Urgent advise in heat transfer fluids.

[lombardo]

I want to calculate energy savings with a high performance heat transfer fluid. This fluid have 50% better thermal conductivity but the specific heat is almost the same compared with the actual lubricant I use, so my questions are:
how can I calculate the savings if the thermal conductivity of the nwe fluid I want to use is 50% higher but the specific heat is almost the same with my actual fluid.?
Is important to use a fluid with a high specific heat or is better to use a fluid with a low specific heat and why?
Really apreciate your help.
Thanks in advance
Alejandro Peterson
Mexico

 

[TD2K]

The higher thermal conductivity will improve the heat transfer coefficient. Higher heat transfer coefficient means a lower dT is needed and thus you need to circulate less fluid. That saves on pumping costs and maybe somewhat on your fired heater costs (I would think on the heater the controlling resistance is on the gas side hence you may not see much improvement).

Ultimately, you are however still transferring the same number of BTUs or kJs from your heater (assuming you have a fired heater).

If the fluid has a low specific heat, you need to circulate more of it for a given temperature change so you piping, pumps, etc increases.

 

[lombardo]

Thanks TD2K,
Thanks a lot for your prompt response,
You mean a low specific heat fluid need to be circulating faster than a high specific heat fluid in order to mantain the same temperature, but the higher specific heat fluid need more heat in order to increase its temperature 8 that means to me more diesel comsuption)so what is the best fluid.
High thermal conductivity and high specific heat?
Or high thermal conductivity and low specific heat?
The oil I want to use is the Royal Purple high therm 707 (high performance fluid) and the manufacturer of this lubricant told me that the high specific heat retains more heat but I think the higher specific heat means more heat in order to heat the fluid so increase the diesel comsuption
or Im wrong?
Thanks again

 

[quark]

Just play with Q = mCpdT and you can find the solution yourself.

Q/dT = mCp and if you have constant Q and dT, if the specific heat of the fluid is high then to receive or reject heat, the fluid flowrate will be lower.

Though you consume more fuel to heat up the fluid with higher specific heat, you are transfering higher heat content to the end user at the same time. So you can obviously reduce the flowrate of the fluid. You should note that the heat capacity (the product of mass and specific heat of a fluid) will always be constant for a constant heat transfer rate and temperature difference.
With low flowrates, you can go for a smaller size pump, smaller size piping and accessories etc.

The advantage with higher heat transfer coefficient is that the bulk temperature tends to near the film temperature and thus better transfer rates of heat. This may not be very significant when compared with the savings you get with a higher specific heat fluids.

 

[25362]

There are many factors to contemplate when comparing heat transfer fluids, besides heat transport and heat transfer performance, which refer to safety, cost and operability, which would take up quite a long response to deal with.

1. Heat transport.

The volumetric flow rate through the heater V (m³/h), to transport a certain quantity of heat Q (kJ/h), is

V = (Q/[ρ]c)(1/[Δ]t)​

where

[ρ] = density, kg/m³
c = specific heat, kJ/kg K
[ρ]c = vol. sp. heat, kJ/m³ K
[Δ]t, (out-in) temp. difference, K

High values of density and specific heat result in reduced flows and improved accompanying economies. Of course, a further reduction in flow rates can be achieved by increasing the [Δ]t; however, a temp. difference of 20-30^oC has proved to be advantageous in most field applications.

2. Heat transfer (HT)

Regarding just HT performance, let's say that besides thermal conductivity and specific heat, two other physical properties are of importance in determining the convective HT coefficient h, density and viscosity.

Simplified formulas can be used for comparison purposes, (e.g. the Sieder and Tate expression) for h, W/m² K, provided flow conditions in heat exchange equipment are in the range of 1 to 4 m/s.
One of them:

h = 0.0235 v^0.8[κ]^0.62([ρ]c)^0.48d^-0.2[μ]^-0.32​

where
[κ] = thermal conductivity, W/m K
[μ] = kinematic viscosity, m²/s
d = internal tube dia., m
v = flow velocity, m/s

Assuming the ratio of bulk and wall viscosity to the 0.14 power is ~1.

As a ROT one can say that good h values can always be expected with low-viscosity HT fluids. Note the effect velocities have on h.

When comparing results (for about equal linear velocities) between synthetic heat transfer media and mineral-oil based fluids it has been found that both give almost equal results at ~300^oC, but differ at lower temperatures at which there is a small advantage for synthetic fluids; this difference becomes greater at lower temperatures.

