a liquid with density of 1.36*10^3 kg/m^3 through two horizontal sections of round tubing joined end to end. in first section, diameter of the tubing is 10cm and the flow speed is 230cm/s and the pressure is 1.3*10^5, the diameter is 6 cm
a the cross sectional area in each tube
b the flow speed in smaller tubing
c the pressure in the smaller tubing
To find the cross-sectional area, flow speed, and pressure in the smaller tubing, you can use the principle of continuity.
The principle of continuity states that the mass flow rate of an incompressible fluid is constant within a closed system. This means that the mass flow rate of the fluid entering one section of the tubing will be the same as the mass flow rate leaving the other section of the tubing.
Let's start with the first section of tubing, where the diameter is 10 cm and the flow speed is 230 cm/s.
a) To find the cross-sectional area of this section, you can use the formula:
A = π * (d/2)^2
where A is the cross-sectional area, π is a mathematical constant (approximately 3.14159), and d is the diameter.
Plugging in the values, we get:
A = π * (10 cm / 2)^2
= π * 5^2
= π * 25 cm^2
Using the approximate value of π as 3.14159, the cross-sectional area is approximately:
A ≈ 3.14159 * 25 cm^2
≈ 78.54 cm^2
b) To find the flow speed in the smaller tubing, we will use the principle of continuity. The cross-sectional area of the tubing in the first section is known (78.54 cm^2), and we can assume that the mass flow rate is constant. From the principle of continuity:
A1 * V1 = A2 * V2
where A1 and V1 are the cross-sectional area and flow speed in the first section, and A2 and V2 are the cross-sectional area and flow speed in the second section.
Plugging in the known values, we get:
(78.54 cm^2) * (230 cm/s) = A2 * V2
Solving for V2, we find:
V2 = (78.54 cm^2 * 230 cm/s) / A2
Since we are given the diameter of the smaller tubing (6 cm), we can use the same formula as before to find its area:
A2 = π * (6 cm / 2)^2
= π * 3^2
= π * 9 cm^2
Plugging in the values, we get:
V2 = (78.54 cm^2 * 230 cm/s) / (π * 9 cm^2)
≈ 2011.7 cm^3/s
Therefore, the flow speed in the smaller tubing is approximately 2011.7 cm/s.
c) To find the pressure in the smaller tubing, we can use Bernoulli's equation, which relates the pressure, flow speed, and height of a fluid at two different points in a streamline. Assuming there are no significant changes in height, we can neglect the height terms and use the simplified form of Bernoulli's equation:
P1 + 0.5 * ρ * V1^2 = P2 + 0.5 * ρ * V2^2
where P1 and V1 are the pressure and flow speed in the first section, and P2 and V2 are the pressure and flow speed in the second section. The density of the liquid is given as 1.36 * 10^3 kg/m^3.
Plugging in the known values, we get:
(1.3 * 10^5 N/m^2) + 0.5 * (1.36 * 10^3 kg/m^3) * (230 cm/s)^2
= P2 + 0.5 * (1.36 * 10^3 kg/m^3) * (2011.7 cm/s)^2
Converting the flow speeds from cm/s to m/s by dividing by 100:
(1.3 * 10^5 N/m^2) + 0.5 * (1.36 * 10^3 kg/m^3) * (2.3 m/s)^2
= P2 + 0.5 * (1.36 * 10^3 kg/m^3) * (20.117 m/s)^2
Simplifying and solving for P2, we find:
P2 ≈ (1.3 * 10^5 N/m^2) + 692.64 N/m^2
≈ 130,692.64 N/m^2
Therefore, the pressure in the smaller tubing is approximately 130,692.64 N/m^2 (or Pascal).
To find the answers to your questions, we can use the equation of continuity and Bernoulli's equation for fluid flow.
a) The equation of continuity states that the product of the cross-sectional area and the flow speed remains constant for an incompressible fluid in steady flow. Therefore, we can set up the following equation:
A1v1 = A2v2
Where A1 and A2 are the cross-sectional areas of the first and second sections of tubing, and v1 and v2 are the flow speeds in the first and second sections of tubing, respectively.
We can convert the given measurements to the appropriate units before substitution:
In the first section:
d1 = 10 cm = 0.1 m (diameter)
r1 = 0.05 m (radius)
v1 = 230 cm/s = 2.3 m/s (flow speed)
In the second section:
d2 = 6 cm = 0.06 m (diameter)
r2 = 0.03 m (radius)
Now we can substitute these values into the equation:
A1 * 2.3 = A2 * v2
We know that the cross-sectional area of a round tube can be calculated using the formula:
A = πr^2
Substituting the values of r1 and r2 into the equation:
π(0.05)^2 * 2.3 = π(0.03)^2 * v2
Simplifying:
0.00715 = 0.0002827 * v2
Now solve for v2:
v2 = 0.00715 / 0.0002827
b) Now that we have found the flow speed in the smaller tubing (v2), we can substitute this value into the equation of continuity to find the cross-sectional area of the smaller tube (A2):
A1 * 2.3 = A2 * 0.00715 / 0.0002827
We already know the value of A1:
A1 = π(0.05)^2
Simplifying:
π(0.05)^2 * 2.3 = A2 * 0.00715 / 0.0002827
Now solve for A2:
A2 = π(0.05)^2 * 2.3 * 0.0002827 / 0.00715
c) Finally, to find the pressure in the smaller tubing, we can use Bernoulli's equation. Bernoulli's equation states that the total pressure of a fluid is constant along a streamline. In this case, we can assume that the fluid is incompressible and the flow is steady, so Bernoulli's equation can be simplified to:
P1 + 1/2 * ρ * v1^2 = P2 + 1/2 * ρ * v2^2
Where P1 and P2 are the pressures at the first and second sections of tubing, ρ is the density of the fluid, and v1 and v2 are the flow speeds at the first and second sections of tubing, respectively.
We already know the value of P1:
P1 = 1.3 * 10^5 Pa
Substituting all the known values into the equation:
1.3 * 10^5 + 0.5 * (1.36 * 10^3) * (2.3)^2 = P2 + 0.5 * (1.36 * 10^3) * (v2)^2
Simplifying:
1.3 * 10^5 + 0.5 * 1.36 * 10^3 * 5.29 = P2 + 0.5 * 1.36 * 10^3 * (v2)^2
Now solve for P2 by substituting the calculated value of v2:
P2 = 1.3 * 10^5 + 0.5 * 1.36 * 10^3 * 5.29 - 0.5 * 1.36 * 10^3 * (v2)^2