# Homework Help: Difference in pressure question

1. Dec 9, 2008

### grog

1. The problem statement, all variables and given/known data

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

2. Relevant 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)

3. 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?

2. Dec 9, 2008

### aerospaceut10

Hm.

Is the question asking for dynamic, static, or stagnation pressure?

3. Dec 9, 2008

### grog

The problem doesn't seem to specify. how would I tell, and how would that affect my answer?

4. Dec 9, 2008

### Andrew Mason

Your error is in determining the speed. You seem to be mixing up flow with speed.

Since $Flow = dV/dt = d/dt(Ax) = vA => v = Flow/A[/tex] For the 60 cm section, [itex]v = Flow/\pi r^2 = 2.74/3.14*.36 = 2.42 m/sec$

AM

Last edited: Dec 9, 2008
5. Dec 9, 2008

Thanks!