Question

**9**. Water at a pressure of 4.50 atm at street
level flows into an office building at a speed of 0.80 m/s through
a pipe 6.00 cm in diameter. The pipes taper down to 3.00 cm in
diameter by the top floor, 27.0 m above. Calculate the water
pressure in such a pipe on the top floor.

Answer #1

**from equation of continuity**

**volume flow rate remains same in a fluid flow**

**A1*v1 = A2*v2**

**A1 = pI*r1^2**

**A2 = pi*r2^2**

**r1 = radius of larger pipe = 6/2 = 3 cm = 0.03
m**

**r2 = radius of smaller pipe = 3/2 = 1.5 cm = 0.015
m**

**v1 = 0.8 m/s**

**v2 = ?**

**pi*0.03^2*0.8 = pi*0.015^2*v2**

**v2 = 3.2 m/s**

**==================**

**from Bernoullis principle**

**P1 + (1/2)*rho*v1^2 + rho*g*h1 = P2 + (1/2)*rho*v2^2 +
rho*g*h2**

**P2 = P1 + (1/2)*rho*(v1^2 - v2^2) + rho*g*(h1 -
h2)**

**h1 = 0**

**h2 = 27m**

**P1 = 4.5 atm = 4.5*10^5 Pa**

**P2 = 4.5*10^5 + (1/2)*1000*(0.8^2-3.2^2) +
1000*9.8*(0-27)**

**P2 = 1.806*10^5 Pa
<<<<<----------ANSWER**

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