The 3.0 -diameter water line in the figure splits into two 1.0 -diameter pipes. All pipes are circular and at the same elevation. At point A, the water speed is 2.0 and the gauge pressure is 50 .What is the gauge pressure at point B?
I know we have to find P in kPa, but am unsure how to set this problem up or what to plug in where. Any help would be great
To solve this problem, we can use the principle of continuity and Bernoulli's equation.
First, let's denote the gauge pressure at point B as P_B. According to Bernoulli's equation, the sum of the pressure, kinetic energy per unit volume, and potential energy per unit volume along a streamline remains constant.
The equation can be written as:
P + 1/2ρv^2 + ρgh = constant
Since the pipes are at the same elevation, the potential energy term, ρgh, cancels out.
Now let's consider the principle of continuity. According to the principle of continuity, the mass flow rate of an incompressible fluid remains constant in a steady-state flow. This can be written as:
A_1v_1 = A_2v_2
where A_1 and A_2 are the cross-sectional areas of the pipes, and v_1 and v_2 are the velocities of the water in the respective pipes.
In this problem, the water line with a diameter of 3.0 splits into two pipes with diameters of 1.0 each. The relationship between the cross-sectional areas of the pipes and their diameters is given by:
A = πr^2
where r is the radius of the pipe.
In the wider pipe with a diameter of 3.0, the radius is 3.0/2 = 1.5. So, the cross-sectional area is:
A_1 = π(1.5)^2
In each of the narrower pipes with a diameter of 1.0, the radius is 1.0/2 = 0.5. So, the cross-sectional area is:
A_2 = π(0.5)^2
Using the principle of continuity, we can rewrite the equation as:
A_1v_1 = A_2v_2
(π(1.5)^2)v_1 = (π(0.5)^2)v_2
(1.5^2)v_1 = (0.5^2)v_2
2.25v_1 = 0.25v_2
v_1 = (0.25/2.25)v_2
v_1 = (1/9)v_2
Now we have expressions for v_1 in terms of v_2, and we can substitute these expressions in Bernoulli's equation to solve for P_B.
P_A + 1/2ρv_A^2 = P_B + 1/2ρv_B^2
We are given that P_A = 50 and v_A = 2.0. We need to find P_B.
Substituting the values into Bernoulli's equation:
50 + 1/2(ρ)(2.0)^2 = P_B + 1/2(ρ)v_B^2
50 + 2.0^2 = P_B + v_B^2
Simplifying:
50 + 4 = P_B + v_B^2
54 = P_B + v_B^2
Since we know that v_1 = (1/9)v_2, we can substitute this expression into the equation:
54 = P_B + [(1/9)v_2]^2
54 = P_B + (1/81)v_2^2
Now we need to find an expression for v_2 in terms of v_1:
v_1 = (1/9)v_2
v_2 = 9v_1
Substituting this expression into the equation:
54 = P_B + (1/81)(9v_1)^2
54 = P_B + (1/81)(81v_1^2)
54 = P_B + v_1^2
Rearranging the equation:
P_B = 54 - v_1^2
Finally, substitute v_1 = 2.0 into the equation to find the gauge pressure at point B:
P_B = 54 - (2.0)^2
P_B = 54 - 4
P_B = 50 kPa
Therefore, the gauge pressure at point B is 50 kPa.
To solve this problem, we can use the principle of continuity of fluids, which states that the product of the cross-sectional area and the fluid velocity is constant along a streamline.
Let's denote the cross-sectional area and velocity at point A as A1 and v1, respectively, and at point B as A2 and v2, respectively. Since the pipes have different diameters, we can use the relationship between the cross-sectional area and the diameter (A = πr^2) to express the areas in terms of the radii.
Given:
- Diameter of the 3.0-diameter water line = 3.0
- Speed at point A (v1) = 2.0 m/s
- Gauge pressure at point A (P1) = 50 kPa
First, we need to find the cross-sectional areas A1 and A2. Considering the formula A = πr^2, the radius of the 3.0-diameter water line is 1.5 (half of the diameter). Thus, the cross-sectional area at point A (A1) is π(1.5)^2 = 7.07 m^2.
Since the 3.0-diameter line splits into two 1.0-diameter pipes, the radius of each 1.0-diameter pipe is 0.5. Therefore, the cross-sectional area at point B (A2) for each 1.0-diameter pipe is π(0.5)^2 = 0.79 m^2.
Using the principle of continuity, we have the equation:
A1v1 = A2v2
Substituting the known values, we have:
(7.07 m^2)(2.0 m/s) = (2 * 0.79 m^2)v2
Simplifying, we find:
14.14 m^3/s = 1.58 m^2v2
Now, we need to find the velocity at point B (v2):
v2 = (14.14 m^3/s) / (1.58 m^2)
v2 = 8.95 m/s
Next, we can use Bernoulli's equation to determine the gauge pressure at point B (P2). Bernoulli's equation relates the fluid's pressure, density, and velocity at different points along a streamline.
The equation is:
P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2
We can assume the fluid is incompressible and at the same elevation, so the term (1/2)ρv^2 cancels out.
Simplifying the equation, we have:
P2 = P1 + (1/2)ρ(v1^2 - v2^2)
To convert the speed from m/s to km/h, we multiply v1 and v2 by 3.6:
v1 = 2.0 m/s * 3.6 = 7.2 km/h
v2 = 8.95 m/s * 3.6 = 32.22 km/h
Assuming the density of water (ρ) is 1000 kg/m^3, we can substitute the values into the equation:
P2 = 50 kPa + (1/2)(1000 kg/m^3)(7.2^2 - 32.22^2 m^2/s^2) (Note: The units need to be consistent throughout.)
Calculating the expression inside the parenthesis, we have:
(1/2)(1000)(51.84 - 1038.5684) m^2/s^2 = (1/2)(1000)(-986.7284) m^2/s^2 = -493,364.2 m^2/s^2
Finally, substituting this value into the equation:
P2 = 50 kPa - 493,364.2 m^2/s^2
Converting kPa to Pa, we have:
P2 = 50,000 Pa - 493,364.2 m^2/s^2
Therefore, the gauge pressure at point B is 50,000 Pa - 493,364.2 m^2/s^2.