I am trying to calculate the energy budget of a water body and the last term I'm trying to calculate is the energy advected from the water body ($Q_v$). This equation is given by:

$Q_v = V ( (T_1 - T_0) / A)$

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where $V$ is the daily inflow (in $\mathrm{m}^3 \mathrm{day}^{-1}$); $T_1$ is the temperature of the inflow ($^\circ\mathrm{C}$); $T_0$ is the temperature of the outflow ($^\circ\mathrm{C}$); $A$ is the surface area of the water body ($1\times10^{10} cm^2$)

and $Q_v$ is given in $\mathrm{cal}\cdot\mathrm{cm}^{-2} \mathrm{day}^{-1}$.

How would I convert $Q_v$ into $\mathrm{W\cdot m}^{-2}$ i.e. Watts per meter squared?

I've seen an example in page 881 in this paper


which shows that

$1 \mathrm{cal\cdot cm}^{–2}\mathrm{h}^{–1} = 4.19\times10^{–4}\mathrm{J\cdot m}^{–2}\mathrm{h}^{–1} = 11.63 \mathrm{W}/\mathrm{m}^2$

Could anyone explain to me the process of calculating this and for the example I provided?

  • $\begingroup$ You cannot convert $\text{cal cm}^{-1} \text{day}^{-1}$ to $\text{W}/\text{m}^2$ because they are not the same physical quantity: $\text{cal cm}^{-1} \text{day}^{-1}$ is energy per length per time, while $\text{W}/\text{m}^2$ is energy per length squared per time. $\endgroup$ – Michiel May 20 '13 at 11:46
  • $\begingroup$ Did you look at the link? They seemed to do it, and that is a published article. Also, I've made a minor change to the post. $\endgroup$ – Emma May 20 '13 at 11:48
  • $\begingroup$ Yes I did, and they don't do that: they convert $\text{cal cm}^{-2} \text{day}^{-1}$ to $\text{W}/\text{m}^2$. Note the -2 on the centimeter $\endgroup$ – Michiel May 20 '13 at 11:49
  • $\begingroup$ OK thanks. Ignoring the equation for Qv, how would you convert 1 cal cm–2 h–1 to W/m2, I'm struggling to work through this step mainly $\endgroup$ – Emma May 20 '13 at 11:57
  • 1
    $\begingroup$ The fact that it is in a published paper doesn't necessarily mean it is correct. It could be a typo, a typesetting error or just an error in itself. I remain convinced that the equation is dimensionally incorrect. Just check the units for yourself: you get $Qv=[\text{m} ^oC/\text{day}]$ $\endgroup$ – Michiel May 20 '13 at 12:31

Converting units is simple if you simply work with the units as algebraic symbols. Thus start with the equalities $$1\textrm{ cal}=4.18\text{ J},$$ $$1\text{ day}=24\times60\times60\text{ s}=86,400\text{ s, and}$$ $$1\text{ cm}=0.01\text{ m}$$ and just plug the numbers in: $$ 1\text{ cal cm}^{-2}\text{ day}^{-1} =\frac{4.18\text{ J}}{(0.01\text{ m})^2 \times 86400\text{ s}}\approx0.48\text{ W m}^{-2}. $$ Easy!


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