25 - 62085 - 1000 * 9.8 * 4.85) / 0.5 * 1000)
Now you can plug in the value of v1 (inlet velocity) and solve for v2 using the equation above.
b) To find the gauge pressure in the outlet pipe, we can use the hydrostatic pressure formula again:
P_gauge = P - P_atm
Where:
P_gauge is the gauge pressure
P is the absolute pressure
P_atm is the atmospheric pressure
Since the question mentions that the water enters the tank at a gauge pressure of 1 atm, the absolute pressure can be calculated by adding the atmospheric pressure to the gauge pressure:
P = P_gauge + P_atm
P = 1 atm + 1.013 * 10^5 Pa
Now substituting the known values into the equation, you can find the gauge pressure in the outlet pipe (P_gauge).