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!