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Faculty of Engineering
Science and
Built Environment
Course: BEng Building Services Engineering
Mode: Part Time
Level: Two
Unit: Thermofluids 2 A
Tutor: Dr A Paurine
Unit Code: DTF-2-257
Date: Semester 1, 2010
Prepared By
Douglas Buchan
2
Question 1 For turbulent flow, the velocity profile inside a duct and a pipe of diameters Dd and
Dp and both with distance y from the centre line is:
, where
and
. Plot the velocity profiles for:
(a) Air flowing in a duct measuring Dd = 0.6m and with Re = 105.
(b) Water flowing in a pipe measuring Dp = 0.5m and with Re = 105. (c) Calculate the and for both cases (a) and (b)
(d) If the Blasius equation f = 0.079Re-0.25 applies, calculate the values of f and
for (a) Question 2
(a) Explain briefly how the Dittus-Boelter equation shown below for forced convection may be obtained by consideration of dimensional analysis, describing the dimensionless groups which the equation contains. Nu = 0.023Re0.8Pr0.4
Where, Nusselt Number,
, Reynolds’ Number,
and Prandtl
Number,
.
(b) A refrigerant line of diameter radius 0.025m must have its internal convective heat transfer co-efficient determined for the conditions below. Determine the convection co-efficient by calculation using the Dittus-Boelter equation and calculate the heat gain to the refrigerant if the temperature difference across the film is 2 K, per metre of pipe. The refrigerant velocity is 0.6 m/s and following data apply: Thermal conductivity, k (kW/mK) = 0.000595 Dynamic viscosity, μ (Pas) = 0.000245 Density, ρ (kg/m3) = 670 Specific heat capacity, Cp (kJ/kgK) = 4.55 Comment upon the size of Reynolds’ Number and the flow regime.
3
Question 1 For turbulent flow, the velocity profile inside a duct of diameters Dd and Dp
and both with distance y from the centre line is:
, where
and
. Plot the velocity profiles for:
(a) Air flowing in a duct measuring Dd = 0.6m and with Re = 105.
In order to ascertain the velocity profile I will need to obtain the mean velocity value
. This can be found using the Reynolds’ Number,
as follows;
Where,
Dynamic Viscosity of Air, 0.000018 kg/msi
D = Duct Diameter, 0.6 m
Density of Air, 1.18 kg/m3
Reynolds Number, 105
Transposed to find :
Volume (Q) of Air flowing in the duct can be calculated using the mean velocity;
Q =
Area A = Q = Q = 0.719
i Concepts of chemical engineering 4 chemists By Stefaan J. R. Simons
4
Now that Vmean has been calculated the value can be used to find as follows;
In order to check that the calculation for is accurate, given that the flow is
turbulent, as Re>2100, the following formula can be used to find Vmax;
Vmax = 1.22 x Vmean
Vmax = 1.22 x 3.11
Vmax = 3.10m/s
Therefore;
Vmax =
I can now calculate the pressure drop (ΔP) using the following formula;
ΔP= 4 f
Where;
ΔP= Pressure Drop
F = Fanning Friction Factor
L= Length, assumed as 1m for this calculation(in order to find ΔP for 1m length).
D = Duct Diameter, 0.6 m
Density of Air, 1.18 kg/m3
V = Vmean
However in order to calculate the equation, I must first find the fanning friction factor
(f). In order to find f the Blassius equation applies;
f =0.79 Re-0.25
Where,
Reynolds Number, 105
f =0.79 (105)-0.25
5
f = 4.44-3
Therefore
ΔP= 4x4.44-3 x
ΔP= 0.113 pa/m
The velocity profile can now be calculated using the following formula;
Vy = Vy=0
Where;
r= radius, 0.3m
Vy=0= Velocity at centre of pipe (Vmax), 3.11m/s (calculated)
The Microsoft Excel programme was used, the data was inputted and formula was
entered with the following results;
Velocity Profile in Pipe
y (m) Vy (m/s) y (m) Vy (m/s)
0.30 0.00 0.14 2.84
0.29 1.91 0.13 2.87
0.28 2.11 0.12 2.89
0.27 2.24 0.11 2.91
0.26 2.33 0.10 2.93
0.25 2.41 0.09 2.96
0.24 2.47 0.08 2.98
0.23 2.53 0.07 2.99
0.22 2.57 0.06 3.01
0.21 2.62 0.05 3.03
0.20 2.66 0.04 3.05
0.19 2.69 0.03 3.06
0.18 2.73 0.02 3.08
0.17 2.76 0.01 3.09
0.16 2.79 0.00 3.11
0.15 2.82 - -
Where; Y(m) = Distance from centre line and Vy = Velocity at distance from centre Line
6
The calculated results were checked for accuracy using the formula, once they were
established as correct I could then plot the points in a graph, this would provide a
graphical representation of the velocity profile. It should be noted that only the results
for one side of the duct are shown, the reason for this is that profile either side of the
centre line is exactly the same. In order to plot the graph all points were calculated.
