2 D ℜ= 2.0 log 3.7 D + 2.51 ℜ + 1 = 0: ∆ P=f Lρ V Ρ Vd μ ε f f
2 D ℜ= 2.0 log 3.7 D + 2.51 ℜ + 1 = 0: ∆ P=f Lρ V Ρ Vd μ ε f f
2 D ℜ= 2.0 log 3.7 D + 2.51 ℜ + 1 = 0: ∆ P=f Lρ V Ρ Vd μ ε f f
A
main script is required to call the function to execute the /m 3, a dynamic viscosity
=1.79 x 10-5Ns/m2, a diameter D = 0.005m, a Velocity of V= 40m/s, and a
Roughness = 0.0000015m. Initial guesses were to be obtained using the Blasius
formula of f = 0.316/Re0.25. In part (b) the same computation was to be executed but
using a different Roughness = 0.000045m. The equations below were provided
within problem 8.12
Pressure Drop
L V 2
P=f
2D
Reynolds Number
VD
2.0 log
Colebrook Equation
2.51
1
+
+ =0
3.7 D f f
Methodology/ Discussion
Before the Pressure Drop can be calculated, the friction factor had to be
found. The method chosen to solve the Colebrook Equation for the friction factor
was the Newton-Raphson method. This method was chosen because the problem
statement provided a formula to find an initial guess for Reynolds Number and only
1
f
2
Equation was found as shown in this equation.
3
2
2.51
f
2.51
+
ln (10 )
3.7 D f
3
2
Next, we calculated the initial guess for f from the equation given in problem 8.12
which is; f = 0.316/Re0.25. Using the values given in part (a) .The calculated value of
the friction factor was 0.02826. With the analytical derivative and our initial guess,
we wrote a MATLAB code to solve for the friction factor using the Newton-Raphson
method. Once we got the code to function correctly, we made it in to a function and
saved it as Colebrook.m with input variables (, D, Re) that can be manipulated by
the user. Next, we integrated this function into a new script to solve for the change
in pressure P. Using the data from part (a), we function and script to see how
changes in Reynolds Number, pipe diameter, and roughness would affect the
pressure drop.
1200
1150
P (Pa)
1100
1050
1000
10000
12000
14000
16000
18000
Reynolds Number
0.005m. The tested values of Re ranged from 10000 to 16000 and were
Figure 1
incremented by 500. These values
were then put into our MATLAB program and the outputs of friction factor and
pressure drop were tabulated in excel and graphed for a better visual understanding
(see Table 1 and Figure 1).
Table 1.
Re
10000
10500
11000
11500
12000
12500
13000
13500
13743.
02
14000
14500
15000
15500
16000
P
(Pa)
1233.
65
1218.
51
1204.
36
1191.
08
1178.
58
1166.
8
1155.
66
1145.
11
1140.
17
1135.
09
1125.
56
1116.
48
1107.
81
1099.
53
f
0.031
33
0.030
96
0.030
60
0.030
26
0.029
94
0.029
64
0.029
36
0.029
09
0.028
98
0.028
83
0.028
59
0.028
36
0.028
16
0.029
73
By examining the data, it shows that Reynolds number has a negative relationship
with both friction factor and pressure drop. The relationship is not quite linear but
the trend is evident. This trend could have been affected due to Reynolds number
and pressure drop being inversely related to Diameter. Another possibility is that,
raising Reynolds number and simulating a less viscous fluid, would make that a less
viscous fluid would have less friction and would maintain a more regulated constant
pressure.
D (m)
0.001
0.002
0.003
0.004
0.005
0.006
0.007
0.008
0.009
P
(Pa)
f
6074.6 0.0308
2
7
2914.9 0.0296
2
2
1914.9 0.0291
4
9
0.0289
1425.4
7
1135.0 0.0288
9
3
0.0287
942.98
9
Table 2
0.0286
806.47
8
0.0286
704.48
3
0.0286
625.98
0
Roughness
P (Pa) 4000
2000
0
0.01
0.01
0.01
Diameter (m)
S
econd, we tested what would happen to pressure drop
and friction factor when we change the diameter of the
pipe and keep all other variables constant. Again, we
simply hard coded the values of Re = 14000 and
=
Figure 2
The values used for diameter ranged from, 0.001m to 0.009m and incremented by
0.001m. Again these values had a negative relationship with pressure drop and
friction factor. On the other hand, their relationship looked much more exponential
in the range of values that we studied when compared to that of Reynolds number
and pressure drop. This relationship also makes intuitive sense, the smaller the pipe
the more surface area touching the liquid flowing through it and this means more
energy lost, slower velocity and less pressure.
10
15
20
25
(m)
(
m
)
0.
5
1
1.
5
P
(Pa)
1120 0.02
.958 847
1128 0.02
.055 866
1135 0.02
.08
884
1142 0.02
3
.06 Table
901
1182 0.03
.619 005
1246 0.03
.027 166
1305 0.03
.102 316
1360 0.03
.628 457
Conclusion: