1.) water is flowing through a horizontal pipe of varying crosection, the diameter is 12.0 cm at a pressure of 8.0x10^4 Pa. at section 2, the pipe has a diameter of 6.0 cm and the pressure is 6.0x10^4 Pa. find the speed of water through section 1 and 2 (HINT: Use the continuity equation and bernoullis eqn.)
To find the speed of water through sections 1 and 2, we can use the continuity equation and Bernoulli's equation.
Step 1: Apply the continuity equation.
The continuity equation states that the product of the cross-sectional area and the speed of water at any section is constant.
A1V1 = A2V2
Step 2: Find the cross-sectional areas of sections 1 and 2.
The cross-sectional area of a pipe can be calculated using the formula A = πr^2, where r is the radius.
Given: Diameter of section 1, D1 = 12.0 cm
Radius of section 1, r1 = D1/2 = 6.0 cm = 0.06 m
Cross-sectional area of section 1:
A1 = πr1^2 = π(0.06 m)^2
Given: Diameter of section 2, D2 = 6.0 cm
Radius of section 2, r2 = D2/2 = 3.0 cm = 0.03 m
Cross-sectional area of section 2:
A2 = πr2^2 = π(0.03 m)^2
Step 3: Use the continuity equation to find the relationship between the speeds of water at sections 1 and 2.
A1V1 = A2V2
V2 = (A1V1) / A2
Step 4: Find the speed of water at section 1.
Given: Pressure at section 1, P1 = 8.0x10^4 Pa
Using Bernoulli's equation:
P1 + (1/2)ρV1^2 + ρgh1 = P2 + (1/2)ρV2^2 + ρgh2
Since the pipe is horizontal, we can ignore the term ρgh.
Also, since the two sections are at the same vertical height, h1 = h2, and the term ρgh cancels out.
This simplifies Bernoulli's equation to:
P1 + (1/2)ρV1^2 = P2 + (1/2)ρV2^2
We can rearrange this equation to solve for V1:
V1 = sqrt[((P2 - P1) / ρ) + V2^2]
Step 5: Calculate the speed of water at section 2.
Given: Pressure at section 2, P2 = 6.0x10^4 Pa
Assume the density of water, ρ = 1000 kg/m^3
Substitute the given values into the equation:
V2 = sqrt[((P2 - P1) / ρ) + V1^2]
Step 6: Substitute the values of A1, A2, and V2 into the continuity equation to find V1.
V2 = (A1V1) / A2
Solving these equations step by step will allow you to find the speed of water through section 1 and section 2.
To solve this problem, we need to use the continuity equation and Bernoulli's equation.
First, let's use the continuity equation. The continuity equation states that the flow rate of an incompressible fluid is constant along a streamline. In other words, the product of the cross-sectional area of the pipe and the velocity of the fluid remains constant.
Q1 * A1 = Q2 * A2
Where Q1 and Q2 are the flow rates at sections 1 and 2, and A1 and A2 are the cross-sectional areas at sections 1 and 2, respectively.
Let's find A1 and A2:
A1 = π * (d1/2)^2 = π * (0.12/2)^2 = 0.01131 m^2
A2 = π * (d2/2)^2 = π * (0.06/2)^2 = 0.00283 m^2
Now, we can rewrite the continuity equation as:
Q1 * 0.01131 = Q2 * 0.00283
Next, let's use Bernoulli's equation. Bernoulli's equation states that the sum of the pressure, kinetic energy per unit volume, and potential energy per unit volume is constant along a streamline. In this case, we'll neglect changes in potential energy.
For section 1:
P1 + (1/2) * ρ * v1^2 = P + (1/2) * ρ * v^2
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2 -- (Equation 1)
For section 2:
P1 + (1/2) * ρ * v1^2 = P2 + (1/2) * ρ * v2^2 -- (Equation 2)
Here, P1 and P2 are the pressures at sections 1 and 2, respectively.
ρ is the density of water.
v1 and v2 are the velocities at sections 1 and 2, respectively.
We know the values of P1, P2, and the densities of water. We need to find v1 and v2.
From Equation 1, we have:
P1 - P2 = (1/2) * ρ * (v2^2 - v1^2)
Plugging in the given values, we get:
(8.0x10^4 Pa) - (6.0x10^4 Pa) = (1/2) * ρ * (v2^2 - v1^2)
Simplifying, we have:
2.0x10^4 Pa = (1/2) * ρ * (v2^2 - v1^2) -- (Equation 3)
Now, from the continuity equation, we have:
Q1 * 0.01131 = Q2 * 0.00283
Since Q = A * v, we can rewrite the equation as:
v1 * 0.01131 = v2 * 0.00283
Now, we have two equations with two unknowns (v1 and v2). We can solve these equations simultaneously to find the values of v1 and v2.