Hewitt and Robert Flow Regime | Assignment

Running Head: HEWITT AND ROBERT FLOW REGIME
HEWITT AND ROBERT FLOW REGIME
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HEWITT AND ROBERT FLOW REGIME 2
Question 1
Calculating the boundary conditions
𝛼𝐿 = 0.25𝑚𝑖𝑛 [ 1.0, ( 𝐷∗
22.22)𝑆
] Equation 1
Where 𝛼 is the void fraction
𝛼𝐵𝑆 = 𝛼𝐿 ( 𝐺𝑚 ≤ 2000 𝑘𝑔/𝑚2𝑠) Equation 2
𝛼𝐵𝑆 = 𝛼𝐿 + 0.001(𝐺𝑚 − 2000)(0.5 − 𝛼𝐿) Equation 3
for (2000 < 𝐺𝑚 < 3000 𝑘𝑔
𝑚2𝑆)
𝛼𝐵𝑆 = 0.5 for 𝐺𝑚 ≤ 3000 𝑘𝑔
𝑚2𝑆 Equation 4
Now mapping the regime
Curve 1
y = 2.376 – 0.454x + 0.13x2 Equation 5
curve 2
y= -2.6746 + 2.59x-0.3817 x2 Equation 6
curve 3
y= -27.895 + 17.253 x – 2.334 x2 Equation 7
x = log (𝜌𝑙𝑖𝑙
2) Equation 8
where, 𝜌𝑙 is the density of the liquid, 𝐺𝑚 is the mass flux and 𝑖𝑙
2 is superficial velocity
of the liquid

HEWITT AND ROBERT FLOW REGIME 3
finding the cross-section area of the pipe;
A= 𝜋 𝑥 𝑑2
4
= 𝜋 𝑥 152𝑥 0.25
= 176.7375 𝑚𝑚2
= 1.767375 x 10-4 m2
Calculating speed of water in m/s
0.5 kg ≡ (1000 𝑘𝑔𝑚−3 / 0.5kg)-1
≡ 0.0005 𝑚−3
The height covered by the above
h= 0.0005 𝑚−3
1.767375 x 10−4 m2
= 2.829 m
So, 2.829 m is covered in one second therefore the speed is 2.829 ms-1 . however,
there is deceleration due to gravity. The speed of particles can be calculated using the
Newtonians equation of motion.
v= √𝑢2 − 2𝑎𝑠
where u is the initial velocity, a is gravitational pull, s is a given distance along the
point and v is velocity at s distance in the pipe. The equation 8 changes to
x= log((√2.829 2 − 2(9.82)𝑠) 𝑖𝑙
2) Equation 9
mapping the flow using the above equations, the following figure was generated using
matlab.