Overall Heat Transfer Co-Efficient Calculation

Hello guys........Hope all are enjoying the start of these weekend !!

Today i wanna add a full stop for many peoples fantasies of 'U' value calculation, the most common line that i heard from many engineers was Consider U as 150 for Reactors and 450 for Condensers, in return if you ask them what are the units for those 150 and 450, many of them will take some strategic time out as in IPL.

An Engineer is the one who holds the ability to deal any of the situation with a smart work, but not the one who follow the Thumb blindly, i'm not criticizing the Thumb, as many cases i too prefer the Thumb rules in practical cases but my intention is An Engineer should be able to derive the U value if needed.

 And for that we need to know some basics, before going into topic.

What are Heat Transfer Co-Efficient?

Heat transfer coefficient is a quantitative characteristic of convective heat transfer between a fluid medium (a fluid) and the surface (wall) flowed over by the fluid.

Also Read:

How to Design a Condenser?  

How to Select a Pump , Motor, Line Sizings ?

What is Overall Heat Transfer Co-efficient?

Overall Heat Transfer Co-efficient is the combination of the possible Individual Heat Transfer Co-efficient, which are due to the following Heat Transfer modes,
1. Rate of convective heat transfer from Fluid to Medium Wall, and 
2. Conductive Heat Transfer from Medium Walls one end to other end,
3. Convective Heat Transfer from Medium wall to inside Fluid. 

Credits: TLV.com

What is Thermal Conductivity?

Thermal Conductivity of a material characterizes the Rate of Heat Transfer of a material, whether the heat is flowing in or out.

The thermal conductivity - k - for some typical materials (varies with temperature)

• Polypropylene PP : 0.1 - 0.22 W/mK
• Stainless steel : 16 - 24 W/mK
• Aluminum : 205 - 250 W/mK
• Glass lining :  1.2 W/mK

What are the units of Overall Heat Transfer Co-efficient and How are they derived? 

British Units for U are Btu/(hr-ft²°F) , SI units for U are Watts/Sq.m .°K or KCal/Sq.m .sec.°K

They are basically derived from the formula Q = U x A x LMTD,

where Q holds the units Watts , A - Sq.m, LMTD - °K or °C.

So, that's how U units were derived.

Credits: engineersedge.com

What will be the Trend of Convective Heat Transfer Co- efficients?

The Convective Heat Transfer Trend will be as follows:

Fluid	Convective heat transfer coefficient (h)
Water	about 1000 W/(m2°C)
Hot Water	1000 – 6000 W/(m2)°C
Steam	6000 – 15000 W/(m2°C)

The convection heat transfer coefficient - h - depends on

1. The type of fluid - gas or liquid,
2. The flow properties such as velocity,
3. Other flow and temperature dependent properties.

Convective heat transfer coefficient for some common fluids:

• Air - 10 to 100 W/m²K
• Water - 500 to 10 000 W/m²K

 So, I think right now you acquired some courage to drive the fantasy of  'U' value calculation from your mind. And, i'll start the main show now.

Recommended Read:

How To Select a Condenser? 

Determine the Power Required for an Operation?  

Industrial Distillation Column Design Steps

U value Calculation:

For calculation of U value we need some basic data regarding the Jacket fluid properties, Vessel Mass properties, Jacket properties, Equipment Physical data  etc.

General Data includes: Vessel Dia, Vessel MOC, Vessel wall thickness.

For Calculating Overall Heat Transfer Coefficient we need Individual Heat Transfer Co-efficients used in the following co-relation.

1 / U = ( 1 / hj ) + ( Fj ) + ( X/K ) + Fv + ( 1/hi )

The hj, hi need to be calculated primarily,

Jacket Side Heat Transfer Co-efficient Calculation:

For Calculation of  hj , we need to know Pipe coil ID, Flowrate in pipe, Fouling factor Fj, Fluid properties[Density, Viscosity, Specific heat, Conductivity],

With an example i'll illustrate here,

Let pipe ID is 2", Flowrate(Q) is 10 Cu.m/hr, Fouling factor(Fj) = 0.0002 h.m².°C/Kcal,

Density (rho) 1000 Kg/Cu.m, Viscosity(μ) 0.01 cP, Specific heat (Cp) 0.9 KCal/Kg.°C, Conductivity(k) 0.2 KCal/m.h.°C,
Viscosity @ wall is (μ_w) = 0.01 cP

Velocity, V = Q/A = 2.7410 m/sec
Reynold's number, Nre = D x V x rho / μ =17638677,
Prandtl's number, Npr = Cp x μ _/ k = 0.2093

using the below co-relation ,

hj x D / K = 0.023 x ( Nre ^ 0.8 ) x ( Npr ^ 0.33 ) x (( μ_/μ_w)^0.14)

hj = 27202.81 Kcal/h.m².°C

Also Read:

Agitation Scale up  

How to find Reactor Heat Transfer Area? 

Vessel Side Heat Transfer Co-efficient Calculation:

For this calculation we need to know the inside fluid properties, which includes Density, Viscosity, Conductivity, Specific Heat, Viscosity @ wall.

Within an example i'll illustrate here,

Agitator RPM 36, Agitator Dia 36", Density (rho) 900 Kg/Cu.m, Viscosity(μ) 2cP, Specific Heat(Cp) 1KCal/Kg.°C, Conductivity (k) 0.5 KCal/m.h.°C, Viscosity @wall (μ_w) 1 cP, Fouling factor(Fv) 0.0002 h.m².°C/Kcal,

Reynold's number, Nre = N x D^2 x rho / μ = 313548,
Prandtl's number, Npr = Cp x μ _/ k = 14.4

Now, Individual Heat Transfer Co-efficient hi is Calculated by following co-relation,

hi x D / K = 0.53 x (Nre ^0.66) x ( Npr ^ 0.33) x (( μ_/μ_w)^0.24)

hi = 1441.16 Kcal/h.m².°C

So, Now our task came to an end,

U value Calculation,

1 / U = 1/hj + Fj + X/K + Fv + 1/hi


1 / U = 1/27202.8 + 0.0002 + 0.0007 + 0.0002 + 1/1441.2

U = 537.019 Kcal/h.m².°C


That's it............Task Completed Cheers....!!

Actually one more thing i should have been included in the basic "What are the factor do the  U value depend?", anyway by the end of this tutorial you should have learned this,

Getting mails asking me to demonstrate with some example data, and that's why today i've illustrated the formulae with data,

If you understood clearly then say cheers, or any queries, feel free to contact me, M happy to help,

Comments are most appreciated.....!!!

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About The Author

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Hi! I am Ajay Kumar Kalva, Currently serving as the CEO of this site, a tech geek by passion, and a chemical process engineer by profession, i'm interested in writing articles regarding technology, hacking and pharma technology.
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