Bernoulli Equation (SI unit)
Bernoulli's law is an important principle used in fluid mechanics, describing the relationship between the velocity and pressure of a fluid. According to this law, as speed increases, pressure decreases, and as speed decreases, pressure increases. This principle is expressed by the following formula:
Energy basis : P1/ρ1 + u1^2/2 + z1 = P2/ρ2 + u2^2/2 + z2 [Pa, kg/m3, m/s, m]Pressure basis : P1 + ρ1*u1^2/2 + ρ1*g*z1 = P1 + ρ2*u2^2/2 + ρ2*g*z2 [Pa, kg/m3, m/s, m]
Head basis : P1/(ρ1*g) + u1^2/2g + z1 = P2/(ρ2*g) + u2^2/2g + z2 [Pa, kg/m3, m/s, m]
Where,
ρ = fluid density (kg/m3)
g = acceleration due to gravity (9.8 m/s)
P1 = pressure at elevation-1 (Pa)
u1 = velocity at elevation-1 (m/s)
h1 = height of elevation-1 (m)
P2 = pressure at elevation-2 (Pa)
u2 = velocity at elevation-2 (m/s)
h2 = height at elevation-2 (m)
Example of Bernoulli Equation (SI unit)
Water at 25 degC enters a pipe at the velocity of 1 m/s and a pressure of 101.3 kPa. The pipe has a constant diameter and friction loss is negligible. A change in the pipe’s elevation changes the downstream pressure in the pipe to 2.1 atm. Most nearly, what is the elevation change?
P1/(ρ1*g) + u1^2/2g + z1 = P2/(ρ2*g) + u2^2/2g + z2u1 = u2, ρ1 = ρ2
P1/(ρ1*g) + z1 = P2/(ρ2*g) + z2
101.3 kPa = 101,300 Pa
2.1 atm = 212,730 Pa
z2 - z1 = (P1 - P2)/ρ*g = (101,300 Pa - 212,730 Pa)/(997 kg/m3 * 9.81 m/sec) = -11.4 m
Bernoulli Equation (Imperial unit)
Pressure basis : P1 + ρ1*u1^2/2gc + ρ1*g*z1/gc = P1 + ρ2*u2^2/2gc + ρ2*g*z2/gc [lbf/ft2, lbm/ft3, ft/s, ft, 32.2 lbm-ft/lbf-sec2, 32.2 ft/sec2]
Head basis : P1*gc/(ρ1*g) + u1^2/2g + z1 = P2*gc/(ρ2*g) + u2^2/2g + z2 [lbf/ft2, lbm/ft3, ft/s, ft, 32.2 lbm-ft/lbf-sec2, 32.2 ft/sec2]
ρ = fluid density (lbm/ft3)
g = acceleration due to gravity (32.2 ft/sec2)
P1 = pressure at elevation-1 (lbf/ft2, = psia * 144 in2/ft2)
u1 = velocity at elevation-1 (ft/s)
h1 = height of elevation-1 (ft)
P2 = pressure at elevation-2 (lbf/ft2, = psia * 144 in2/ft2)
u2 = velocity at elevation-2 (ft/s)
h2 = height at elevation-2 (ft)
Example of Bernoulli Equation (Imperial unit)
The pump supplies water at a flow rate of 4.0 ft3/sec from an open reservoir through a horizontal 8 inch pipe. the head loss from the reservoir to the suction of the pump is 2 ft-lbf/lbm. and the discharge is at 90 psi in to 6 inch pipe (5.761 inch). The pump has an efficiency of 65%. The head that must be delivered by the pump to water is?
ρ1 = ρ2 = 62.4 lbm/ft3
u1 = 0
z1 = 0
z2 = 0
hf = 0
P2 - P1 = 90 psi = 90 lbf/in2 = 12,960 lbf/ft2
6 inch pipe diameter = 3.14*(5.761 inch * (1 ft / 12 inch)/2)^2 = 0.181 ft2
u2 = 4.0 ft3/sec / 0.181 ft2 = 22.1 ft/sec
gc = 32.2 lbm-ft/lbf-sec2,
g = 32.2 ft/sec2
hpump = (P2-P1)*gc/(ρ*g) + u2^2/2g = 12,960 lbf/ft2 / 62.4 lbm/ft3 + 22.1^2 / (2 * 32.2) = 215 ft
