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/m^{2}/^{ o}C or Kw/m^{2}/^{ o}C, 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 )
In this article calculate the HTC by taking one example.
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 | ^{o}C | 84 | |
Hot water (heating media) outlet temperature | to | ^{o}C | 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-^{o}C | 0.91 | Handbook of cane sugar engineering by E.Hugot pg. no. 449 |
Density of juice | ρj | kg/M^{3} | 1048 | Cane Sugar Engineering by peter rein page no. 745 |
Juice inlet temperature | Ti | ^{o}C | 45 | |
Juice outlet temperature | To | ^{o}C | 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/^{o}C | 0.507 | Principles of sugar technology by Pieter Honig Vol.1 pg. no.27 |
Collect the physical properties for condensate water at 85 ^{o}C | ||||
Density of water | ρw | kg/M^{3} | 968 | |
Specific heat of hot water | Cw | Kcal/kg-^{o}C | 1 | |
Thermal conductivity of water | Kw | W/m ^{o}C | 0.67281 | |
Kcal/hr/m/^{o}C | 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/ft^{2}/hr/^{o }F | ||||
Heat transfer coefficient | Uo ass | Btu/ft^{2}/hr/^{o}F | 225 | |
Kcal/hr/m^{2}/^{o}C | 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 | M^{3}/sec | 0.06096 | Qj / ( ρj x 3600) |
Cross sectional area of one tube | At | M^{2} | 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/m^{2}/^{o}C/hr | 1098.45 | |
∆T = To-Ti | ^{o}C | 20.00 | ||
∆Ti = ti – To | ^{o}C | 19.00 | ||
∆Te = to – Ti | ^{o}C | 17.00 | ||
Ln(∆Ti/∆Te) | ^{o}C | 0.11 | ||
∆Tm = ∆Ti -∆ Te / ( ln(∆Ti/∆Te)) | 17.98 | |||
Heating Surface | S | m^{2} | 211.93 | |
i .e | m^{2} | 212 | ||
Number of tubes | N_{T} | 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 | N_{T} | 391.23 | ||
i.e | no.s | 392 | ||
Tube plate Area = 0.866 x P^{2} x N_{T} / 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) | ||||
H_{f} = 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 | M^{2} | 0.034 | |
Qj | Kg/hr | 230000.00 | ||
G | kg/hr/m^{2} | 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/m^{2}/^{o}C | 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 Re^{0.55} 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 | m^{2} | 0.1495 | (P – D) x Ds x Lb / P |
Fluid flow rate on the shell | m | kg/hr | 52.854 | W/3600 |
Qw | M^{3}/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 Re^{0.55} x Pr^{ 0.33} x (μj /μw)^{0.14} | ||||
Hs | Kcal/hr/m^{2}/^{o}C | 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/^{o}F | 26 | |
Kcal/hr/m/^{o}C | 13.96 | |||
Material heat transfer coefficient | Kt / Tk | Kcal/hr/m^{2}/^{o}C | 11633 | |
Fouling factor for light hydrocarbons | w/m^{2}/^{o}C | 5000 | Culson&Richanrdsons “Chemical engineering” volume 6 page no.640 | |
Kcal/hr/m^{2}/^{o}C | 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/m^{2}/^{o}C | 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/m^{2}/^{o}C
Some Related Articles
For the condensate heater calculation of heating surface, number of tubes , juice inlet and outlet lines dia. ..etc, purpose go through the below link.
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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.
Please calculation send my mail