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
