tkfn
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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.
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.

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