This isn’t really homework, but it might as well be with all the calculations involved.(adsbygoogle = window.adsbygoogle || []).push({});

I am trying to figure out the difference in pressures inside a pipe carrying a fluid when the pipe goes from a ¼” diameter to 1/8” diameter when the constant flow rate is 9 Liter per minute. I got an answer, but it seems way to large to be reasonable.

I am using these two equations,

A_1 * V_2 = A_2 * V_2

P_1 + ½ (V_1)^2 = P_2 + ½ (V_2)^2

If I have some fluid flowing through a system of pipes at a constant rate of 9 Liters per minute, this would be equal to 1.5 E-4 m^3 / s.

The starting diameter of the pipe is ¼ inches (radius of 3.2 E-4 m^2) and the smallest diameter is 1/8 inches (radius of 1.6 E-4 m^3).

I know that A_1 * V_1 = 1.5 E-4 m^2/s and that A_1 = 3.17 E-7 m^2,

So that means that V_1 must equal 473.2 m/s. This seems very fast, but OK, I will accept it. When the pipe diameter is halved, the velocity of the fluid is quadrupled. So V_2 must equal 1894 m/s. This seems extremely high (over 5 times the speed of sound), but I keep going.

Now that I know the two velocities of the fluid in both parts of the pipe, I want to find the pressure difference.

I know,

P_1 + ½ (V_1)^2 = P_2 + ½ (V_2)^2, plugging in gives me,

P_1 + ½ (473.2)^2 + P_2 + ½ (1894)^2, solving for the difference in pressure,

P_1 – P_2 = ½ (473.2)^2 - ½ (1894)^2 = 1.69 E6 N/m^2

A pressure difference of 1.69 E6 N/m^2 = 1690 kPa = 16.7 atmospheres, this has got to be wrong, right? Where did I screw up?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Bernoulli's Principle Question

**Physics Forums | Science Articles, Homework Help, Discussion**