Question

Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed

of 0.50 m/s through a 4.0-cm diameter pipe in the basement under a pressure of



5 3.03 10 Pa,

the

velocity is

11.83m /s

in a 2.6-cm diameter pipe on the second floor 5.0 m above, its preesure is

_______

5 10 Pa​

Answer 1

where is the ans I want ans

Answer 2

Given:-

In a Hot water heating system , water is circulating  and having Speed of water pumping (v1)= 0.50 m/s , through a 4cm diameter pipe in the basement under a pressure (p1)=53.0310 Pa, the velocity (v2) is 11.83m/s through a 2.6 cm diameter pipe on the second floor 5 m above

To find:-

The Pressure p2

Solution:-

For solving this problem we have to apply "Bernoulli's principle equation"

• p1+ 1/2 ρv1^2 + ρgh1  =  p2 + 1/2 ρv2^2 + ρgh2    (let say eq1)

here , v1= 0.50m/s , v2= 11.83 m/s , p1= 53.0310Pa , h1= 0m (basement)

h2= 5m (second floor) , d1= 4cm = 0.04m , d2= 2.6cm = 0.026 m ,

density of water(ρ) = 1* 10^3 Kg/m^3

• Now by putting all the values in the above equation1 we get

           p2= p1 +  1/2 ρv1^2 + ρgh1 - 1/2 ρv2^2 - ρgh2

p2 = (53.0310Pa) + 1/2 (1*10^3 Kg/m^3) [ (0.50m/s)^2 - (11.83m/s)^2] - ( 1* 10^3 Kg/m^3) ( 9.8 m/s^2) [ 5m-0]

p2 = 53.0310 - 69.84 *10^3 - 49 * 10 ^3

   =>    53.0310 - 118.84 *10^3

   =>    -118786.969 Pa

  =>    - 118.786 KPa

So the required pressure value p2 =  - 118.786 KPa