# Heat Transfer Coefficient of Liquid -Liquid Tubular Heater Calculation | HTC

## How to find Condensate Juice Heater Heat Transfer Coefficient Calculation

The fundamental expressions relating to heat transfer coefficient in tubular heat exchanger are available in many standard texts on heat transfer and chemical engineering unit operations.

For basic concepts of overall heat transfer coefficient explained in the below following link

Fundamental  Concepts of overall heat transfer coefficient.

The Overall heat transfer coefficient ( U ) , Kcal / hr/m2/ oC or Kw/m2/ oC, of the heater determines the heat flux generated in the heater for a given temperature force, and it is most important performance factor for the unit. The overall coefficient is related to the various individual transfer coefficients encountered in the process by the expression.

1/U = 1/Hf + 1/Hs + Kt/Tk + 1/Usc

Which gives the reciprocal of  U as an overall resistance which is the sum of the individual resistance.

a) Juice film resistance (Cold media),

b) Condensate water(Hot media),

c) Tube resistance – According to MOC (material of construction) of the tube

d) Resistance of scale. ( fouling factor) Steps in calculating the heat transfer coefficient of tubular juice heater heater

Step1 : Defined the duty and calculate energy balance and flow rates of temperatures.

Step2 : Collect the physical properties of both hot and cold media.

Step3 : Assume the value of the overall coefficient Uo ass from the standard author books.

Step4 : As per the assumed HTC value from standard books  and calculate the heater parameters like heating surface , number of the tubes and its shell dia, . etc

Step5 : Estimate the tube side heat transfer coefficient.

Step6: Decide the baffle spacing and estimate shell side heat transfer coefficient.

Step7 : Calculate the Material coefficient  and Fouling factor coefficient

( Tube resistance – According to MOC (material of construction) of the tube, Resistance of scale. )

Step8: Calculating Overall heat transfer coefficient  Uo cal.

Calculate the percent of variation from Uo ass to Uo cal . The percent of variation is less than 30% then U0 cal. is Ok otherwise will be going to step 3. (i.e Again assume the Uo ass and calculate Uo cal )

Step1 : Sugar cane juice heated by excess condensate (Hot water) available in its sugar process by shell and tube heat exchange.

