A horizontal pipe carries oil whose coefficient of viscosity is 0.700 N·s/m2. The diameter of the pipe is 46.0 cm and its length is 51 km.
(a) What pressure difference is required between the ends of this pipe if the oil is to flow with an average speed of 1.5 m/s? Pa
(b) What is the volume flow rate in this case? m3/s
(c) If both the length of pipe and its diameter were each doubled in size, what would the pressure difference required be then? Pa
(d) What is the volume flow rate in this case?
m3/s
I feel i understand how to do b and D but i am unsure if it is correct please can someone help
b)
Cross section area of the pipe
= pi * r^2 = pi * (0.46 / 2)^2
= 0.16619025 m^2
Speed = 1.2 m/s, so the flow rate
1.2 * 0.16619025 = 0.1994283 m^3/s
D.
Would you take the answer from b and multiply by 4 since the diameter is doubled, the cross sectional area is increased by 2^2?
To solve these problems, we can use the equation that relates flow rate (Q) to the pressure difference (∆P), viscosity (η), and the dimensions of the pipe.
The equation is Q = (∆P * π * r^4) / (8 * η * L), where Q is the volume flow rate, ∆P is the pressure difference, π is a mathematical constant (approximately 3.14159), r is the radius of the pipe, η is the coefficient of viscosity, and L is the length of the pipe.
(a) To find the pressure difference (∆P), we rearrange the equation to solve for ∆P: ∆P = (Q * 8 * η * L) / (π * r^4). Plugging in the given values, Q = 1.5 m/s, η = 0.700 N·s/m^2, L = 51,000 m, and r = 0.46 m / 2, we can calculate:
∆P = (1.5 * 8 * 0.700 * 51,000) / (π * (0.46/2)^4) ≈ 2,002,070 Pa
Therefore, the pressure difference required is approximately 2,002,070 Pa.
(b) To find the volume flow rate, we can directly use the formula provided in the question, which is Q = 1.5 m/s times the cross-sectional area of the pipe. Plugging in the values, we have:
Q = 1.5 m/s * π * (0.46/2)^2 ≈ 0.625 m^3/s
Therefore, the volume flow rate in this case is approximately 0.625 m^3/s.
(c) If both the length of the pipe and its diameter are doubled in size, we need to calculate the pressure difference required using the new dimensions. Let's call the new length L' = 2 * 51,000 m and the new diameter d' = 2 * 0.46 m. The new radius would be r' = d' / 2.
Using the same formula as in (a), we can calculate the new pressure difference:
∆P' = (1.5 * 8 * 0.700 * 2 * 51,000) / (π * (2 * 0.46/2)^4) ≈ 500,525 Pa
Therefore, the pressure difference required with the new dimensions would be approximately 500,525 Pa.
(d) To find the volume flow rate in the new case, we can use the same formula as in (b), but with the new dimensions:
Q' = 1.5 m/s * π * (2 * 0.46/2)^2 ≈ 2.500 m^3/s
Therefore, the volume flow rate in the new case would be approximately 2.500 m^3/s.