Question

Water is flowing at the rate of 1.676 m/s in a 3.068-inch diameter horizontal pipe at a pressure p1 of 68.9 kPa. It then passes to a pipe having an inside diameter of 2.067 inches. The density of the water is 998 kg/m^3. a) Calculate the new pressure p2 in the 2.067-inch pipe. Assume no friction losses. b) If the piping is vertical and the flow is upward, calculate the new pressure p2. The pressure tap for p2 is 0.457 m above the tap for p1. c) Compare and discuss your answers in 2(a) and 2(b).

          Water is flowing at the rate of 1.676 m/s in a 3.068-inch diameter horizontal pipe at a pressure p1 of 68.9 kPa. It then passes to a pipe having an inside diameter of 2.067 inches. The density of the water is 998 kg/m^3.

a) Calculate the new pressure p2 in the 2.067-inch pipe. Assume no friction losses.
b) If the piping is vertical and the flow is upward, calculate the new pressure p2. The pressure tap for p2 is 0.457 m above the tap for p1.
c) Compare and discuss your answers in 2(a) and 2(b).
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Water is flowing at the rate of 1.676 m/s in a 3.068-inch diameter horizontal pipe at a pressure p1 of 68.9 kPa. It then passes to a pipe having an inside diameter of 2.067 inches. The density of the water is 998 kg/m^3. a) Calculate the new pressure p2 in the 2.067-inch pipe. Assume no friction losses. b) If the piping is vertical and the flow is upward, calculate the new pressure p2. The pressure tap for p2 is 0.457 m above the tap for p1. c) Compare and discuss your answers in 2(a) and 2(b).
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Transcript

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00:01 All right, so let's say we have a horizontal pipe where the diameter in one section is 3 .068 inches, and the velocity is 1 .676 meters per second, and the pressure is 68 .9 ,000 pascals.
00:21 And then it narrows to a diameter of 2 .067 inches.
00:29 And we want to know what is the new pressure at that side.
00:34 So the change in pressures is basically going to, since it's a horizontal pipe, it's going to be one -half row v -prime squared minus v -squared, because it only comes from the difference in the flow rates.
00:47 And v -prime is going to be d over d -prime squared times v.
00:53 So we can write the change in pressures in total is one -half row times v squared, times d over d prime square sorry not squared fourth minus one so if we plug in the numbers we have 3 .068 divided by 2 .067 to the fourth minus one times 1 .676 squared divided by 2 .676 squared by 2 times 998 the difference is going to be 5 ,401 .4 pascals and so the pressure at 0 .2 is going to be that much smaller than the pressure on the other side.
01:34 So it'll be about 62 .6 ,000 pascals for rounding.
01:40 And then part b asks, if the pipe is vertical and the flows upwards, calculate the new pressure...
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