Microsoft Excel graph of plotted points for the duct.
0.00
0.50
1.00
1.50
2.00
2.50
3.00
3.50
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31
Velocity profile in duct
Vy (m/s)
Full Velocity Profile for the duct.
Centre line Y=0
Duct Wall
Duct Wall
Y
RVelocity Profile
7
Question1
For turbulent flow, the velocity profile inside a pipe of diameters Dd and Dp
and both with distance y from the centre line is:
, where
and
. Plot the velocity profiles for:
(b) Water flowing in a pipe measuring Dp = 0.5m and with Re = 105.
In order to ascertain the velocity profile I will need to obtain the mean velocity value
. This can be found using the Reynolds’ Number,
as follows;
Where,
Dynamic Viscosity of Water, 0.001 kg/msii
D = Pipe Diameter, 0.5 m
Density of Water, 1000 kg/m3
Reynolds Number, 105
Transposed to find :
Volume (Q) of Water flowing in the pipe can be calculated using the mean velocity;
ii Concepts of chemical engineering 4 chemists By Stefaan J. R. Simons
8
Q =
Area A = Q = Q = 0.0392
Now that Vmean has been calculated the value can be used to find as follows;
In order to check that the calculation for is accurate, given that the flow is
turbulent, as Re>2100, the following formula can be used to find Vmax;
Vmax = 1.22 x Vmean
Vmax = 1.22 x 2.00
Vmax = 2.44m/s
Therefore;
Vmax =
I can now calculate the pressure drop (ΔP) using the following formula;
ΔP= 4 f
Where;
ΔP= Pressure Drop
F = Fanning Friction Factor
L= Length, assumed as 1m for this calculation(in order to find ΔP for 1m length).
D = Duct Diameter, 0.5 m
Density of water, 1000 kg/m3
V = Vmean
However in order to calculate the equation, I must first find the fanning friction factor
(f). In order to find f the Blassius equation applies;
9
f =0.79 Re-0.25
Where,
Reynolds Number, 105
f =0.79 (105)-0.25
f = 4.44-3
Therefore
ΔP= 4x4.44-3 x
ΔP= 7.10 pa/m
The velocity profile can now be calculated using the following formula;
Vy = Vy=0
Where;
r= radius, 0.25m
Vy=0= Velocity at centre of pipe (Vmax), 2.45 m/s (calculated)
The Microsoft Excel programme was used, the data was inputted and formula was
entered with the following results;
Velocity Profile in Pipe
y (m) Vy (m/s) y (m) Vy (m/s)
0.25 0.00 0.12 2.23
0.24 1.55 0.11 2.26
0.23 1.71 0.10 2.28
0.22 1.81 0.09 2.30
0.21 1.89 0.08 2.32
0.20 1.95 0.07 2.34
0.19 2.00 0.06 2.36
0.18 2.04 0.05 2.37
0.17 2.08 0.04 2.39
0.16 2.12 0.03 2.41
0.15 2.15 0.02 2.42
0.14 2.18 0.01 2.44
0.13 2.21 0.00 2.45
Where; Y(m) = Distance from centre line and Vy = Velocity at distance from centre Line
10
The calculated results were checked for accuracy using the formula, once they were
established as correct I could then plot the points in a graph, this would provide a
graphical representation of the velocity profile. It should be noted that only the results
for one side of the pipe are shown, the reason for this is that profile either side of the
centre line is exactly the same. In order to plot the graph all points were calculated.
