calculate Q:

Q(kW)=C_hot(T_hot,in-T_hot,out)=C_cold(T_cold,out-T_cold,in)

C(kW/K)=m'(kg/s)*c(kJ/kg K)

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calculate U:

You will need to calculate U(W/m^2 K) so you will need h_hot and h_cold (same units)

1/U=1/h_hot+1/h_cold here you find U

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calculate DT_lm:

DT_lm=(DT_in-DT_out)/ln(DT_in/DT_out) DT meaning temp difference

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calculate A:

A=Q/U*DT_lm (m^2)

See please

Heat Exchanger Design Handbook

https://books.google.com.ua/books/about/Heat_Exchanger_Design_Handbook.html?id=G52EfFF4uQYC&redir_esc=y

Enjoy

Use NTU method for counter fliow heat exchanger.

The Heat Mass design data book will guide you in this concern.

Dear Anand

You can also do this using Comsol Multiphysics. It is in its 5.3 version and can be found on comsol.com.

Regards

Heat transfer coefficients depend in general on

* Geometry

* Fluid properties

* Fluid mass flux (e.g. Reynolds number

The relations quantifying the heat transfer coefficients depend on the stuation at hand (single phase laminar/turbulent flow, boiling, condensation…) and may be found in many available basic textbooks on heat transfer. In addition, one needs to calculate the overall heat transfer resistance R=1/U and the weighed overall temperature driving force for heat transfer (if applicable). You should consult a basic textbook, e.g. Holman: Heat Transfer, Incropera et al: Fundamentals of Heat Transfer, Cengel: Heat and Mass Transfer etc, where you will find the methodoligies you need for all of this. Good Luck!

Dear colleague, every answer will be correct under certain conditions. It is a good answer to give a heat engineer familiar with these conditions and having more experience in that direction. Thus, the error will be reduced to 5-6%, otherwise to 50%.

The respective heat transfer coefficients (the convection coefficients) depend on the flow rates and specific geometries (shell-and-tube, plate type, etc). If you assume bulk fluid properties can be used, then the heat transfer coefficients can be calculated directly using textbook equations or those published by researchers.

Iraklis Zahos-Siagos already provided the basic thermal resistance and mean temperature difference equations in an earlier response. They are also in any first-semester heat transfer textbook. In my experience running identical heat exchangers in concurrent and counterflow, respectively shows approximately 30% lower thermal capacity for the concurrent as a rule of thumb.

All previous answers are correct. The only addition would be a possible thermal resistance (in series) due to scaling after a period of use.

How can we determine the heat transfer coefficients of two fluids in the concurrent flow heat viewer? -Departing from the calculation methodology of the heat exchangers either by the NTU method or by the logarithmic mean temperature difference. By this last one knowing the temperatures of entrances and exits of the fluids, the mass flows and the area of transfer of exchanger can be cleared of the equation U

U= Q/(A·F·DTML)

How is it against flow

T2=Thi-Tco

T1=Tho-Tci

F = is a coefficient that depends if the exchanger if it has one or more steps if it is a single step can be assumed as 1 The other form answers the previous questions. What are the heat transfer coefficients dependent on? U depends on the transfer modes that intervene in the heat exchangers its plates or tubes and shells If it's Tubes and shells

, heat is transferred by convection of the hot substance to the tube then by conduction and finally by convection from the outside of the tube to the cold substance

It can be determined by the following equation

U= (1/hi·Ai)+(ln(Di/Do)/(2·3.14·K·L))+(1/ho·Ao)

The heat transfer coefficients h inside and outside of the tube must be estimated. For this, you need to determine if it is free or forced convection. For each case there are expressions that can be found in the heat transfer fundamentals book

Dear Anand,

Note use of correlations depend on specific application parameters such as type of flow: Single phase /two phase, Laminar/Turbulent, value of dimension less numbers such as reynolds number, Reyleigh Number, prandtl number, Boiling Number, Froude Number, Lockhart Martineil Parameter, Heat Exchanger configuration say Tube in Tube , Shell Tube or Plate Heat Exchanger,etc.

Here i will be showing for a Tube in Tube Heat Exchanger application

Overall heat transfer coefficient) is given by, (1/UA) = (1/Uo*Ao) = (1/Ui*Ai) = { (1 / (hr * Ar)) + (Rf,r / Ar) + ( ( ln(Do/Di)) / (2 * PI * Kwall * Ltube)) + (Rf,liq / Aliq) + ( 1/ (hliq * Aliq))}

Required, Overall heat transfer coefficient, UA(W/K), UA = Desired Q /LMTD

Finding the required heat transfer area or Heat transfer coefficient for predefined heat exchanger, Involves Iterative steps, you need to design it for minimum pressure drop & optimum Heat Transfer Coefficient.

