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

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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/