Hydraulic question- A pipe line 60cm diameter bifurcates at a Y-juncti...
Diameter pipelines. If the fluid flow rate in the main pipeline is 1000 liters per second, what is the flow rate in each of the branch pipelines?
Assuming the fluid is incompressible and the flow is steady, we can apply the principle of conservation of mass, which states that the mass (and thus volume) of fluid entering a junction must equal the mass (and volume) of fluid leaving the junction. In other words, the total flow rate in the main pipeline must equal the sum of the flow rates in the two branch pipelines.
Let Q1 be the flow rate in the 60cm diameter pipeline, and Q2 and Q3 be the flow rates in the two 40cm diameter pipelines. Then we have:
Q1 = 1000 L/s (given)
Q1 = Q2 + Q3 (conservation of mass)
The flow rate in a pipe is proportional to the square of its diameter (assuming the same velocity and viscosity), so we can use the following formula to relate the flow rates:
(Q1/Q2) = (D1/D2)^2
(Q1/Q3) = (D1/D3)^2
where D1, D2, and D3 are the diameters of the main pipeline and the two branch pipelines, respectively.
Substituting the given values, we get:
(Q1/Q2) = (60/40)^2 = 2.25
(Q1/Q3) = (60/40)^2 = 2.25
Multiplying both sides of the first equation by Q2, we get:
Q2 = Q1/2.25 = 444.4 L/s (approximately)
Similarly, we can obtain Q3:
Q3 = Q1/2.25 = 444.4 L/s (approximately)
Therefore, the flow rate in each of the branch pipelines is approximately 444.4 L/s.
Hydraulic question- A pipe line 60cm diameter bifurcates at a Y-juncti...
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