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ashwinanand
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Homework Statement
Gingerale (ρ=1100 kg/m3) moves through a system of pipes in a beverage factory. Initially, it travels through a pipe with a cross-sectional area of 3.80 cm2 at a speed of 4.77 m/s. Gradually, the gingerale descends by 8.49 m while the pipe's cross-sectional area increases to 8.15 cm2. The speed of the gingerale at the lower level is 2.22m/s. What is the pressure at the lower level if the pressure in the upper level is known to be 1.50x10^5Pa?
Homework Equations
delta P = P2-P1
P1+(1/2)*rho*(v^2)+rho*g*y
upper level pressure=1.50x10^5Pa
The Attempt at a Solution
P1 + 0.5(1100kg/m^3)(4.77m/s^2)+(1100)(9.8m/s^2)(8.49m)=P2 + 0.5(1100)(2.22m/s^2)+(1100)(9.8m/s^2)(8.49m)
P1 + 1.04x10^5=P2 + 9.42x10^4
(deltaP)P2-P1=9.80x10^3Pa
upper level pressure- deltaP
1.50x10^5Pa-9.80x10^3Pa = 1.40x10^5Pa
I am confused of if I had used bernoulli's equation right and if the proper height was used. I can't seem to get the right pressure.