Asoft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 cans, 0.355 - l each, per minute. at point 2 in the pipe, the gauge pressure is 152 kpa and the cross-sectional area is 8.00 cm2. at point 1, 1.35m above point 2, the cross-sectional area is 2.00 cm2. find the gauge pressure at point 1 (in kpa).

Flow rate = 220*0.355 l/m = 78.1 l/min = 1.3 l/s = 0.0013 m^3/s

Point 2:
A2= 8 cm^2 = 0.0008 m^2
V2 = Flow rate/A2 = 0.0013/0.0008 = 1.625 m/s
P1 = 152 kPa = 152000 Pa

Point 1:
A1 = 2 cm^2 = 0.0002 m^2
V1 = Flow rate/A1 = 0.0013/0.0002 = 6.5 m/s
P1 = ?
Height = 1.35 m

Applying Bernoulli principle;
P2+1/2*V2^2/density = P1+1/2*V1^2/density +density*gravitational acceleration*height
=>152000+0.5*1.625^2*1000=P1+0.5*6.5^2*1000+1000*9.81*1.35
=> 153320.31 = P1 + 34368.5
=> P1 = 1533210.31-34368.5 = 118951.81 Pa = 118.95 kPa

Earthplanet yplanet y5=earth     *many different equationsplanet x1.5=earthplanet x is two times planet yobject 5,2500 mass5,1,3,2,4object 1 is four times obect 4