Water is pumped at a rate of 24 cubic meter per second through a 0.50 cm radius pipe on the main floor of the house to a 0.35 cm radius pipe on the solar hot water collector 4.0 m higher on the roof. If the pressure in the pipe on the roof is 120000 Newton per square meter, what is the pressure in the large pipe on the main floor?
To determine the pressure in the large pipe on the main floor, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a pipe.
Bernoulli's equation states:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures at two different points in the fluid flow.
ρ is the density of the fluid.
v1 and v2 are the velocities at the two points.
g is the acceleration due to gravity.
h1 and h2 are the heights at the two points.
In this case, we want to find the pressure in the large pipe on the main floor (P1), given the following information:
P2 = 120000 Newton/m^2 (pressure in the pipe on the roof)
v1 = ? (velocity in the large pipe on the main floor)
v2 = ? (velocity in the smaller pipe on the roof)
h1 = 0 (since the large pipe is on the main floor)
h2 = 4.0 m (height difference between the two pipes)
First, let's find the velocities at the two points using the continuity equation, which states that the flow rate is constant through a pipe:
A1v1 = A2v2
Where:
A1 and A2 are the cross-sectional areas of the pipes.
Given:
A1 = πr1^2 (area of the large pipe)
A2 = πr2^2 (area of the smaller pipe)
r1 = 0.50 cm (radius of the large pipe)
r2 = 0.35 cm (radius of the smaller pipe)
Convert the radii from centimeters to meters:
r1 = 0.50 cm = 0.50 * 0.01 m = 0.005 m
r2 = 0.35 cm = 0.35 * 0.01 m = 0.0035 m
Substitute the values into the equation to find the velocities:
A1v1 = A2v2
πr1^2v1 = πr2^2v2
π(0.005)^2v1 = π(0.0035)^2v2
v1 = (π(0.0035)^2v2) / (π(0.005)^2)
v1 = (0.0035^2 / 0.005^2) * v2
v1 = (0.0035 / 0.005)^2 * v2
v1 = (0.7/0.5)^2 * v2
v1 = (1.4)^2 * v2
v1 = 1.96 * v2
Now we can substitute the values of v1, v2, h1, and h2 into Bernoulli's equation to solve for P1:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
P1 + (1/2)ρ(1.96v2)^2 + ρg(0) = 120000 + (1/2)ρv2^2 + ρg(4.0)
Since h1 = 0, the term ρgh1 = 0.
Simplifying the equation:
P1 + (1/2)ρ(1.96v2)^2 = 120000 + (1/2)ρv2^2 + ρg(4.0)
P1 + (1/2)ρ(1.96)^2v2^2 = 120000 + (1/2)ρv2^2 + 4ρg
Since ρ is the density of water and g is the acceleration due to gravity, we can substitute their respective values (ρ = 1000 kg/m^3, g = 9.8 m/s^2) into the equation:
P1 + (1/2)(1000)(1.96)^2v2^2 = 120000 + (1/2)(1000)v2^2 + 4(1000)(9.8)
Simplifying further:
P1 + 1960v2^2 = 120000 + 500v2^2 + 39200
P1 + 1960v2^2 - 500v2^2 = 159200
P1 + 1460v2^2 = 159200
Now, we need to recognize that the water is incompressible, so ρv1^2 = ρv2^2. This allows us to cancel out the density term:
P1 + 1460v2^2 = 159200
P1 = 159200 - 1460v2^2
Since we don't know the velocity in the smaller pipe (v2), we need more information to determine the pressure in the large pipe on the main floor.