The volumetric flow rate if friction is negligible = 0.00197 m³/sec
Given data :
P₁ = 7 atm = 709.275 kPa
P₂ = 1.5 atm = 151.988 kPa
d₁ = 0.74 cm = 0.0074 m
d₂ = 1.20 cm = 0.0120 m
h = 82 m
density of water ( p ) = 1000 kJ/m³
Estimate the volumetric flow rate
given that friction losses are negligible
V₁² - V₂² = [tex]\frac{2(P_{2} - P_{1}) }{p} + 2g(h )[/tex]
= 2( 151.988 - 709.275 ) + 2 * 9.81 * 82
= 2 ( - 557.287 ) + 1608.84
= - 1114.574 + 1608.84 = 494.266
∴ V₁² - V₂² = 494.266 ----- ( 1 )
Also given that:
volumetric flow rate ( Q ) = A₁V₁ = A₂V₂
= π / 4 ( 0.0074 )² * V₁ = π / 4 ( 0.0120 )² * V₂
= 0.000043 V₁ = 0.000113 V₂
∴ V₁ = 2.63 V₂
Back to equation ( 1 )
( 2.63 v₂ )² - V₂² = 494.266
V₂² = ( 494.266 ) / 1.63
∴ V₂ = √303.23 = 17.41 m/sec
Therefore the volumetric flow rate ( Q )
Q = A₂V₂
= π / 4 * ( 0.0120 )² * 17.41
= 0.00197 m³/sec
Learn more about volumetric flow rate : https://brainly.com/question/24356835
