Question

Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of 0.50m/s through a 4.0-cm diameter pipe in the basement under a pressure of 3.03x10^5 Pa, what will be the velocity and pressure in a 2.6-cm diameter pipe on the second floor 5.0m above?

Answer 1

Answer:

velocity and pressure in a 2.6-cm:

P2 = 2.53x10^5Pa, v2 = 1.18m/s

Explanation:

Pressure = P, Velocity = v, Height = h, Diameter = d, Radius= r, Area = A

Area = πr^2

From the question:

v1 = 0.5m/s

d1 = 4cm = 0.04m

r1 = d1/2 = 0.04/2 = 0.02m

Since water was pumped from basement, h1 = 0m

P1 = 3.03x10^5 Pa

A1 = π×0.02×0.02

A1 = 0.0004πm^2

v2 = unknown

d2 = 2.6cm = 0.026m

r2 = d2/2 = 0.026/2 = 0.013m

h2 = 5m

P2 = unknown

A2 = π×0.013×0.013

A2 = 0.000169πm^2

Using continuity equation:

A1v1 = A2v2

0.0004π * 0.5 = 0.000169π * v2

v2 = (0.0004π * 0.5)/(0.000169π)

v2 = 1.18m/s

Applying a Bernoulli principle

P + 1/2*density*v^2 + density*g*h =C

C = constant

P1 + 1/2*density*v1^2 + density*g*h1

= P2 + 1/2*density*v2^2 + density*g*h2

Let g = 9.81m/s

density of water = 1000kg/m^3

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

(P1-P2) = 1/2* density(v2^2 - v1^2) +

density* g(h2-h1)

(3.03x10^5 - P2)= 1/2*1000 (1.18^2-0.5^2) + 1000(9.81(5-0))

(3.03x10^5 - P2) = 500(1.3924-0.25) + 49050

3.03x10^5 - P2 = 571.2 + 49050

3.03x10^5 - P2 = 49621.2

3.03x10^5 - 49621.2 = P2

P2 = 253378.8

P2 = 2.53x10^5Pa

P2 = 2.53x10^5Pa, v2 = 1.18m/s