Answer to Question #234072 in Mechanical Engineering for Randal Rodriguez

Question

8. A vertical composite fluid column whose upper end is open to the atmosphere is composed of 20

inches of mercury (S.G = 13.45), 30 inches of water (density = 62.4 lb m /ft 3 ) and 35 inches of oil (S.G =

0.825). Determine the absolute pressure in psia: a) at the base of the column, b) at the oil-water

interface, c) at the water-mercury interface.

Answer 1

Part a

Let P₃abs = absolute pressure at the bottom of column $= P_3 + Patm$

$P_3abs = P_3 + Patm = 81698 Pa + 101325 Pa = 183023 Pa\\ P_3abs = 183023 Pa = (183023 * 0.000145038) psai = 26.545 psai\\ P_3abs = 26.5 psai$

Part b

Let $P_1$ = gauge pressure at oil-water interface = $d_3 g h_3$

$P_1 = d_3 g h_3 = 825 x 9.81x 0.889 = 7195 Pa\\ P_1 = 7195 Pa$

Let $P_2$ = gauge pressure at water-mercury interface =$P1 + d_2 g h_2$

$P_2 = P1 + d_2 g h_2 = 7195 Pa + (1000 x 9.81 x 0.762 ) = 14670 Pa\\ P_2 = 14670 Pa$

Let P₃ = gauge pressure at the bottom of column = $P_2 + d_1 g h_1$

$P_3 = P_2 + d_1 g h_1 =14670 Pa + (13450 x 9.81 x 0.508 ) = 81698 Pa\\ P_3 = 81698 Pa$

Let P₁abs = absolute pressure at oil-water interface = P₁ + P_atm

$P_{1}abs = P_1 + P_{atm} = 7195 Pa + 101325 Pa = 108520 Pa\\ P_{1}abs = 108520 Pa = (108520 * 0.000145038) psai = 15.739 psai\\ P_1abs = 15.7 psai$

Part c

Let P₂abs = absolute pressure at water-mercury interface = $P_2 + Patm$

$P_2abs = P_2 + Patm = 14670 Pa + 101325 Pa = 115995 Pa\\ P_2abs = 115995 Pa = (115995 * 0.000145038) psai = 16.824 psai\\ P_2abs = 16.8 psai$