solution:-
given data
density of glycerin (ρ ) =1252 kg/m^3
viscosity of glycerin(μ )=0.3073 kg/m-s
diameter of pipe (D) =0.04 m
length of pipe (L)=70 m
maximum velocity(umax)=6 m/s
the velocity profile in fully developed laminar flow is
u(r)=umax(1−R2r2)
=6(1−(0.02)2r2)
u(r)=6(1−2500r2)m/s
average velocity
Va=2umax=26=3m/s
flow rate
Q=VaA
Q=3×π×4(0.04)2
Q=3.77×10−3m3/s
renolds number can be given
Re=μρVaD
=0.30731252×3×0.04
Re=488.9
Re<2300
so its laminar flow.
therefore friction factor
f=Re64=488.964=0.1309
and head loss
H=fD2gLVa2
H=0.04×2×9.80.1309×70×32
H=105.1m
by using the Bernoulli equation pressure difference can be written as
ΔP=P1−P2=ρg(Z2−Z1+H)
pipe is horizontal so Z1 and Z2 will be zero.
ΔP=1252×9.8(0+105.1)
ΔP=1291×103Pa
pumping power can be given as
pumping power=Q×ΔP
=3.77×10−3×1291×103
pumping power=4.87×103W
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