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9. Water at a pressure of 4.80 atm at street level flows into an office building

ID: 2032933 • Letter: 9

Question

9. Water at a pressure of 4.80 atm at street level flows into an office building at a speed of 0.85 m/s through a pipe 6.20 cm in diameter. The pipes taper down to 2.30 em in diameter by the top floor, 23.0 m above. Calculate the water pressure in such a pipe on the top floor. Pa 10. The blood speed in a normal segment of a horizontal artery is 0.16 m/s. An abnormal segment of the artery is narrowed by an arteriosclerotic plaque down to one-eighth its normal cross-sectional area. What is the difference in blood pressures between the normal and constricted segments of the artery? (The density of the blood is 1060 kg/m) Pa

Explanation / Answer

9)

From equation of continuity

A1V1 = A2V2

V2 = (A1/A2)V1

A = pi*d2/4

V2 = (d1/d2)2V1

V2 = (6.2/2.3)2*0.85

V2 = 2.291 m/s

From bernoulli's theorem

P1 + (1/2)pV12 + pgh1 = P2 + (1/2)pV22 + pgh2

P1 + (1/2)*1000*0.852 + 1000*9.81*0 = P2 + (1/2)*1000*2.2912 + 1000*9.81*23

P1 - P2 = 2.279 x 105 pa

10)

As we know that

AV = Constant

Therefore

A*0.16 = (A/8)*V

V = 1.28 m/s

According to the Bernoulli Equation

P + pgh + 1/2*p*v^2 = Constant

Therefore

Pressure Difference = 0.5*p*(v^2 - v1^2)

P = 1/2*1060*(1.28^2 - 0.16^2)

P = 854.784 Pa