This isn’t really homework, but it might as well be with all the calculations involved. 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?