I have no idea what to do, and the teacher wants the 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?
To find the diameter of the pipe required to satisfy the 1 m/s rule, you need to use the formula for volumetric flow rate and rearrange it to solve for the diameter.
The volumetric flow rate (Q) can be calculated using the formula:
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)
To convert the volumetric flow rate from m^3/day to m^3/s, you need to divide the flow rate value by the number of seconds in a day (24 hours * 60 minutes * 60 seconds = 86,400 seconds).
Q = (6,000 m^3/day) / (86,400 s/day)
Now, considering the 1 m/s rule of thumb, you can solve for the diameter (D) using the rearranged formula:
Q = (π/4) * D^2 * V
Since the velocity (V) is given as 1 m/s, plug in the known values:
(6,000 m^3/day) / (86,400 s/day) = (π/4) * D^2 * 1
Simplify the equation and solve for the diameter:
D^2 = (4 * (6,000 m^3/day) / (86,400 s/day)) / π
D^2 = 0.027 m^2
Taking the square root of both sides, you can find D:
D = √0.027 m^2
Since the question asks for the diameter in cm, you can convert the result to cm by multiplying it by 100:
D = √0.027 m^2 * 100 cm/m
Therefore, the required diameter to satisfy the 1 m/s rule is approximately 16.43 cm.