TABLE OF CONTENTS A piping system is conveying 10 ft 3/s (0.28 m 3/s) of ethanol. At a particular cross section of the system, section 1, the pipe diameter is 12 in (0.30 m), the pressure is 18 lb/in 2 (124 kPa), and the elevation is 140 ft (42.7 m). At another cross section further downstream, section 2, the pipe diameter is 8 in (0.20 m), and the elevation is 106 ft (32.3 m). If there is a head loss of 9 ft (2.74 m) between these sections due to pipe friction, what is the pressure at section 2? Assume that the specific gravity of the ethanol is 0.79.
1. Compute the velocity at each section. Use the equation of continuity
where Q is volumetric rate of flow, A is cross-sectional area, V is velocity, and the subscripts refer to sections 1 and 2. Now, A=( ?/4 )d 2 , where d is (inside) pipe diameter, so A 1=( ?/4)(1ft) 2 =0.785 ft 2, and A 2=( ?/4)(8/12 ft) 2=0.349ft 2; and Q=10 ft 3/s. Therefore, V 1=10/0.785=12.7 ft/s, and V 2=10/0.349=28.7 ft/s.
2. Compute the pressure at section 2. Use Bernoulli s theorem, which in one form can be written as
where g is the acceleration due to gravity, 32.2 ft/s 2; p is pressure;
