Thermal conductivity of any fluid is a property of the fluid (similar to viscosity, specific heat capacity), and not related to the flow. You may only determine the thermal conductivity from Your Experiment only if you are able to measure the temperature gradient very near the wall, which, in practice, is impossible. (Note: Fourier's Law - 'heat flux through tube wall=fluid conductivity multiplied by fluid temperature gradient' only applies at the wall.)

If you have a heat transfer correlation you trust 100%, then you may extract the thermal conductivity from the measured heat transfer coefficient and the Nu-correlation. (However, remember that even the best correlations for turbulent tube flows typically have an error band of at least 10%. Also note that all uncertainties in Your Experiment will propagate to the value obtained for the thermal conductivity…..)

My advice is to measure the conductivity using an appropriate Method, alternatively see if you can find it presented in numerical form somewhere.

Thermal conductivity can be calculated by dividing the heat flux applied to internal tube to the temperature distribution within the fluid flowing inside the tube. Therefore by using thermocouples you measure the radial temperatures in the fluid, and so calculate thermal conductivity.

Thermal condictivity is calculated from heat flux dividing to the temperature difference.

This is not as simple as has been suggested. You may not be able to accurately estimate the liquid thermal conductivity with this data since not only are you dealing with a liquid but that liquid is also flowing in the tubes and convection heat transfer is governing.

Thermal conductivity of any liquid is mainly a function of temperature and pressure. If pressure variations are less then thermal conductivity will be the function of temperature only. So, If you are to able to measure the temperature distribution in the fluid then thermal conductivity can be estimated as a function of temperature.

The thermal conductivity of liquid dipends on viscosity and velocity of liquid inside the tube.it behave like a shell type heat exchanger so you can use the lmtd method to calculate the desired parameters

Thermal conductivity of any fluid is a property of the fluid (similar to viscosity, specific heat capacity), and not related to the flow. You may only determine the thermal conductivity from Your Experiment only if you are able to measure the temperature gradient very near the wall, which, in practice, is impossible. (Note: Fourier's Law - 'heat flux through tube wall=fluid conductivity multiplied by fluid temperature gradient' only applies at the wall.)

If you have a heat transfer correlation you trust 100%, then you may extract the thermal conductivity from the measured heat transfer coefficient and the Nu-correlation. (However, remember that even the best correlations for turbulent tube flows typically have an error band of at least 10%. Also note that all uncertainties in Your Experiment will propagate to the value obtained for the thermal conductivity…..)

My advice is to measure the conductivity using an appropriate Method, alternatively see if you can find it presented in numerical form somewhere.

Design methods for heat exchangers may appear somewhat complex, but typically ultimately rely on simple empirical correlations for heat transfer, relating dimensionless numbers, such as Nu, Re, and Pr. For instance; for forced turbulent convection of fluids through pipes, it can be often accepted that: Nu = 0.023(Re)0.28(Pr)0.4. Here, the Prandtl number is just characteristic of the fluid flowing through the pipe. The Nusselt number is chiefly a design dependent parameter. Re and Nu not only depend on fluid properties, but also on a flow related parameter; namely, the velocity of the fluid (flow rate) ― for Re ― and the heat transfer coefficient ― for Nu. Because both Nu and Pr are dependent on fluid thermal conductivity, we may find convenient to combine both. The mentioned correlation could then be given as Nu/Pr0.4 = 0.023(Re)0.28, with Nu/Pr0.4 = (h·d/k)/ (cp·μ/k)0.4 = (h·d/k0.6)/ (cp·μ)0.4. Here, h is the convective heat transfer coefficient, d is the pipe internal diameter, k is the thermal conductivity of the fluid, cp is the fluid specific heat, and μ is its dynamic viscosity.

For a shell and tube heat exchanger as understood, temperatures (inlet and wall temperatures), heat flow rate (generally average of shell side and tube side), overall heat transfer coefficient (inside and outside heat transfer coefficients and tube resistance) are calculated. This means that thermal analysis (final) has been done (that means after performing preliminary analysis). Hence seeing this, it is assumed that name of the liquid flowing inside the tube of shell and tube heat exchanger is known. Generally, the heat transfer from shell to tube is generally by forced convection.

In order to determine thermal conductivity of known liquid flowing inside tube at a particular temperature, bulk temperature is to determined i.e.( Tw+(Ti+To)/2)/2, where, Tw is average wall temperature which is the average of temperature of thermocouples fixed on the surface of tube and Ti, To are inlet and outlet temperatures of fluid flowing inside a tube. After this determination, taking heat transfer data book or handbook of heat transfer at that particular bulk temperature for a known liquid, thermal conductivity can be estimated. For this thermal conductivity of liquid determination, I think there is no need of using the above determined values for back calculation.

Second Mehod: If thermal conductivity is to determined for unknown liquid flowing inside tube of shell and tube heat exchanger, having known inlet (Ti) and outlet temperature (To) of fluid and assuming tube vertical of length (L) and as input at inlet (Q) is more so that temperature at inlet (Ti) is more, heat transfer rate (Q) is known and taking heat flow is along vertical, Applying Fourier's law of heat conduction, Q = -K A (Ti-To)/L, If A is cross sectional area, thermal conductivity (K) can be calculated.

Good details.

@ Pisipaty . sir your 1st answer is looking close to solution. I think 2nd answer is wrong bcoz heat tnsfr in the heat exchanger along the length is not by conduction mainly rather it is forced convection.So how can Fourier's law be applied there?

The second method described as like this: Coming to determination of thermal conductivity of liquids experiment, heat (Q), here also it is Q, is given underneath and on either side of the surface of small circular cavity (inside liquid) temperatures are measured, here it is taken as Ti and To, cross sectional area there is circular here also the same as tube cross section circular. As said forced convection exists but under steady state conditions all are constant and temperatures can be taken on either sides. Hence tube taking vertical, length, L. Thermal conductivity of liquid (K) as in the case of experiment

can be determined if unknown liquid is taken. We can draw the diagram and check. This is an appropriate method for unknown liquid flowing inside tube.

You can try measure thermal conductivity using a know heat flow in your sample, employing a lineat heat source techinique.

Yes, may be. Thanks Jorge Zazueta-Niebla.

If you can find heat transfer coefficient and Nusselt number, you can find the fluid thermal conductivity. I don't think that you will be able to find the accurate values for them, so I recommend for you to use the linear heat source method for a fluid between two close walls in order to avoid the effect of natural convection.

Thermal conductivity is a property of fluid not of flow, and you can measure it bu K2D meter

Good answer Dear.Kadhum A.

You can measure the heat rate for the two streams, hot and cold, by using thermodynamic relation: Q = m dot.cp. delta-T. Then Q = h.A.LMTD. If you want to make correlation with Nusselt number, then you must define the evaluation temperature to determine k. Basically, you may choose this evaluation temperature as you like. However, if you'd like to compare your correlation to another ones then you should use the same definition of the evaluaiton temperature. So, please check this one.

Regards

This a good and simple experimental method.