The water flowing through a 1.9 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the three smaller pipes are 24, 18, and 13 L/min, what is the flow rate in the 1.9 cm pipe? (b) What is the ratio of the speed of water in the 1.9 cm pipe to that in the pipe carrying 24 L/min?
Now I know A1v1=A2v2 and i have been converting to SI units but I am still not getting the right answer. Would it be something like A1v1=A2(v1+v2+v3)?
vbig/vsmall = (55/24)(1.3^2/1.9^2)
His answer was backwards.
No water is lost or stored, just add.
call Q the flow per second in each pipe.
Q big pipe = 24+18+13 = 55 L/min
Q little pipe = 24 L/min
Vbig pipe pi rBig^2 = 55
Vsmall pipe pi rsmall^2 = 24
so Vsmall = 24/[pi(1.3^2 /4)]
vbig = 55 /[pi(1.9^2/4)]
vbig/vsmall = (55/24)(1.9^2/1.3^2)
Doubt
🐍🐍🐍
Yes
To solve this problem, you need to use the principle of conservation of mass. According to this principle, the mass of water entering a section of the pipe should be equal to the mass of water exiting that section. Since density is constant, you can also say that the volume flow rate (Q) into a section is equal to the volume flow rate out of that section.
Let's start by calculating the flow rate in each of the three smaller pipes (A2v2) using the formula A1v1 = A2v2.
Given:
Inside diameter of the 1.9 cm pipe (A1) = πr1^2 = π(1.9/2)^2 = 2.83 cm^2 = 0.000283 m^2.
Inside diameter of the smaller pipes (A2) = πr2^2 = π(1.3/2)^2 = 1.33 cm^2 = 0.000133 m^2.
Flow rates in three smaller pipes being 24, 18, and 13 L/min, we need to convert them to m^3/s:
24 L/min = (24/1000) L/s = 0.0004 m^3/s
18 L/min = (18/1000) L/s = 0.0003 m^3/s
13 L/min = (13/1000) L/s = 0.000217 m^3/s
Now, let's calculate the flow rate in the 1.9 cm pipe (Q1):
A2v2 = (0.000133 m^2) (0.0004 m^3/s + 0.0003 m^3/s + 0.000217 m^3/s) = 0.000133 m^2 × 0.000917 m^3/s = 0.000000122 m^3/s
So, the flow rate in the 1.9 cm pipe (Q1) is 0.000000122 m^3/s.
For part (b), we need to calculate the ratio of the speed of water in the 1.9 cm pipe to that in the pipe carrying 24 L/min.
The speed of water in the 1.9 cm pipe can be calculated as v1 = Q1 / A1:
v1 = (0.000000122 m^3/s) / (0.000283 m^2) = 0.000431 m/s
The speed of water in the pipe carrying 24 L/min can be calculated directly as v2 = Q2 / A2:
v2 = (0.0004 m^3/s) / (0.000133 m^2) = 3.008 m/s
Therefore, the ratio of the speed of water in the 1.9 cm pipe to that in the pipe carrying 24 L/min is:
v1 / v2 = 0.000431 m/s / 3.008 m/s = 0.000143
So, the ratio is approximately 0.000143.