Microsoft Excel graph of plotted points for the duct.
0.00
0.50
1.00
1.50
2.00
2.50
3.00
1 3 5 7 9 11 13 15 17 19 21 23 25
Velocity Profile in Pipe
Vy (m/s)
Full Pipe Velocity Profile
Centre line Y=0
Pipe Wall
Pipe Wall
Y
RVelocity Profile
11
Question 1
(c) and (d)
The values for Vmean = and for pressure drop have been stated in my
calculations for question1(a) and(b)
Question 2 (a) Explain briefly how the Dittus-Boelter equation shown below for forced convection may be obtained by consideration of dimensional analysis, describing the dimensionless groups which the equation contains. Nu = 0.023Re0.8 Pr0.4
Where, Nusselt Number,
, Reynolds’ Number,
and Prandtl
Number,
.
The Dittus-Boelter equation relates the heat transfer coefficient of a cylindrical pipe
to the Reynolds number (Re) and the Prandtl (Pr) number of the fluid flowing in the
pipe:
Nu = hD/k = 0.023 Re0.8 Pr0.4
where: h = heat transfer coefficient, D = the pipe diameter, and k = the thermal
conductivity of the fluid. Using dimensional analysis I can show how the Dittus-
Boelter equation can be obtained.
The Dittus-Boelter equation contains 3 different groups of equations;
The Nusselt number;
Nu =
Where; h : is the convective heat coefficient ,W/m2K
D: is the characteristic length, m
k : is the thermal conductivity, W/m2K
12
The Reynolds number;
Where;
: is the density, kg/m3
: is the velocity, m/s
D : is the diameter, m
: is the dynamic viscosity, kg ms
The Prandlt Number;
Where;
: is the specific heat, kJ/kgK
: is the dynamic viscosity, kg/ms
k : is the thermal conductivity, KW/mK
The Dittus-Boelter equation can now be expressed using the Si units in each set of equations;
As the units cancel each other out the Dittus-Boelter equation is classed as
dimensionless.
13
Question 2 (b) A refrigerant line of diameter radius 0.025m must have its internal convective heat transfer co-efficient determined for the conditions below. Determine the convection co-efficient by calculation using the Dittus-Boelter equation and calculate the heat gain to the refrigerant if the temperature difference across the film is 2 K, per metre of pipe. The refrigerant velocity is 0.6 m/s and following data apply: Thermal conductivity, k (kW/mK) = 0.000595 Dynamic viscosity, μ (Pas) = 0.000245 Density, ρ (kg/m3) = 670 Specific heat capacity, Cp (kJ/kgK) = 4.55 Comment upon the size of Reynolds’ Number and the flow regime.
Nu = hD/k = 0.023 Re0.8 Pr0.4
In order to find h which is the convective heat coefficient , I will transpose the Nusselt
group of equations;
Transposed;
I need to find Nu so I will calculate the Reynolds number and the Prandtl number using the values given in the question;
Kg/m3
kg/ms
= 2 x 0.025m = 0.05m
kJ/kgK
kW/mK
14
Reynolds Number;
Prandlt Number;
4
These results for the Reynolds number and the Prandlt number can now be used to calculate the Nusselt number;
2.999 kW/m2K The Heat transferred to the refrigerant can now be assessed using the following formula;
15
In order to calculate this I must first find the area (A). This can be found using the following formula;
Area = 2 rL (Where L is taken as 1m length)
Area = 2 0.025 x 1 Area = 0.157m2
The heat transferred can now be calculated;
Therefore it can be concluded that will be transferred per metre of
pipe under these conditions.
16
Bibliography
Concepts of chemical engineering 4 chemists
By Stefaan J. R. Simons
Thermodynamic and transport properties of fluids: SI units
By G. F. C. Rogers, Yon Richard Mayhew
Thermodynamics and Thermal Engineering
By J.Selwin Rajadurai
Fundamentals of engineering thermodynamics
By Michael J. Moran, Howard N. Shapiro
Websites Consulted
http://web.mit.edu/
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