A 6.0-cm diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32 kPa and 24 kPa, respectively. What is the volume rate of flow?
I know to use Bernoulli's Equation and the Equation of Continuity, but I can't figure out how to solve for either velocity variable.
horrible
To solve this problem, you can use Bernoulli's equation and the equation of continuity. Let's start by using the equation of continuity.
The equation of continuity states that the volume flow rate (Q) of an incompressible fluid is constant along a pipe. Mathematically, it can be expressed as:
Q1 = Q2
where Q1 and Q2 are the volume flow rates at two different sections of the pipe.
We can express the volume flow rate as the product of the cross-sectional area (A) and the velocity (v) of the fluid:
Q = A * v
Now, let's apply this equation to our problem. We'll use subscripts 1 and 2 to represent the two sections of the pipe.
Since the diameter of the pipe at the first section is 6.0 cm, its radius (r1) can be calculated as r1 = 6.0 cm / 2 = 3.0 cm = 0.03 m.
Similarly, the diameter of the pipe at the second section is 4.0 cm, so its radius (r2) can be calculated as r2 = 4.0 cm / 2 = 2.0 cm = 0.02 m.
Now, we can calculate the cross-sectional areas A1 and A2:
A1 = π * r1^2
A2 = π * r2^2
Next, let's consider Bernoulli's equation:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
where P1 and P2 are the gauge pressures at the two sections, ρ is the density of water, v1 and v2 are the velocities at the two sections, g is the acceleration due to gravity, and h1 and h2 are the heights at the two sections.
In this problem, the pipe is horizontal, so the heights h1 and h2 are the same and cancel out in the equation.
Also, the water density ρ is approximately constant, so we can simplify the equation to:
P1 + 1/2 * ρ * v1^2 = P2 + 1/2 * ρ * v2^2
Now, we have equations for the volume flow rates (Q1 = A1 * v1 and Q2 = A2 * v2) and the pressures (P1 and P2).
Combining these equations, we get:
A1 * v1 = A2 * v2
P1 + 1/2 * ρ * v1^2 = P2 + 1/2 * ρ * v2^2
We can now solve these two equations simultaneously to find the velocity variables v1 and v2.
After you've found the values of v1 and v2, you can substitute those values back into the equation Q = A * v to calculate the volume flow rate Q.
To solve for the velocity variables in this problem, we can use the principles of fluid dynamics, specifically Bernoulli's equation and the equation of continuity.
First, let's consider the equation of continuity, which states that the volume rate of flow (Q) is constant at any point along the pipe. Mathematically, this can be expressed as:
A1v1 = A2v2
Where:
A1 and A2 are the cross-sectional areas of the pipe at sections 1 and 2, respectively, and
v1 and v2 are the velocities of the water at sections 1 and 2, respectively.
Here, we know the diameter of the pipe at both sections, so we can use the formula for the cross-sectional area of a circle to find A1 and A2:
A = πr^2
Where r is the radius of the circle. Using this formula, we can find the areas A1 and A2.
Given that the diameter of the pipe at section 1 is 6.0 cm, the radius (r1) can be calculated by dividing the diameter by 2:
r1 = 6.0 cm / 2 = 3.0 cm = 0.03 m
Similarly, the diameter at section 2 is 4.0 cm, so the radius (r2) would be:
r2 = 4.0 cm / 2 = 2.0 cm = 0.02 m
Using the formula for the cross-sectional area of a circle, we have:
A1 = π(0.03 m)^2 and A2 = π(0.02 m)^2
Now that we have the areas A1 and A2, we can solve the equation of continuity for the velocities v1 and v2:
v1 = (A2/A1) * v2
Next, let's consider Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a system. Mathematically, it can be expressed as:
P1 + 0.5ρv1^2 + ρgh1 = P2 + 0.5ρv2^2 + ρgh2
Where:
P1 and P2 are the pressures at sections 1 and 2, respectively,
ρ is the density of the fluid (in this case, water),
g is the acceleration due to gravity,
v1 and v2 are the velocities of the fluid at sections 1 and 2, respectively,
and h1 and h2 are the heights of the fluid at sections 1 and 2, respectively.
In this problem, both sections are at the same height, so the terms ρgh1 and ρgh2 cancel out. Given the values of the pressures (32 kPa and 24 kPa) and knowing ρ (the density of water, which is approximately 1000 kg/m^3), we can solve Bernoulli's equation for the velocity terms:
0.5ρv1^2 = P2 - P1 + 0.5ρv2^2
Now, we have two equations:
A1v1 = A2v2 and 0.5ρv1^2 = P2 - P1 + 0.5ρv2^2
By substituting the first equation into the second equation, we can eliminate v1 and solve for v2:
0.5ρ((A2/A1)v2)^2 = P2 - P1 + 0.5ρv2^2
After simplifying and isolating v2, we can solve for the velocity at section 2:
v2 = sqrt((2(P1 - P2))/(ρ((A2/A1)^2 -1)))
Substituting the given values (ρ = 1000 kg/m^3, P1 = 32 kPa - convert to Pa, P2 = 24 kPa - convert to Pa, A1 and A2 calculated earlier), we can calculate the value of v2.
Finally, to find the volume rate of flow (Q), we can substitute the values of v2 and A2 into the equation of continuity:
Q = A2 * v2
By calculating this expression, you will get the volume rate of flow through the pipe.
P1 - P2 = (rho/2)(V2^2 - V1^2)= 8000 N/m^2 (Bernoulli equation)
(rho is the fluid density)
6^2*V1 = 4^2*V2, so V2 = (4/9)V1
(from the continuity equation)
Substitute (4/9)V1 for V2 in the first equation and solve for V1
The volume flow rate is (pi/4)D^2*V1
where D = 0.06 m