Start Your Calculation By approximately Selecting Sizes of Tubes(OD & Length) Involved & Check in the Final Stage, whether they Meet the Criteria. If Not Again Calculate With Different Selection of Sizes Involved.

3 Things You Need To Find Out First: 1. Required Cooling / Heating Capacity 2. Overall Heat Transfer Coefficient of Heat Exchanger Based On Thermal & Fluids Properties As Well Materials Of Heat Exchangers Invloved. 3. Log Mean Temperature Difference of Both Fluids.

Once, You have Been obtained All of Them, For Required Heat Transfer Area, A = (Q capacity) / (LMTD * Overall Heat Transfer Coefficient).

1. For LMTD equation, Depending On Phase(Two Phase or Single Phase), Concurrent or counter flow & Extreme End Temperatures of Both Fluid Involved, You Can Get Equations From Textbooks of Heat & mass Transfer.

" Thermodynamics an engineering approach & Heat & Mass Transfer by Younus Cengel"

LMTD = {(ΔT1- ΔT2)/ (ln(ΔT1 / ΔT2))}K

2. For Finding Overall Heat Transfer Coefficient,

Hereby i am discussion the procedure for Two phase heat transfer coefficient estimation for a typical tube in tube configuration:

1. Convection Resistance from Refrigerant to Refrigerant Wall, Need to Find Refrigerant Side in case of Two phase heat transfer, Convective Heat Transfer Coefficient, find h refrigerant, will be Using Correlation for Two Phase for Evaporation Based on Laminar or Turbulent Flow & Dimensionless Numbers Parameters.

For example depending on my flow conditions & Thermophysical parameters, for finding convection heat transfer coefficient on refrigerant side, i used following correlation:

"Ginger & Winterton Correlation"

h twp phase = h ref = hl * [ ( 1 ) + ((3000) * (Bo^0.86)) + (1.12)(( x / (1-x))^0.75)((ρl/ρv)^0.41)]

Go through the links:

http://web2.clarkson.edu/projects/subramanian/ch330/notes/Heat%20Transfer%20in%20Flow%20Through%20Conduits.pdf

https://en.wikipedia.org/wiki/Nusselt_number

(http://nptel.ac.in/courses/112105129/pdf/R&AC%20Lecture%2023.pdf)

2. Fouling Resistance Due to Outside of Wall (i.e. Refrigerant Side).

3. Conductive Resistance from Refrigerant Tube Wall, Find R Refri. Wall, Simply Will be Using Correlation for Resistance Provided by Cylindrical Wall.

4. Fouling Resistance due to inside of Wall (i.e. Liquid Wall)

5. Convective Resistance from Liquid Flowing Inside Liquid Tube, Find Liquid Side Convective Heat Transfer Coefficient, will be Using Correlation for Single Phase Based on Laminar or Turbulent Flow & Dimensionless Numbers Parameters.

In this case , all thermodynamic properties need to be evaluated at Mean bulk wall temperature

For single phase turbulent flow,

Nu(Gleneiski Correlation) = https://en.wikipedia.org/wiki/Nusselt_number

After Estimating All Thermal Resistances, Overall Convective Heat Transfer Coefficient Will be Found.

Hence, By Placing in Required Heat Transfer Area Can be Found by A = Q / (U) * LMTD.

Once You Have Obtained Required Size of heat Exchanger, 1. It is Very Important to Check Total Pressure Drop Involved

Total pressure drop = frictional pressure drop due to flow +static pressure drop + momentum pressure drop(accelkerational)

Estimation of frictional pressure loss by Friedel Correlation

Frictional pressure drop:

ΔP fric = Δpliq. * (Mfri.^2), Psig

Δpliq(Pa) = 4 * Fl0 * (L / di)(Mdot refrigerant ^2)( 1/ 2ρl) = 2 * Fl * G^2 / (ρl * D) * L

Liquid Only Two phase Multiplier as per Friedel correlation, Mfri^2 = ( (E ) + ( ( 3.24 * F * H) / ( ( Frh^0.045) * (WeL^0.035)) ) )

Accelerational pressure drop:

Acceleration pressure drop is given by, P acc. Loss(Pa) =

L * [ ( ( 16 * ( Mdot refrigerant ^2 ) ) / (Pi^2 )( D^4) ) * { ( ( (x2 ^2 ) / ( ρv * e2 ) ) + ( ( ( 1- x2)^2) / ( ρl * ( 1- e2) ) ) ) + ( (x1 ^2 ) / ( ρv * e1 ) ) + ( ( ( 1- x1)^2) / ( ρl * ( 1- e1) ) ) )

If Total Pressure Drop Exceeds Design Criteria, Try Another Iteration by Selecting Pipes of Different OD's.

Hope This Helps You. Thanks.