Designing Steps for a Heat Exchanger

Designing Steps for a Heat Exchanger

Reetika Saxena 1

M.Tech. Student in I.F.T.M. University, Moradabad

Sanjay Yadav 2

Asst. Prof. in I.F.T.M. University, Moradabad

Distillation is a common method for removing dissolved solids and to obtain pure water for drinking and for the purpose like battery water, electroplating etc. Distillation of water have certain problems and operational issues and too like as it is an energy consuming process.

Multiple effect distillation is a distillation process generally used for sea water distillation. It consists of stages (effect). In first stage feed water is heated by steam in tubes. Some of the water evaporates and this fresh steam flows into the tube of the next stage. But for better heat transfer, it is necessary to design a heat exchanger which fulfills the requirements of MED unit.

The most common problems in heat exchanger design are rating and sizing. The rating problem is concerned with the determination of the heat transfer rate, fluid outlet temperature, inlet temperature, heat transfer area and the sizing problem involves determination of the dimension of the heat exchanger. An heat exchanger (shell and tube type) is being designed here with proper dimension for 45 Kg/hr steam. This method is also used for other temperature range and increased mass flow rate of steam. Using this design procedure of heat exchanger we can also increases the boiler efficiency which produces steam for multi effect distillation unit.

Related input data for the design of heat exchanger, which is used for multiple distillation unit is given below-

Input steam pressure = 4 bar gauge Standard atmospheric pressure = 1.013 bar

Absolute pressure = Atmospheric pressure + gauge Pressure Therefore Input steam pressure = 1.013 + 4 = 5.013 bar

Inlet Temperature of Steam, Thi = 152.7 C at 5.013 bar Outlet Temperature of Steam, Tho = 117 C

Inlet Temperature of Water, Tci = 110 C Specific Heat for Steam, Cph = 2.5 kJ/kg K Specific Heat for Water, Cpc = 4.18 kJ/kg K Mass flow rate of Steam, mh = 45 kg/h

= 45/3600

= 0.0125 kg/s Mass flow rate of Water, mc = 250 kg/h

= 250/3600

= 0.0694 kg/s

1. Energy Balance Eq.,

   mc * Cph * ( Tco Tci ) = mh * Cph * (Thi Tho )

   0.0694 * 4.18 * ( Tco 110 ) = 0.0125 * 2.5 * (152.7 117 )

   0.290092 * Tco 31.91012 = 1.1156

   0.290092 * Tco = 33.025745 Tco = 113.85 C

   Outlet Temperature of Water, Tco = 113.85 C

2. Heat Transfer Rate,

   Q = mc*Cph* ( Tco Tci )

   = 0.0694 * 4.18 * (113.85 110 )

   = 1.17 kW

3. Log Mean Temperature Difference, LMTD for Parallel Flow,

   Tho Tco

   T1	=	Thi – Tci			Thi
   	=	152.7 110	=	42.7C
   T2	= =	Tho – Tco 117 113.85	=	3.15C	Tci

   T1	=	Thi – Tci			Thi
   	=	152.7 110	=	42.7C
   T2	= =	Tho – Tco 117 113.85	=	3.15C	Tci

   LMTD, Tm = T1 T2

   ln (T1 /T2)

   = 42.73.15

   ln (42.7)

   ln (42.7)

   3.15

   Tm = 39.55 / 2.6067 = 15.17C

4. Temperature Correction Factor,

   The selection of parameters R and P should be such that the value of correction factor, Ft is more than 0.75

   Capacity Ratio, R = Tci Tco

   Tho Thi

   = (110 113.85) / (117 152.7)

   = 0.12

   Temperature Ratio, P = Tho Thi Tci Thi

   = (117 152.7) / (110 152.7)

   = 0.84

   The value may taken from the chart:-

   Ft =

   2+1

   1

   Fig.1: Relation between R, P and Ft

   1

   1

   2 +1 2+1

   2 +1+ 2+1

   Ft =

   0.122+1

   0.121

   10.84

   10.840.12

   20.84 0.12+1 0.122+1

   20.84 0.12+1+ 0.122+1

   Ft = 0.904

   Both result are same, therefore Ft = 0.904

5. Mean Temperature Difference, DTm = Ft * LMTD

   = 0.904 * 15.17

   DTm = 13.71C

6. Overall Heat Transfer Co-efficient,

   Study based on, Steps for design of Heat Exchanger by Dr. Reyad Shewabkeh, Dept. of Chemical Engineering, King Fahd University of Petroleum & Minerals, The range of overall heat transfer co-efficient for water is 800 1500 w/m2 C.