3. General

It is worthy to note that the fluids' comparative influence on heat transport and heat transfer is well below the undesirable effects caused by deposits, decomposition and contamination of either, the heat-transfer medium or the consumer side.

Thus the decision on what high-temperature fluid to use shifts predominantly to lower-temperature behaviour (including pumpability during start up in cold seasons) on the one hand and thermal stability (ie, service life) considerations on the other.

 

[mauner]

To calculate energy savings you need to consider how effective is the heat transfer equipment.

Less energy is wasted when you have lowest exit temperature from the hot fluid (assuming mass flow is constant).

On the hot side you have m_hot*cp*dt = C_hot*dt

and on the cold side: m_cold*cp*dt = C_cold*dt.

Q_max = C_min * (T_hot - T_cold)

Where:

C_min = lower value of the product of the mass flow and the specific heat. (C_hot or C_cold)
T_hot = hot fluid in
T_cold = cold fluid in.

Eff = Q_actual/Q_max

The bigger C_min the greater the maximum heat transfer.
If the new fluid is the maximum, the equations used to calculate Eff uses C_min/C_max. Eff approachs 100% as C_min/C_max approachs zero.

You can read more about the Epsilon-NTU method in any heat transfer book to learn how to calulate the effective heat transfer for a given UA, Cmin, and Cmax.

As for the change in the overall heat transfer coefficient, U, increasing the thermal conductivity, k, increases convective heat tranfer coefficient, h, which increases U.

This is mainly due to Nusselts number Nu = h*d/k or h=k*Nu/d

Tube side flow is typically a function of Reynolds number and Prandlt number ( Nu = .023*Re^.8* Pr^.3). The minor effect of increasing K is to decrease Prandlt's number Pr = Cp*viscosity/k which decreases h.

Increasing Cp offsets the decrease in h resulting in a larger U and more effective heat transfer.

Assuming your not changing mass flow rate, the bottom line is that you want a:

Larger Cp and a Larger K

 

[lombardo]

Really Really apreciate your help guys, I read a couple of thechnical sheets of heat transfer fluids and all the shhets said "this product save energy because the higher specific heat and the greater thermal conductivity that means low energy costs (low diesel comsuption) but wwhat I do not understand is why is desired a high specific heat if the higher specific means technically more heat requiered in order to increase the temperature of the fluid (for me that means more diesel comsuption)
Thanks Again

 

[IRstuff]

You're confused. The point is to NOT increase the temperature of the coolant. This increases the heat flow from the object being cooled.

TTFN

 

[lombardo]

could you explain me??

 

[25362]

Lombardo, consider the well-known expression Q = mC[Δ]t. For a constant heat load Q to be taken from the heater and transferred to the consumers, a higher heat capacity of the thermal fluid C means either a lower flow rate m or a narrower [Δ]t.

Now, when considering a constant flow rate, this translates into a narrower [Δ]t, not only on the heater but also at the consumers' end.

Besides, the heater efficiency would be improved because the greater convective heat transfer coefficient (about proportional to C^0.4) would mean a cooler tube temperature, a more efficient heat transfer, and probably less thermal degradation and fouling.

 

[mauner]

The greater specific heat does require more energy to start up the equipment.

Thermal performance is measured at steady state. If your equipment will be turned off every night and started up every day then you need to consider the transient heat added and lost.

Consider the fluid contained in a system of piping, pumps and heat exchangers was equal to 500 gallons of water.

The assumed start up required heating the system up from 68F to 168F. The amount of heat needed is

415,000 Btu = 500 gal * 8.3 lb/gal * 1 Btu/(lb*F)*100F

This equates to 3.5 gallons of No. 2 fuel oil
(assumed to be burned at 82% efficiency, heating value of 19,000 Btu/lb & a density of 7.6 lb/gallon).

If the new fluid Cp is 30% greater you would consume 4.5 gallons to start up your equipment.

The cost of this extra gallon would need to be subtracted from your potential savings. If you consumed 100 gallons a day at 82% efficieny you will need to improve your efficiency to 83% to compensate.

Higher efficiencies will be needed to compensate for the added cost to buy the new fluid.

If your plant uses make-up water due to equipment operation, leaks, or frequent maintenance, you would need to subtract the higher cost to store, handle, and replace the fluid.

Marketing departments and application engineers selling a product don't know this or offer this information.

That is why Engineers like myself have a job.

 

[willard3]

It will also be useful to consider the higher kinematic viscosity of most heat transfer fluids.

If the kinematic viscosity of the fluid is higher, the velocity will be higher than water and the pumping hp will be higher.