- #1

tkfn

- 2

- 0

We are asked to transform this equation:

h=1147(1+0.05625T)[itex]\frac{V^{0.6}}{D^{0.4}}[/itex]

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

h->[itex]\frac{kcal}{h·m^{2}·ºC}[/itex]

T->ºC

V->[itex]\frac{m}{s}[/itex]

D->cm

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

h->[itex]\frac{Btu}{h·ft^{2}·ºF}[/itex]

T->ºF

V->[itex]\frac{ft}{s}[/itex]

D->inches

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[itex]\frac{V^{0.6}}{D^{0.4}}[/itex]

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)[itex]\frac{V^{0.6}}{D^{0.4}}[/itex]

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

h->[itex]\frac{kcal}{h·m^{2}·ºC}[/itex]

T->ºC

V->[itex]\frac{m}{s}[/itex]

D->cm

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

h->[itex]\frac{Btu}{h·ft^{2}·ºF}[/itex]

T->ºF

V->[itex]\frac{ft}{s}[/itex]

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[itex]\frac{V^{0.6}}{D^{0.4}}[/itex]

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: