I'm not sure how to approach this problem. The teacher wants volumetric flow rate (in cm).
A commonly used rule of thumb is that average velocity in a pipe should be about 1 m/s or less for "thin"(viscosity about water). If a pipe needs to deliver 6,000 m^3 of water a day, what diameter is required to satisfy the 1 m/s rule
if the diameter is d meters, the cross-section of the pipe has area pi/4 d^2 m^2
In one second, at 1 m/s, then, pi/4 d^2 m^3/s of water will flow through the pipe
now just plug in the numbers:
6000m^3/day * 1day/86400s = pi/4 d^2 m^3/s
d = 1/3 √(5/2pi) = 0.297m
To find the required diameter of the pipe, we can use the formula for volumetric flow rate:
Q = A * V
Where:
Q is the volumetric flow rate (in m^3/s),
A is the cross-sectional area of the pipe (in m^2),
V is the average velocity of the fluid (in m/s).
We are given that the volumetric flow rate needs to be 6,000 m^3/day, which can be converted to m^3/s by dividing by the number of seconds in a day.
1 day = 24 hours * 60 minutes * 60 seconds = 86,400 seconds
Therefore, the volumetric flow rate is:
Q = 6,000 m^3/day / 86,400 s = 0.0694 m^3/s
The average velocity is limited to 1 m/s. We can rearrange the formula to solve for the cross-sectional area:
A = Q / V
A = 0.0694 m^3/s / 1 m/s = 0.0694 m^2
Since we want the diameter of the pipe in cm, we can convert the area to cm^2 by multiplying by 10,000 (1 m^2 = 10,000 cm^2):
A = 0.0694 m^2 * 10,000 = 694 cm^2
The area of a circle is given by the formula:
A = π * r^2
Where:
A is the area of the circle (in cm^2),
π is a mathematical constant approximately equal to 3.14159,
r is the radius of the circle (in cm).
We can rearrange the formula to solve for the radius:
r = √(A / π)
r = √(694 cm^2 / 3.14159) ≈ √(221.22) ≈ 14.88 cm
Finally, the diameter (D) of the pipe is twice the radius:
D = 2 * r = 2 * 14.88 cm ≈ 29.76 cm
Therefore, to satisfy the 1 m/s rule, a pipe with a diameter of approximately 29.76 cm is required to deliver 6,000 m^3 of water per day.
To approach this problem, we can use the relationship between volumetric flow rate, cross-sectional area, and average velocity.
1. First, let's convert the volumetric flow rate from cubic meters per day to cubic centimeters per second.
- 6,000 m^3/day = 6,000,000,000 cm^3/day
- Since there are 24 hours in a day, we divide by 24 to get the flow rate per hour: 6,000,000,000 cm^3/day / 24 hours = 250,000,000 cm^3/hour
- Since there are 3600 seconds in an hour, we divide by 3600 to get the flow rate per second: 250,000,000 cm^3/hour / 3600 seconds = 69,444.44 cm^3/s
2. Now, let's calculate the cross-sectional area of the pipe using the formula: area = volumetric flow rate / velocity.
- We are given that the average velocity should be 1 m/s.
- Convert the velocity to centimeters per second: 1 m/s = 100 cm/s
- Divide the volumetric flow rate (69,444.44 cm^3/s) by the velocity (100 cm/s) to get the cross-sectional area: 69,444.44 cm^3/s / 100 cm/s = 694.44 cm^2
3. Finally, we can calculate the diameter of the pipe using the formula: diameter = 2 * sqrt(area / π).
- Substitute the calculated area (694.44 cm^2) into the formula: diameter = 2 * sqrt(694.44 cm^2 / π)
- Calculate the square root: diameter = 2 * sqrt(221.22 cm^2)
- Simplify: diameter = 2 * 14.86 cm
- Result: The required diameter of the pipe is approximately 29.72 cm (rounded to two decimal places).
Therefore, to satisfy the 1 m/s rule with a flow rate of 6,000 m^3 of water per day, you would require a pipe with a diameter of approximately 29.72 cm.