# Heat Transfer Coefficient Problem

## The Problem

An unusual reference gives the formula below for the heat transfer coefficient of an emulsion.

$$H = 0.0061\left(X^{0.6} + 0.038T\right)\cdot V^{0.35}$$

In the formula $$T$$ is the temperature in $$\pu{^\circ F}$$, $$V$$ is the flow rate of the emulsion in $$\pu{\frac{ft}{min}}$$, $$X$$ is a characteristic dimension in inches and $$H$$ is the heat transfer coefficient in $$\pu{BTU h^-1 ft^2 ^\circ F}$$.

Calculate the heat transfer coefficient in $$\pu{W m-2 K-1}$$ when emulsion at $$\pu{30 ^\circ C}$$ flows at $$\pu{1.25 m s-1}$$ and the characteristic dimension is $$\pu{0.8 mm}$$.

## My Attempt

1. I converted the characteristic dimension ($$X$$) from $$\pu{mm}$$ to $$\pu{in}$$.

$$\pu{0.8 mm} ⋅ \frac{\pu{39.37 in}}{\pu{1000 mm}} = \pu{0.03 in}$$

1. I converted the temperature ($$T$$) from $$\pu{^\circ C}$$ to $$\pu{^\circ F}$$.

$$\frac{(n \ \pu{^\circ F} - 32)}{1.8} = 30 \ \pu{^\circ C}$$

$$n \ \pu{^\circ F} = \pu{86 ^\circ F}$$

1. I converted the flow rate of the emulsion ($$V$$) from $$\pu{m/s}$$ to $$\pu{ft/min}$$.

$$\frac{\pu{1.25 m}}{\pu{s}} ⋅ \frac{\pu{3.28 ft}}{\pu{m}} ⋅ \frac{\pu{60 s}}{\pu{min}} = \frac{\pu{246 ft}}{\pu{min}}$$

1. I plugged the values of $$X$$, $$T$$, and $$V$$ into the equation to solve for $$H$$.

$$H = 0.0061\left(0.03^{0.6}\ \mathrm{in} + 0.038(86\ \mathrm{^\circ F})\right)\left(\frac{246\ \mathrm{ft}}{\mathrm{min}}\right)^{0.35}$$

Is my work correct so far? If yes, I'm a little confused as to what I should do next. How do I simplify the units in my last step in order to then convert to a value with the desired units $$\pu{W m-2 K^-1}$$?

• there is a little typo in the last equation , the mm should be ft. Numerically then you have H in BTU /hr ft$^2$F. Now convert this into SI unit by substituting with 3600 s /hour, 0.0254 *12m /ft etc and convert temperature in F into Kelvin. – porphyrin Jun 26 '20 at 17:05
• Shouldn't the mm actually be inches not ft (the value I calculated in step 1 is in inches because the problem defines X as the 'characteristic dimension in inches')? – air_nomad Jun 26 '20 at 21:38
• however, you want values in ft as used in the other part of the equation and to get consistent values to get BTU/hr/ft$^2$ then you change each of these to SI units. – porphyrin Jun 27 '20 at 6:19

Is my work correct so far?

Yes, your conversions to the units for the inputs to the equation (#1 to #3) look correct to me.

How do I simplify the units in my last step in order to then convert to a value with the desired units...?

You don't.

This sort of expression embodies a purely empirical relationship -- it has no particular basis of connection to the underlying physical relationships between these various quantities. The reason the units of each of the input quantities are specified ($$^\circ\mathrm{F}$$, $$\mathrm{\frac{ft}{min}}$$, etc.) is because you're supposed to use just the numerical values of those quantities, after conversion to the indicated units, and substitute those into the equation. So, instead of the equation you have in #4, it would just be:

$$H = 0.0061 \cdot \left(0.03^{0.6} + 0.038 \cdot 86\right) \cdot 246^{0.35}$$

Then, you take the purely numerical result you obtain for $$H$$ from this equation and tack on the indicated units; in this case, $$\mathrm{BTU \cdot {hr}^{-1} \cdot {ft}^{-2} \cdot \ {^\circ F}^{-1}}$$. From there, you can carry out a straightforward conversion to the metric units asked for in the final answer.

These sorts of empirical correlations are extremely common in chemical engineering practice, where one doesn't especially care if an equation has solid theoretical underpinnings, as long as it performs reasonably well in predicting quantities of interest.

To note, while you can construct internally consistent units for the various constants in the formula, they usually don't end up having any particular useful physical interpretation. For example, since it's only possible to add two quantities together if they have the same units, the $$0.6$$ exponent on $$X$$ means that the $$0.038$$ coefficient on $$T$$ would have to have a dimensional value of $$0.038\ \mathrm{{in}^{0.6} \over {^\circ F}}$$, which has no particular physical significance that I can determine. I'm not even sure what sense there is to be made of units such as "$$\mathrm{{in}^{0.6}}$$", "$$\left(\mathrm{ft\over min}\right)^{0.35}$$", etc.

• That clears it up a lot since this problem is for an assignment in a chemical engineering course I'm taking. =) But does that mean that if that was done for more theoretical purposes in chemistry, I should simplify the units in the last step? – air_nomad Jun 26 '20 at 21:36
• @air_nomad See edit. tldr; No. :-) – hBy2Py Jun 27 '20 at 18:20