   U = 945 w/m2 C

   Table 1: Overall heat transfer coefficient for different combination

7. Provisional Area,

   A = Q

   U T

   = 1170 94513.71

   = 0.090 m2 = 900 cm2

8. Tube Outer Diameter,

   Case I- Number of Tubes, Nt = 7

   Length of Tubes, L = 100 cm do = 900 / (3.14*7*100) = 0.41 cm

   = 4.1 mm

   Case II- Number of Tubes, Nt = 5

   Length of Tubes, L = 100 cm do = 900 / (3.14*5*100) = 0.57 cm

   = 5.7mm

   Case III- Number of Tubes, Nt = 7

   Length of Tubes, L = 50 cm do = 900 / (3.14*7*50) = 0.82 cm

   = 8.2 mm

   Case IV- Number of Tubes, Nt = 5

   Length of Tubes, L = 50 cm do = 900 / (3.14*5*50) = 1.15 cm

   = 11.5 mm

   Above design calculation are not feasible because calculations are based on LMTD 13.71C. MED Unit will provide best result when condensation of steam will take place with minimum temperature difference.

   Study based on paper Porteous,A. (1975), Saline water distillation Process (1st Ed) Longman UK, London, 150p, it is clear that condensation can take place with a temperature difference of 2C.

   Therefore our design calculation will be based on 4C. DTm = 4C

   A = Q

   U DT m

   = 1170 9454

   = 0.309523 m2

   = 3095.23 cm2

   Case I-	Number of Tubes, Nt Length of Tubes, L	= =	7 100 cm
   do	= 3095.23 / (3.14*7*100)	= =	1.408 cm 14.08 mm
   Case II-	Number of Tubes, Nt Length of Tubes, L	= =	5 100 cm
   do	= 3095.23 / (3.14*5*100)	= =	1.8207 cm 18.21 mm
   Case III-	Number of Tubes, Nt Length of Tubes, L	= =	7 50 cm
   do	= 3095.25 / (3.14*7*50)	= =	2.816 cm 28.16 mm
   Case IV-	Number of Tubes, Nt Length of Tubes, L	= =	5 50 cm
   do	= 3095.25 / (3.14*5*50)	= =	3.942 cm 39.42 mm

   Case V-	Number of Tubes, Nt Length of Tubes, L	= =	5 75 cm
   do	= 3095.25 / (3.14*5*75)	= =	2.628 cm 26.28 mm
   Case IV-	Number of Tubes, Nt Length of Tubes, L	= =	7 75 cm
   do	= 3095.25 / (3.14*7*75)	= =	1.877 cm 18.77 mm

   Sr. No.	NUMBER OF TUBES, Nt	LENGTH OF TUBES, L (mm)	OUTER DIAMETER, do (mm)
   1.	7	1000	14.08
   2.	5	1000	18.21
   3.	7	500	28.16
   4.	5	500	39.42
   5.	5	750	26.28
   6.	7	750	18.77

   Table 2: Different configuration for tubes

   We get best result in case III, therefore outer diameter of tube is 28.16 mm, And the configuration of the tube is-

