# Dimensional analysis, conversion of units in an equation

We are asked to transform this equation:

h=1147(1+0.05625T)$\frac{V^{0.6}}{D^{0.4}}$

where the units of each variable h, T, V and D are:

h->$\frac{kcal}{h·m^{2}·ºC}$
T->ºC
V->$\frac{m}{s}$
D->cm

into another equation where the units of each variable h, T, V and D now are:

h->$\frac{Btu}{h·ft^{2}·ºF}$
T->ºF
V->$\frac{ft}{s}$
D->inches

Data:
1 Btu <> 0.252 kcal
1 ft <> 0.3048m
1 ft <> 12 inches
1 ºC <> 1.8 ºF
(ºF) = 1.8(ºC)+32

The professor said the solution is

h=2.479T$\frac{V^{0.6}}{D^{0.4}}$

however I've tried the very same solution method which had worked for me in the past to solve other conversion problems but I can't seem to get the answer right, I'm confused as to which temperature conversion "factor" should we use, 1 ºC<>1.8 ºF or (ºF)=1.8(ºC)+32? I have really given up on this. Last edited:

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SteamKing
Staff Emeritus
Homework Helper

In your unit conversions, you have something per degree C. You are not trying to find a particular temperature, so F = 1.8C + 32 is not the correct conversion.

Well what i did first is find out the units of the constant 1147 in the original equation so the principle of dimensional homogeneity at both sides of the = sign is satisfied:

$\frac{kcal}{h·m^{2}·ºC}$=$\frac{kcal·cm^{0.4}·s^{0.6}}{h·m^{2.6}·ºC^{2}}$·ºC·$\frac{m^{0.6}}{cm^{0.4}·s^{0.6}}$

Then I repeated the same process for the units of the new desired equation:

$\frac{Btu}{h·ft^{2}·ºF}$=$\frac{Btu·inches^{0.4}·s^{0.6}}{h·ft^{2.6}·ºF^{2}}$·ºF·$\frac{ft^{0.6}}{inch^{0.4}·s^{0.6}}$

Then I made an attempt to convert the constant with its units from the original equation into a constant with the units from the desired equation:

1147$\frac{kcal·cm^{0.4}·s^{0.6}}{h·m^{2.6}·ºC^{2}}$·$\frac{Btu}{0.252kcal}$·$\frac{0.394^{0.4}inches^{0.4}}{cm^{0.4}}$·$\frac{0.3048^{2.6}m^{2.6}}{ft^{2.6}}$· This is where I don't know how to proceed in order to convert the temperature units from ºC$^{2}$ to ºF$^{2}$, the professor said we should use the factor 1ºC <> 1.8ºF when we have ΔT in an equation BUT when we have a specified T we should use (ºF)=1.8(ºC)+32, but SteamKing disagrees so I don't know how to proceed at all. I have a hunch my professor made a mistake but I'm not sure.