Water is flowing down through the pipe shown in the drawing. point a is 0.410 m higher than
b. the speed of the water at a and b are va = 5.00 m/s and vb = 3.24 m/s. determine the difference pb - pa in pressures between b and
a. the density of water is 1.00 × 103 kg/m3.

Bernoulli’s principle can be stated in the form:

(1/2)v^2 + gh + p/rho = constant

where v = velocity

 g= 9.8 m/s^2

h = height = 0.41 m

p = pressure

rho = density  = 1.00 × 103 kg/m3

In this case we can equate points A and B:

(1/2)vA^2 + ghA + pA/rho =(1/2)vB^2 + ghB + pB/rho

Rearranging the equation to get PB-PA and substituting the given values:

pB – pA = ρ (0.5vA^2 - 0.5vB^2 + g(hA-hB))

pB – pA = 1000 (0.5*5^2 – 0.5*3.24^2 + 9.8*0.41)

pB-pA = 11,269.2 Pa = 11.3 kPa

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