flow rate = Q = pi r^2 v
= amount per second that flows through the pipe
so
Q = pi (.563)^2 (13.5)
if no oil has been added or subtracted or compressed then Q is the same everywhere
so
pi (D^2/4) (5.65) = Q
A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead the pipe's diameter is 56.3 cm (0.563 m) and the flow speed of the petroleum is 13.5 m/s. At the refinery the petroleum flows at 5.65 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe\'s diameter at the refinery?
3 answers
round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.7 cm
and the flow speed of the petroleum is 11.1 m/s.
At the refinery, the petroleum flows at 5.51 m/s.
What is the volume flow rate of the petroleum along the pipe, and what is the pipe's diameter at the refinery?
and the flow speed of the petroleum is 11.1 m/s.
At the refinery, the petroleum flows at 5.51 m/s.
What is the volume flow rate of the petroleum along the pipe, and what is the pipe's diameter at the refinery?
Using the same formula as before, we have:
Volume flow rate at wellhead = Q = pi (0.577/2)^2 * 11.1 = 0.1391 m^3/s
Since the volume flow rate is constant along the pipe, we can set up another equation:
pi (D/2)^2 * 5.51 = 0.1391
Solving for D, we get:
D/2 = sqrt(0.1391/(pi*5.51)) = 0.236
So the diameter of the pipe at the refinery is:
D = 0.472 m = 47.2 cm
Therefore, the volume flow rate of the petroleum is 0.1391 m^3/s and the pipe's diameter at the refinery is 47.2 cm.
Volume flow rate at wellhead = Q = pi (0.577/2)^2 * 11.1 = 0.1391 m^3/s
Since the volume flow rate is constant along the pipe, we can set up another equation:
pi (D/2)^2 * 5.51 = 0.1391
Solving for D, we get:
D/2 = sqrt(0.1391/(pi*5.51)) = 0.236
So the diameter of the pipe at the refinery is:
D = 0.472 m = 47.2 cm
Therefore, the volume flow rate of the petroleum is 0.1391 m^3/s and the pipe's diameter at the refinery is 47.2 cm.