   do	= =	28.16 mm 0.02816 m
   Nt	=	7
   L	=	50 mm

9. Tube Pitch,

   Pt = 1.25 * do

   = 1.25 * 28.16

   = 35.2 mm

10. Bundle Diameter,

    Db = do [ Nt / K1 ]1/n

    Table 3: Relation between constant K1 and n1

    For Square pitch,

    Pt = 1.25 * do

    K1 = 0.0366

    N = 2.63

    Db = 28.16 * [ 7 / 0.0366 ]1/2.63

    = 207.57 mm

    = 0.207 m

11. Bundle diameter clearance,

    For fixed floating head, BDC = 10 mm

    Fig. 2: Bundle diameter clearance

12. Shell Diameter,

    Ds = Db + BDC = 207.57 + 10

    = 217.57 mm

13. Baffle Spacing,

    Bs = 0.4 * Ds

    = 0.4 * 217.57

    = 87.03 mm

14. Area for cross flow,

    As = (Pt do ) Ds Bs

    Pt

    = 35.228.6 217.5787.03

    35.2

    = 3550.33 mm2

    = 3.55 * 10-3 m2

15. Shell side mass velocity,

    Gs = Shell Side flow rate [kg /s]

    As

    Shell side flow rate = 0.0694 kg/s

    3

    3

    Gs = 0.0694

    3.55 10

    = 19.55 kg/m2 s

    = 1.955 * 10-5 kg/mm2-s

16. Shell equivalent diameter for a square pitch arrangement,

    1.27 [ P2 0.785 d2 ]

    de =

    t o

    do

    = 1.27 [ 35.2 2 0.785 28.162 ]

    28.16

    = 27.81 mm = 0.02781 m

17. Shell side Reynolds number,

    Properties of water at 110C temp.
    Density ,	=	951 kg/m3
    Kinematic viscosity,	=	0.273 * 10-6 m2/s
    Fluid thermal Conductivity, kf	=	0.62 W/m-K
    Specific heat, Cp	=	4233 J/kg-K

    e

    e

    R = Gs de

    = Gs de

    19.55 0.02781

    =

    =

    9510.273106

    = 2094.1

    Re > 2000 therefore flow inside shell side is Transition and Turbulent.

18. Prandtle number,

    r

    r

    P = µ Cp

    kf

    = Cp

    kf

    = 9510.273 106 4233

    0.62

    = 1.77

19. Nusselt number,

    Nu	=	0.023 * (Re)0.8 * (Pr)n
    n	=	0.4 for heating
    	=	0.3 for cooling
    Nu	= =	0.023 * (Re)0.8 * (Pr)0.4 0.023 * (2094.1)0.8 * (1.77)0.4
    	=	13.11

20. Heat transfer coefficient,

    o

    o

    h = Nu kf

    de

21. Tube inside diameter, di = do – t

    = 13.11 0.62

    0.02781

    = 292.35 W/m2 -K

    Thickness of tube metal = 6 mm

    di = 28.16 6

    = 22.16 mm

    = 0.02216 m

22. Tube side Reynolds number,

    Properties of steam at 152.7C temp.
    Dynamic viscosity, µ	=	1.408 * 10-5 N-s/m2
    Fluid thermal Conductivity, kf	=	0.0311 W/m-K
    Specific heat, Cp	=	2335.2 J/kg-K

    e

    e

    R = Gs di

    19.55 0.02216

    =

    =

    1.4085 105

    = 100000

    Re > 2000 therefore flow inside tube is Transition and Turbulent.

23. Prandtle number,

    r

    r

    P = µ Cp

    kf

    = 1.4085 10 5 2335 .2

    0.0311

    = 1.057

24. Nusselt number,

    Nu = 0.023 * (Re)0.8 * (Pr)0.3

    = 0.023 * (100000)0.8 * (1.057)0.3

    = 233.86

25. Heat transfer coefficient,

    o

    o

    h = Nu kf

    di

    = 233 .86 0.0311

    0.02216

    = 328.21 W/m2 – K

26. Overall heat transfer coefficient in Shell and tube heat exchanger,

The heat transfer is in radial direction, firstly the heat is transferred by hot fluid to inner wall of tube by convection . then through the wall of the tube by conduction and finally from the outer wall of tube to cold fluid by convection.