 Description Sign UOM Values Formula STEP 2 : Collection of the required data. Juice Flow rate Qj T/hr 230 Hot water (heating media) Inlet temperature ti oC 84 Hot water (heating media) Inlet temperature to oC 62 Velocity of juice in tubes Vj m/sec 1.8 Collect the physical properties for Juice Specific heat of the juice Cj Kcal/kg-oC 0.91 Handbook of cane sugar engineering by E.Hugot pg. no. 449 Density of juice ρj kg/M3 1048 Cane Sugar Engineering by peter rein page no. 745 Juice inlet temperature Ti oC 45 Juice outlet temperature To oC 65.00 Viscosity of juice μj Poice 0.00835 Cane Sugar Engineering by peter rein page no. 745 kg/m.hr 3.006 (1centi Poice = 1mPa.s) Thermal conductivity of juice Kj Kcal/hr/m/oC 0.507 Principles of sugar technology by Pieter Honig Vol.1 pg. no.27 Collect the physical properties for condensate water at 85 oC Density of water ρw kg/M3 968 Specific heat of hot water Cw Kcal/kg-oC 1 Thermal conductivity of water Kw W/m oC 0.67281 Kcal/hr/m/oC 0.578482 Viscosity of water μw μpa.s 333 kg/m.hr 1.1988 STEP 3 : Assume OHTC and heater design parameters Overall heat transfer coefficient given in Perry hand book for the given conditions 200 to 250 BTU/ft2/hr/o F Heat transfer coefficient Uo ass Btu/ft2/hr/oF 225 Kcal/hr/m2/oC 1098.45 Tube OD OD mm 45 Tube Thk Tk mm 1.2 Tube plate thk Tpk mm 25 Tube ID D mm 42.6 OD – Tk Legment Lg mm 12 Tube length L mm 4000 STEP 4 : Calculate heat balance and required quantity of how water(heating media) and then find the heater parameters. Fundamental formula for heat balance Heat received by Juice = Heat rejected by condensate water Qj x Cj x ∆T (to – ti) = W x Cw x ∆T(Ti-To) Quantity of Juice Qj Kg/hr 230000 Weight of hot water W kgs/hr 190273 Calculating number of tubes per pass Volume of juice Qj M3/sec 0.06096 Qj / ( ρj x 3600) Cross sectional area of one tube At M2 0.00142 (π/4) x (ID)2 Number of tubes per pass 23.77 Qj / (At x Vj ) i.e tubes/pass 24 Heating Surface  = S :  Qj x Cj x ∆T = Uo ass x  S x ∆Tm Heat transfer coefficient Uo ass Kcal/m2/oC/hr 1098.45 ∆T = To-Ti oC 20.00 ∆Ti = ti – To oC 19.00 ∆Te = to – Ti oC 17.00 Ln(∆Ti/∆Te) oC 0.11 ∆Tm = ∆Ti -∆ Te / ( ln(∆Ti/∆Te)) 17.98 Heating Surface S m2 211.93 i .e m2 212 Number of tubes NT S = π x Dm x Ef x  NT Mean Dia of the tube Dm mm 43.8 OD – Tk Effective tube length Ef mm 3940 L – 2(Tpk) – 2(5) Number of tubes NT 391.23 i.e no.s 392 Tube plate Area = 0.866 x P2 x NT / β Pitch = OD+legment+tube tolerance+hole tolerance P 57.6 Proportional factor β 0.7 Tube plate area A 1.61 Tube plate dia / heater shell dia Ds mtrs 1.43 Sqrt (A / 0.785) STEP 5 : Calculate the tube side heat transfer coefficient. Tube side Juice coefficient Hf Sieder and Tate formula  (1974) Hf = 0.023(Kj/D) x (DG/μj)0.8 x (Cj μj/Kj)0.333 x (μj/μw)0.14 Tube side mass flux G = Qj /Aj Area of the all tubes in pass Aj M2 0.034 Qj Kg/hr 230000.00 G kg/hr/m2 6727098 Re =Reynolds number = G D/μj Re Taken SI units 95334.12 Pr = Prandtl number = Cj μj /Kj Pr 5.395 Generally The value of “(μj/μw)0.14 “is negligible for low viscous materials Tube side Juice coefficient Hf Kcal/hr/m2/oC 5170.5 STEP 6 : Decide the baffle spacing and estimate shell side heat transfer coefficient. Shell side heat transfer coefficient (Hs) Hs Kern  “Process Heat transfer” page no.137 Hs = 0.36 (k/De) x Re0.55 x Pr 0.33 x  (μ /μw)0.14 Consider baffles in shell side for hot water passing B no.s 3 Re = de  Vw ρw / μw Shell side mass velocity Vo = m/As Cross flow As = (P – D)Ds Lb / P Pitch P m 0.0576 OD of tube D m 0.045 Shell inside dia Ds m 1.4317 Baffle spacing Lb m 0.4772 Ds / B Cross flow As m2 0.1495 (P – D) x Ds x  Lb / P Fluid flow rate on the shell m kg/hr 52.854 W/3600 Qw M3/sec 0.055 m /  ρw Velocity of water in shell side Vw m/sec 0.365 Qw / As m/hr 1315.20 de for square pitch = 1.27( Pt 2 – 0.785 OD 2) / OD de for triangular pitch = 1.1( Pt 2 – 0.917 OD 2) / OD de ( here consider triangular pitch) m 0.0357 Re = de  Vw ρw/ μw 37923.04 Pr = Cw x μw / Kw 2.072 Hs = 0.36 (kj / De) x Re0.55 x Pr 0.33 x  (μj /μw)0.14 Hs Kcal/hr/m2/oC 2447.03 STEP 7 :  Calculate the Material coefficient  and Fouling factor coefficient Material coefficient Thickness of the tube Tk m 0.0012 Thermal conductivity SS tubes Kt Btu/Ft/hr/oF 26 Kcal/hr/m/oC 13.96 Material heat transfer coefficient Kt / Tk Kcal/hr/m2/oC 11633 Fouling factor for light hydrocarbons w/m2/oC 5000 Culson&Richanrdsons “Chemical engineering” volume 6 page no.640 Kcal/hr/m2/oC 4299 1/Usc 0.000233 STEP 8 :  Calculating Overall heat transfer coefficient  Uo cal. Uo cal = Overall heat transfer coefficient 1/Uo cal = 1/Hf + 1/Hs + Kt/Tk + 1/Usc 1/Hf Tube side 0.000193 1/Hs Shell side 0.000409 Kt/Tk Wall 8.6E-05 1/Usc Fouling factor 0.000233 1/Uo cal 0.0009206 Uo cal Kcal/hr/m2/oC 1086.2

In this calculation the percent of  variation in between  Uo ass & Uo cal is less-than 30%. So Uo cal. value is correct.

Conclusion : In sugar industry the condensate heater (Duplex heater) heat transfer coefficient will be take around 1000±100 Kcal/hr/m2/oC

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## 3 thoughts on “Heat Transfer Coefficient of Liquid -Liquid Tubular Heater Calculation | HTC” #### Subramanyam Reddy

(October 30, 2017 - 6:39 am)

Excellent work which helps industry and technical person. #### siva alluri

(October 30, 2017 - 1:16 pm)

thank you #### UDAY PRATAP SINGH

(August 7, 2019 - 9:29 am)

Sir,liquid to liquid cooler heat transfer rate against Syrup at 60 degree Brix (inside tube) and water is outside.50 tonne syrup cooling ,cooling area required……M2.