What is the pressure difference between points A and B (point where glycerol and air heads meet) in the picture, if d1 = 0.50 meters, d2 = 0.25 meters, and d3 = 0.71 meters and d4 = 0.86 meters. (ρH2O= 1.00 x 10^3 kg/m^3, ρgly = 1.25 x 10^3/m^3)
I know that pressure difference can be found through
P2-P1=ρ2gh2 - ρ1gh1
The Attempt at a Solution
I am confused because there are 3 different substances involved, glycerol, air and water.
What I did was I took the pressure at the bottom of the system to be the h0=0 (reference height) and calculated Pressure at A and Pressure at B independently and then subtracted them.
Pressure at point A = ρH2O*g*d1 + ρglycerol*g*d2
Pressure at point A = (1000kg/m^3 * 9.8 m/s^2 * .5m) + (1250kg/m^3 * 9.8m/s * .25m)
Pressure at point A = 7960 Pa
Pressure at point B = ρglycerol*g*d2 + ρglycerol*g*d3 + ρglycerol*g*d4sin45
Pressure at point B = (1250kg/m^3 * 9.8m/s^2 * .25m) + (1250kg/m^3 * 9.8m/s * .71m) + (1250kg/m^3 * 9.8m/s^2 *.86m*sin45)
Pressure at point B = 19200 Pa
Pressure difference = Pressure at point B - Pressure at point A
Pressure difference = 11200 Pa
Does this seem like a reasonable approach and answer?
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