Hot fluid in

Cold fluid in ri

ro

Cold fluid out

Hot fluid out

i To

i To

T

Ri Rwall Ro

Fig. 3: Heat transfer in shell and tube heat exchanger

Total Thermal resistance,

R = Ri + Rwall + Ro

R =

1 +

+

2

1

1

1

= 2 +

+

2

1

2

Q =

= UAT = U A T = U A T

R i i o o

Where, U = Overall heat transfer coefficient in W/m2K

R = 1

= 1

= 1

Overall heat transfer coefficient based on outside surface area of tube can be expressed as:

Uo =

1 =

R Ao

1

+

+ 1

= 1

2

+ + 1

=

=

1

0.01408 + 0.01408 0.01408 + 1

0.01108 328 .21 225 0.01108 292 .35

= 136.89 W/m2-K

Overall heat transfer coefficient based on inner surface area of tube can be expressed as:

Uo =

1 =

R Ai

1

+

+ 1

= 1

2

+

+ 1

=

=

1

0.01108 + 0.01108 0.01408 + 1

0.01408 292 .35 225 0.01108 328 .21

= 136.89 W/m2-K

After comparing the overall heat transfer coefficient, I obtained from previous step with that I assumed in step 6. It is smaller to what I assumed, then I have a valid assumption, that tabulate my results such as total surface area of tubes, number of tubes, exchanger length and diameter and other design specification.

Following conclusion can be made after study:

1. Six different combination of No. of tubes and length of tubes were tried. Above design of heat exchanger was best, because our calculated dimensions are verified very accurately.

2. Due to agronomic consideration of design, the tube of very large diameter is not selected.

3. Smallest diameter is not selected because it is not practically feasible.

4. An appropriate heat exchanger is designed for multiple effect distillation unit to condense 45 Kg/hr steam. Dimension of heat exchanger is given below-

   Number of tubes, Nt	=	7
   Length of tube, L	=	500 mm
   Outer diameter of tube, do	=	28.16 mm
   Thickness of the tube, t	=	6 mm
   Inner diameter of tube, di	=	22.16 mm
   Overall heat transfer coefficient,U	=	136.89 W/m2K
   Tube pitch, pt	=	35.2 mm
   Bundle diameter, Db	=	207.57 mm
   Bundle diameter clearance, BDC	=	10 mm
   Shell Diameter, Ds	=	217.57 mm
   Baffle spacing, Bs	=	87.03 mm
   Area for cross flow, As	=	3550.33 mm2
   Material selected	=	Aluminum

   Following Modification can be done for future work:

   1. Other type of heat exchanger can be design.

   2. Flow of steam and water can be reverse. That means the feed water may be taken inside tubes and condensing steam outside tubes.

   3. Pressure drop inside shell and tube can be calculated.

   4. Counter flow arrangement can be tried.

   5. Design may done for different types of floating head.

   6. We can use some other material than aluminum for better heat transfer.

   7. Effect of corrosion can be considered.

1. Sen P.k,Padma Vasudevan Sen,S.K.Vyas,A.Mudgal,A small-scale Multiple Effect Water Distillation System for Rural sector, international conference on Mechanical Engineering 2007(ICME2007).

2. Kister, Henry Z. (1992). Distillation Design (1st Ed.). McGraw-Hill. ISBN 0-

   07-034909-6

3. Walker, G., Industrial Heat Exchanger. A Basic Guide. Hemisphere, Washington, D. C., 1990.

4. Shah, R. K., Classification of heat exchangers, in heat exchangers Thermo- Hydraulic Fundamentals and Design, Kakac, S. Bergles, A. E. and Mayinger, F., Eds., John Wiley & Sons, New York, 1981.

5. Shah, R. K. and Mueller, A. C., Heat exchangers, in Handbook of Heat Transfer Applications, Rohsenow, W. M., Hartnett, J. P., and Gani, E. N., Eds., McGraw-Hill, New York, 1985, Ch. 4.

6. Kakac, S., Shah, R.K., and Aung. W., Eds., Handbook of Single phase convective heat transfer, John Wiley & Sons, New York, 1987, Ch. 4,18.

7. Allchin, F.R. (1979), India: The Ancient home of Distillation? Man(1): 55- 63. Doi: 10.2307/2801640.

8. Porteous, A. (1975), Saline water distillation processes (1st Ed) Longman UK,

   London, 150p.

9. Bowman, R. A., Mueller, A. C., and Nagle, W. M., Mean temperature difference in design, Trans. ASME, 62, 283, 1940.