Question
Fluid flows through pipe that has two sections. The fluid travels through a larger section # 1 and then through a smaller section # 2. Each section of the pipe has a device which allows us to monitor the pressure in that section of the pipe.
Determine the following:
a) Volume flow rate of fluid through the pipe
b) Speed which the fluid travels through section # 1 and section # 2 of the pipe
c) Difference in pressure between the two sections of the pipe
d) Difference in height of liquid in the two pressure monitoring devices
Explanation / Answer
Given Diameter of section1 of the pipe, d1 = 1.00 *10^-1 m Section 2 diameter , d2 = 2.50 *10^-2 m Mass flow rate , M/ t = 9.42 kg /s Density of the fluid in the pipe , = 1.2 * 10^3 kg /m^3 a) Volume flow rate, Q = Volume / time = (Mass / density )/ time = (M/ t) *(1/) = (9.42 kg /s) * (1 / 1.2 *10^3 kg / m^3) = 7.85 *10^-3 m^3 / s b) From the equation of continuity , Q = A1 v1 ( 7.85 *10^-3 m^3 / s) = ( d1^2 / 4) v1 v1 = 4 * ( 7.85 *10^-3 m^3 / s) / 3.14 *(1.00 *10^-1 m)^2 v1 = 1 m/s is the speed of the fluid in section 1 Similarly, Q =( d2^2 / 4) v2 v2 = 4 * ( 7.85 *10^-3 m^3 / s) / 3.14 *(2.50 *10^-2 m)^2 v2 = 16 m/s is the speed of the fluid in section 2 c) Using Bernoulli's equation, P1 - P2 = 1/2 (v2^2 - v1^2) [ treat the pipe as horizontal] = 1/2 * (1.2 * 10^3 kg /m^3) [(16 m/s)^2 - (1 m/s)^2] P1 - P2 = 1.53 *10^5 Pa d) Hieght of the liquid is given as P1 - P2 = g h h = (1.53 *10^5 Pa) / ((1.2 * 10^3 kg /m^3) (9.8 m/s^2) h = 13.0 m d) Hieght of the liquid is given as P1 - P2 = g h h = (1.53 *10^5 Pa) / ((1.2 * 10^3 kg /m^3) (9.8 m/s^2) h = 13.0 m