A large water tank has an inlet pipe and an outlet pipe. The inlet pipe has a diameter of 2.19 cm and is 2.13 m above the bottom of the tank. The outlet pipe has a diameter of 5.49 cm and is 4.85 m above the bottom of the tank. A volume of 0.359 m3 of water enters the tank every minute at a gauge pressure of 1 atm.
What is the gauge pressure in the outlet pipe?
of you divide the volume of water/min by the areas, you immediately know the velocities in the inlet and outlet pipes.
Assume the inlet pressure is atmospheric, use Bernoulli's equation then to find outlet pressure. You are given the two heights.
To find the gauge pressure in the outlet pipe, we need to consider the difference in height and the difference in pipe diameters between the inlet and outlet.
First, let's calculate the hydrostatic pressure due to the difference in height between the two pipes. The water pressure at any point in a fluid column is given by the formula:
P = ρ * g * h
Where P is the pressure, ρ is the density of water, g is the acceleration due to gravity, and h is the height of the column.
Since the difference in height between the inlet and outlet pipe is 4.85 m - 2.13 m = 2.72 m, we can calculate the difference in pressure due to height:
ΔP_height = ρ * g * Δh
Where ΔP_height is the pressure difference due to height, Δh is the height difference, and ρ and g are constants.
The density of water, ρ, is approximately 1000 kg/m^3, and the acceleration due to gravity, g, is approximately 9.8 m/s^2. Substituting these values into the equation:
ΔP_height = 1000 kg/m^3 * 9.8 m/s^2 * 2.72 m
Calculating this expression gives us the pressure difference due to height, which equals 26432 Pa.
Next, let's consider the pressure difference due to the difference in pipe diameters. The flow rate of water can be related to the pipe diameter according to the principle of continuity, which states that mass flow rate is constant in an incompressible fluid:
A1 * v1 = A2 * v2
Where A1 and A2 are the cross-sectional areas of the inlet and outlet pipes respectively, and v1 and v2 are the velocities of water in the pipes.
The cross-sectional area of a pipe is given by the formula:
A = π * r^2
Where A is the cross-sectional area and r is the radius of the pipe.
Given that the diameters of the inlet and outlet pipes are 2.19 cm and 5.49 cm respectively, we can calculate the cross-sectional areas:
A1 = π * (2.19 cm / 2)^2
A2 = π * (5.49 cm / 2)^2
Substituting these values into the equation, we get:
A1 * v1 = A2 * v2
Now, we know that the volume of water entering the tank every minute is 0.359 m^3. To find the velocity of water in the inlet pipe, we can rearrange the formula:
v1 = (0.359 m^3) / A1
Similarly, to find the velocity of water in the outlet pipe, we can rearrange the formula:
v2 = (0.359 m^3) / A2
Now that we have the velocities, we can calculate the pressure difference due to the difference in pipe diameters using Bernoulli's equation, which states that the sum of pressure, kinetic energy, and potential energy per unit volume is constant along a streamline:
ΔP_diameter = ρ * (v2^2 - v1^2) / 2
Plugging in the values for the densities and velocities we calculated earlier, we get:
ΔP_diameter = 1000 kg/m^3 * ((0.359 m^3 / A2)^2 - (0.359 m^3 / A1)^2) / 2
After calculating this expression, we find that the pressure difference due to the difference in pipe diameters is approximately 2751 Pa.
To find the gauge pressure in the outlet pipe, we need to sum up the pressure difference due to height and the pressure difference due to the difference in pipe diameters:
ΔP_total = ΔP_height + ΔP_diameter
Substituting the calculated values, we find:
ΔP_total = 26432 Pa + 2751 Pa
Calculating this expression, we get the total pressure difference, which is approximately 29183 Pa.
Finally, to find the gauge pressure in the outlet pipe, we need to convert the pressure difference to units of atm. There are 101325 Pa in 1 atm, so we divide the pressure difference by this conversion factor:
Pressure_in_outlet_pipe = ΔP_total / 101325
Calculating this expression, we find the gauge pressure in the outlet pipe to be approximately 0.29 atm.