A horizontal pipe is shown and has an area of 40.0 cm2 in wide areas and 10.0 cm2 in narrow areas. It is also connected to a U shaped tube filled with mercury. water is flowing through the pipe at a rate of 5.00 L/s. Find (a) the flow speeds at the wide and narrow portions (b) the pressure difference between the two points (c) the height difference, h, between the mercury columns (density of mercury is 13.534 g/cm3.)
To find the flow speeds at the wide and narrow portions of the pipe, we can use the equation for flow rate:
Q = A * v
Where Q is the flow rate, A is the cross-sectional area, and v is the flow speed.
Given that the flow rate is 5.00 L/s, we need to convert it to cm^3/s:
1 L = 1000 cm^3
So, 5.00 L/s = 5000 cm^3/s
(a) For the wide portion of the pipe with an area of 40.0 cm^2:
40.0 cm^2 * v_wide = 5000 cm^3/s
v_wide = (5000 cm^3/s) / (40.0 cm^2)
v_wide ≈ 125 cm/s
Similarly, for the narrow portion of the pipe with an area of 10.0 cm^2:
10.0 cm^2 * v_narrow = 5000 cm^3/s
v_narrow = (5000 cm^3/s) / (10.0 cm^2)
v_narrow ≈ 500 cm/s
(b) To find the pressure difference between the two points, we can use Bernoulli's equation:
P1 + 0.5 * ρ * v1^2 + ρ * g * h1 = P2 + 0.5 * ρ * v2^2 + ρ * g * h2
Where P1 and P2 are the pressures at the two points, ρ is the density of the fluid, v1 and v2 are the flow speeds, g is the acceleration due to gravity, and h1 and h2 are the heights of the fluid columns.
In this case, the fluid is water, so we can assume that the pressure is atmospheric pressure at both points (P1 = P2 = Patm). The acceleration due to gravity (g) can be taken as 9.81 m/s^2.
The flow speeds at the wide and narrow portions are already calculated as v_wide = 125 cm/s and v_narrow = 500 cm/s.
Substituting these values into the equation, we can simplify it to find the pressure difference:
0.5 * ρ * v1^2 + ρ * g * h1 = 0.5 * ρ * v2^2 + ρ * g * h2
Since we only need the pressure difference, we can ignore the atmospheric pressure term:
0.5 * ρ * v1^2 = 0.5 * ρ * v2^2 + ρ * g * h2
Simplifying further:
0.5 * ρ * v1^2 - 0.5 * ρ * v2^2 = ρ * g * h2
0.5 * ρ * (v1^2 - v2^2) = ρ * g * h2
Cancelling ρ from both sides:
0.5 * (v1^2 - v2^2) = g * h2
Now, we need to convert the flow speeds from cm/s to m/s:
v_wide = 125 cm/s = 1.25 m/s
v_narrow = 500 cm/s = 5.0 m/s
Substituting the values:
0.5 * (1.25^2 - 5.0^2) = 9.81 * h2
Simplifying:
0.5 * (1.5625 - 25) ≈ -11.71875 = 9.81 * h2
Solving for h2:
h2 ≈ -11.71875 / 9.81 ≈ -1.19 m
However, height difference (h) cannot be negative. So, we take the absolute value:
h ≈ |h2| ≈ 1.19 m
(c) The height difference (h) between the mercury columns is approximately 1.19 meters.
To find the flow speeds at the wide and narrow portions of the pipe, we can use the equation of continuity. The equation states that the product of the cross-sectional area and the flow speed at any point in a pipe is constant. Mathematically, it can be represented as:
A1v1 = A2v2
Where A1 and A2 are the cross-sectional areas of the wide and narrow portions of the pipe, respectively, and v1 and v2 are the flow speeds at those points.
In this case, A1 = 40.0 cm^2 and A2 = 10.0 cm^2. We need to convert these values to square meters since the SI unit of flow rate is cubic meters per second (m^3/s).
A1 = 40.0 cm^2 = 40.0 × 10^(-4) m^2 = 0.004 m^2
A2 = 10.0 cm^2 = 10.0 × 10^(-4) m^2 = 0.001 m^2
Now we can solve for v1 and v2:
v1 = (A2v2)/A1
v1 = (0.001 m^2 × v2)/(0.004 m^2)
v1 = 0.25 × v2
Hence, the flow speed at the wide portion is 0.25 times the flow speed at the narrow portion.
Next, let's calculate the pressure difference between the two points using Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a system.
Bernoulli's equation: P1 + 1/2ρv1^2 + ρgh1 = P2 + 1/2ρv2^2 + ρgh2
In this equation, P1 and P2 are the pressures at the wide and narrow portions, respectively, ρ is the density of water, v1 and v2 are the flow speeds as calculated above, and h1 and h2 are the heights of the fluid columns (unknown).
Since the pipe is horizontal, the height difference (h1 - h2) between the two points is zero. Therefore, the last term (ρgh1 - ρgh2) cancels out, leaving us with:
P1 + 1/2ρv1^2 = P2 + 1/2ρv2^2
To find the pressure difference (ΔP = P1 - P2), we can rearrange the equation:
ΔP = 1/2ρv2^2 - 1/2ρv1^2
Now, let's calculate the pressure difference using the given values:
Density of water, ρ = 1000 kg/m^3
Flow speed at the narrow portion, v2 = 5.00 L/s = 5.00 × 10^(-3) m^3/s
Flow speed at the wide portion, v1 = 0.25 × v2 = 0.25 × 5.00 × 10^(-3) m^3/s
Substituting these values into the equation, we get:
ΔP = 1/2 × 1000 kg/m^3 × (0.25 × 5.00 × 10^(-3))^2 - 1/2 × 1000 kg/m^3 × (5.00 × 10^(-3))^2
Now we can simplify and calculate the numerical value of the pressure difference.
To find the height difference, h, between the mercury columns in the U-shaped tube, we can use the hydrostatic pressure equation. The hydrostatic pressure at a point in a fluid is given by the equation:
P = ρgh
Where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column.
Considering the U-shaped tube is filled with mercury, we use the density of mercury, ρ = 13.534 g/cm^3 = 13.534 × 10^3 kg/m^3, and the acceleration due to gravity, g = 9.8 m/s^2.
Since the pressure at the wide portion is higher than at the narrow portion, we can conclude that the pressure difference is due to the difference in height between the mercury columns.
So, ΔP = ρgh
ΔP = 13.534 × 10^3 kg/m^3 × 9.8 m/s^2 × h
Now, we have the value of ΔP from the previous calculation, so we can solve for h.
h = ΔP / (13.534 × 10^3 kg/m^3 × 9.8 m/s^2)
After substituting the value of ΔP, we can calculate the height difference, h.
I hope this explanation helps you to understand how to find the flow speeds, pressure difference, and height difference in this scenario.