A horizontal pipe 17.00 cm in diameter has a smooth reduction to a pipe 8.50 cm in diameter. If the pressure of the water in the larger pipe is 8.00 104 Pa and the pressure in the smaller pipe is 6.00 104 Pa, at what rate does water flow through the pipes?
Use bernoulli's equation
p+ρV²+ρgh=constant
for each of the sections.
p=pressure
ρ=density
V=velocity
The ρgh term can be neglected if the pipe is horizontal.
The continuity equation gives
A1V1=A2V2, where
V1, V2, A1 and A2 are the velocities and cross sectional areas of the corresponding sections.
So we have two unknowns V1 and V2, and two equations
p1+ρV1²=p2+ρV2² ...(1)
V1*A1=V2*A2 ....(2)
Solve for V1 and V2.
how do i convert the answer to kg/s
Flow is given by the product of the water velocity (m/s) and cross sectional area (m²), multiplied by the density ρ (1000 kg/m³).
or,
Q=A1*V1*ρ
To find the rate at which water flows through the pipes, we can use the principle of continuity, which states that the product of the cross-sectional area of a pipe and the velocity of the fluid flowing through it remains constant.
Let's denote the diameter of the larger pipe as D1 = 17.00 cm and the diameter of the smaller pipe as D2 = 8.50 cm.
First, we need to find the cross-sectional areas of the pipes. The area of a circular pipe is given by the formula:
A = πr^2,
where A is the cross-sectional area and r is the radius of the pipe.
For the larger pipe, the radius is given by:
r1 = D1/2 = 17.00 cm / 2 = 8.50 cm.
For the smaller pipe, the radius is given by:
r2 = D2/2 = 8.50 cm / 2 = 4.25 cm.
Now we can calculate the cross-sectional areas of the pipes:
A1 = πr1^2 = π(8.50 cm)^2,
A2 = πr2^2 = π(4.25 cm)^2.
Next, we need to find the velocities of the water in the pipes. According to the principle of continuity, the ratio of the velocities is the inverse of the ratio of the cross-sectional areas. That is:
v1/v2 = A2/A1.
We can rearrange this equation to solve for v1:
v1 = (A2/A1) * v2.
Now let's substitute the values we have:
v1 = (A2/A1) * v2 = (π(4.25 cm)^2) / (π(8.50 cm)^2) * v2 = (4.25 cm / 8.50 cm) ^ 2 * v2.
Simplifying:
v1 = (0.5) ^ 2 * v2 = 0.25 * v2.
Now we have the relationship between the velocities.
The next step is to consider the pressure difference between the two pipes. The pressure difference drives the flow of water.
ΔP = P1 - P2 = 8.00 * 10^4 Pa - 6.00 * 10^4 Pa = 2.00 * 10^4 Pa.
According to Bernoulli's principle, the pressure difference is related to the difference in velocity by the equation:
ΔP = (1/2) * ρ * (v2^2 - v1^2),
where ρ is the density of water.
Rearranging the equation to solve for the velocity of the smaller pipe, v2:
v2^2 - v1^2 = (2 * ΔP) / ρ,
v2^2 = v1^2 + (2 * ΔP) / ρ,
v2 = √(v1^2 + (2 * ΔP) / ρ).
Finally, substitute the relationship we found earlier for v1:
v2 = √((0.25 * v2)^2 + (2 * ΔP) / ρ).
Now we can solve this equation for v2. Since the equation is not linear, we will have to solve it iteratively using numerical methods or use trial and error.
Alternatively, we can simplify the problem by assuming the water is incompressible and neglecting any losses due to friction or other factors. In this case, the flow rate through both pipes is the same, and we can calculate it using the formula:
Q = A1 * v1 = A2 * v2,
where Q is the flow rate, A1 and A2 are the cross-sectional areas, v1 is the velocity in the larger pipe, and v2 is the velocity in the smaller pipe.
Now we can substitute the values we have:
Q = A1 * v1 = A2 * v2 = (π(8.50 cm)^2) * v2.
Solving for v2:
v2 = Q / (A2 * π(8.50 cm)^2).
Now we can calculate the flow rate by substituting the known values:
Q = (8.00 * 10^4 Pa - 6.00 * 10^4 Pa) / ((π(4.25 cm)^2) * π(8.50 cm)^2).
Evaluate this expression to find the flow rate.
Note: It is important to double-check the units and convert them if necessary to ensure consistency throughout the calculations.