Water discharges from a horizontal, cylindrical pipe at the rate of 5.00x10^-3 m^3/s. At a point in the pipe where the cross-section is 1.000x10^-3 m^2, the absolute pressure is 1.60x10^5 Pa. What is the pipe radius at a constriction if the pressure there is reduced to 1.20*10^5 Pa?
I know that we are to find the velocity first, which I got 8.944 m/s approximately. Then after I found the radius and the answer I got was 0.0000236, is this correct ?
To solve this problem, we can start by using Bernoulli's equation, which relates the pressure and velocity of a fluid in a pipe.
Bernoulli's equation is given by:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
where:
P1 and P2 are the pressures at points 1 and 2,
v1 and v2 are the velocities at points 1 and 2,
ρ is the density of the fluid, and
g is the acceleration due to gravity.
Since the fluid is water, which has a constant density, we can simplify the equation to:
P1 + 1/2 * ρ * v1^2 = P2 + 1/2 * ρ * v2^2
We can assume that the velocity of the water at the constriction (point 2) is larger than at the initial point (point 1) since the pressure is reduced. Therefore, we can ignore the terms related to the height or gravitational potential energy.
First, let's calculate the velocity at the initial point (point 1):
P1 = 1.60x10^5 Pa
v1 = ? (unknown)
P2 = 1.20x10^5 Pa
v2 = 8.944 m/s (given)
Now, rearranging the equation, we have:
1/2 * ρ * v1^2 = P2 - P1 + 1/2 * ρ * v2^2
We know the density of water, which is approximately 1000 kg/m^3. Substituting the known values, we get:
1/2 * 1000 kg/m^3 * v1^2 = (1.20x10^5 - 1.60x10^5) Pa + 1/2 * 1000 kg/m^3 * (8.944 m/s)^2
Simplifying and solving for v1, we find:
v1^2 = [(1.20x10^5 - 1.60x10^5) Pa + 1/2 * 1000 kg/m^3 * (8.944 m/s)^2] / 1000 kg/m^3
v1^2 ≈ [(-0.40x10^5) Pa + 1/2 * 1000 kg/m^3 * (79.952 m^2/s^2)] / 1000 kg/m^3
v1^2 ≈ (-0.40x10^5) Pa / 1000 kg/m^3 + 1/2 * 1000 kg/m^3 * (79.952 m^2/s^2) / 1000 kg/m^3
v1^2 ≈ -0.40x10^2 m^2/s^2 + 1/2 * 79.952 m^2/s^2
v1^2 ≈ -0.40x10^2 m^2/s^2 + 39.976 m^2/s^2
v1^2 ≈ 39.576 m^2/s^2
Taking the square root of both sides, we find:
v1 ≈ sqrt(39.576 m^2/s^2)
v1 ≈ 6.288 m/s
Now, let's calculate the radius at the constriction (point 2) using the equation for the volumetric flow rate:
Q = A * v
where:
Q is the volumetric flow rate (given as 5.00x10^-3 m^3/s),
A is the cross-sectional area (given as 1.000x10^-3 m^2), and
v is the velocity at the constriction (point 2) (known as 8.944 m/s).
We can rearrange the equation:
v = Q / A
Substituting the known values, we can solve for the radius (r2):
8.944 m/s = 5.00x10^-3 m^3/s / (π * r2^2)
Rearrange and solve for r2:
r2^2 = (5.00x10^-3 m^3/s) / (8.944 m/s * π)
r2^2 ≈ 0.000560885 m^2 / (28.0796 m^2/s)
r2^2 ≈ 1.999x10^-5
Taking the square root of both sides, we find:
r2 ≈ sqrt(1.999x10^-5 m^2)
r2 ≈ 0.00447 m (rounded to 5 significant figures)
So the calculated radius at the constriction is approximately 0.00447 m or 4.47 mm.
Your answer of 0.0000236 seems incorrect. Please check your calculations and make sure to use consistent units throughout the calculations.