Consider a fully developed laminar flow of a fluid through a 8027 m long and 4 cm diamete horizontal and circular pipe. The dynamic density and the viscosity of the fluid are 1252 kg/m³ and 0.3073 kg/(m.s). The velocity profile at a cross-section is given by:

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6
Consider a fully developed laminar flow of a fluid through 8027 m long and 4 cm diameter
horizontal and circular pipe. The dynamic density and the viscosity of the fluid are 1252 kg/m³ and
0.3073 kg/(m.s). The velocity profile at a cross-section is given by:
u(r)= 6[1 − ( 7 )²]
-
m/s
Where r is the axial distance from the centre and R is the radius of the pipe.
Determine the following:
(i)
(ii)
the maximum velocity at a cross-section of the pipe, Umax
the average velocity at a cross-section of the pipe, Vave
the volume flow rate, Q
(iv)
Reynolds number, Re
(v)
friction factor, f
(vi)
head loss, hi
(vii)
pressure loss, AP
(viii)
pumping power required,
(ix)
for the same pumping power, the percentage decrease of the flow rate if the pipe is
inclined 10° upward (assume the head loss, hò, calculated in part (vi) does not change)
Useful formulae:
64
f
Re
h₂ = f ( =) ( 2² )
2g
=
Transcribed Image Text:6 Consider a fully developed laminar flow of a fluid through 8027 m long and 4 cm diameter horizontal and circular pipe. The dynamic density and the viscosity of the fluid are 1252 kg/m³ and 0.3073 kg/(m.s). The velocity profile at a cross-section is given by: u(r)= 6[1 − ( 7 )²] - m/s Where r is the axial distance from the centre and R is the radius of the pipe. Determine the following: (i) (ii) the maximum velocity at a cross-section of the pipe, Umax the average velocity at a cross-section of the pipe, Vave the volume flow rate, Q (iv) Reynolds number, Re (v) friction factor, f (vi) head loss, hi (vii) pressure loss, AP (viii) pumping power required, (ix) for the same pumping power, the percentage decrease of the flow rate if the pipe is inclined 10° upward (assume the head loss, hò, calculated in part (vi) does not change) Useful formulae: 64 f Re h₂ = f ( =) ( 2² ) 2g =
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