Question

Water at a gauge pressure of P = 4.2 atm at street level flows into an office building at a speed of 0.78 m/s through a pipe 5.2 cm in diameter. The pipe tapers down to 2.2 cm in diameter by the top floor, 16 m above.

Calculate the flow velocity in the pipe on the top floor.

Calculate the gauge pressure in the pipe on the top floor.

Answer 1

Using equation of continuity

A1 v1 = A2 v2

v2 = v1 ( r1 / r2) ^2

v2 = 0.78 ( 5.2/2.2)^2

v2 = 4.36 m/s

========

b)

Considering Bernoulli's principle

P1 + rho * g h1 + 0.5 rho* v1^2 = P2 + rho* g h2 + 0.5 rho * v2^2

P1 - P2 = 1000*9.8* 10 + 0.5* 1000* (4.36^2 - 0.78^2)

4.2*10^5 - P2 = 1.07*10^5

P2 = 3.128*10^5 Pa

========

Do comment in case any doubt, will reply for sure . Goodluck

Answer 2

SOLUTION :

Flow rate (m^3 /s) remains same at the inlet and outlet to maintain the flow.

So,

A1 v1 = A2 v2

So, v2 

= (A1/A2) v1 

= (d1/d2)^2 v1 

= (5.2/2.2)^2 * 0.78

= 4.36 m/s (ANSWER)

As per energy conservation, 

pressure head + potential head + velocity head = constant

So,

p1/w + 0 + v1^2 /2g = p2/w + 16 + v2^2 /2g

=> 4.2*1.01325*10^5 / 10^3 + 0 + 0.78^2 / (2*9.8)) = p2/10^3 + 16 + 4.36^2 / (2*9.8)

=> 425.60 = p2*10^(-3) + 16.97

=> p2 = (425.60 - 16.97)*10^3 Pa = 408630 Pa

=> p2 = 4.03 atm. gauge  (ANSWER)