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2020-12-29

A 6cm diameter horizontal pipe gradually narrows to 4cm. Whenwater flows throught this pipe at a certain rate, the guagepressure in these two sections is 32.0kPa and 24kPa respectively.What is the volume rate of flow?

Answer & Explanation

Tasneem Almond

Skilled2020-12-30Added 91 answers

diameters d = 6 cm
d' = 4 cm
radii $r = \frac{d}{2} = 3 c m = 0.03 m$
$r^{'} = \frac{d^{'}}{2} = 2 c m = 0.02 m$
Gauge pressures P ' = 32 kPa $= 32 \times 10^{3} P a$
P = 24 kPa $= 24 \times 10^{3} P a$
from bernoulli's equation $(\frac{1}{2}) D [v^{' 2} - v^{2}] = P^{'} - P$
$(\frac{1}{2}) D [v^{' 2} - {(A^{'} v \frac{^{'}}{A})}^{2}] = P^{'} - P$
$(\frac{1}{2}) D [v^{' 2} - (r^{' 2} v \frac{^{'}}{r^{2}}] = P^{'} - P$ since Area $A = π \times r^{2}$
where D = density of water $= 1000 k \frac{g}{m^{3}}$
$500 [v^{' 2} - 0.444 v^{' 2}] = 8 \times 10^{3} P a$
$0.5555 v^{' 2} = 16$
v ' = 5.366 m / s
volume flow rate = A ' v ' $= Π \times r^{' 2} \times v^{'} = 6.74 \times 10^{- 3} m^{4} = 6.74 l i t$

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