Difference in pressure question

In summary, the conversation discusses the calculation of pressure at a point in a pipeline with varying radii. The pipeline has a capacity of 2.37*105m3/day of oil per day and the radius varies from 60 cm to 35 cm. The solution involves determining the difference in pressure between the two sections and then converting it to kPa. The attempt at a solution involves finding the speed of the flow, but the error lies in mixing up flow with speed. The correct calculation takes into account the flow and area of the sections to determine the speed and then the difference in pressure.
  • #1
grog
23
0

Homework Statement



The alaskan pipeline has a capacity of 2.37*105m3/day of oil per day. Along most of the pipeline the radius is 60 cm. Find the pressure at a point where the pipe has a 35 cm radius. Take the pressure in the section with radius 60 cm to be 160 kPa and the density of oil to be 800kg/m3. Answer in units of kPA


Homework Equations



.5 * rho * v12+ rho * g * y + p1 = .5 * rho * v22 + rho * g* y + p2

since y = 0, we end up with:


.5 * rho * v12+ p1 = .5 * rho * v22 + p2

v2 = v1 ( A1 / A2)



The Attempt at a Solution




v1 = 2.37x10^5 m^3/day = 2.743055556 m^3/sec
A1 = 60^2*pi = 11309.73355 cm^2
A2 = 35^2 * pi = 3848.451001 cm^2

v2 = v1 * (A1/A2)

p1-p2 (difference in pressure) = (.5)(rho)(v2^2) - (.5)(rho)(v1^2)

substitute:

= (.5)(rho) ((v1*A1/A2)^2) - (.5)(rho)(v1^2)

= (.5)(rho)(v1^2) [(a1/a2)^2 - 1]

plugging in:

= (0.5)(800)(2.743055556)^2 (7.636401493) = 22983.59459 Pa = p1-p2

since we're given p1 = 160 kPa, converting the difference in pressure to kPa and subtracting yields

22.98359459 = 160 - p2
p2 = 137.0164054 kPa

However, this is an in correct answer. Does anyone see where I'm making my mistake?
 
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  • #2
Hm.

Is the question asking for dynamic, static, or stagnation pressure?
 
  • #3
The problem doesn't seem to specify. how would I tell, and how would that affect my answer?
 
  • #4
Your error is in determining the speed. You seem to be mixing up flow with speed.

Since [itex]Flow = dV/dt = d/dt(Ax) = vA => v = Flow/A[/tex]

For the 60 cm section, [itex]v = Flow/\pi r^2 = 2.74/3.14*.36 = 2.42 m/sec[/itex]AM
 
Last edited:
  • #5
Thanks!
 

1. What is the difference in pressure in a gas?

The difference in pressure in a gas refers to the variation in pressure between two points within the gas. This can be caused by factors such as temperature, volume, and the number of gas molecules present.

2. How can the difference in pressure be measured?

The difference in pressure can be measured using a device called a manometer. This device typically consists of a U-shaped tube filled with a liquid, such as mercury, and is used to measure the difference in height of the liquid on each side of the tube.

3. What causes a difference in pressure in a liquid?

The main cause of a difference in pressure in a liquid is the force of gravity. The weight of the liquid exerts pressure on the bottom of the container, which increases as the depth of the liquid increases.

4. How does the difference in pressure affect the behavior of gases and liquids?

The difference in pressure affects the behavior of gases and liquids by causing them to flow from areas of high pressure to areas of low pressure. This is known as the principle of fluid dynamics and is the reason why air and water move in currents and currents.

5. Can the difference in pressure be used for practical applications?

Yes, the difference in pressure has many practical applications. It is used in devices such as barometers, which measure atmospheric pressure for weather forecasting, and in turbines, which use the difference in pressure to generate electricity. It is also used in various industrial processes, such as filtration and distillation.

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