• Support PF! Buy your school textbooks, materials and every day products Here!

Dimensional analysis, conversion of units in an equation

  • Thread starter tkfn
  • Start date
  • #1
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

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. :confused:
 
Last edited:

Answers and Replies

  • #2
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,666
Show your work.

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.
 
  • #3
2
0
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:

[itex]\frac{kcal}{h·m^{2}·ºC}[/itex]=[itex]\frac{kcal·cm^{0.4}·s^{0.6}}{h·m^{2.6}·ºC^{2}}[/itex]·ºC·[itex]\frac{m^{0.6}}{cm^{0.4}·s^{0.6}}[/itex]

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

[itex]\frac{Btu}{h·ft^{2}·ºF}[/itex]=[itex]\frac{Btu·inches^{0.4}·s^{0.6}}{h·ft^{2.6}·ºF^{2}}[/itex]·ºF·[itex]\frac{ft^{0.6}}{inch^{0.4}·s^{0.6}}[/itex]

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[itex]\frac{kcal·cm^{0.4}·s^{0.6}}{h·m^{2.6}·ºC^{2}}[/itex]·[itex]\frac{Btu}{0.252kcal}[/itex]·[itex]\frac{0.394^{0.4}inches^{0.4}}{cm^{0.4}}[/itex]·[itex]\frac{0.3048^{2.6}m^{2.6}}{ft^{2.6}}[/itex]· This is where I don't know how to proceed in order to convert the temperature units from ºC[itex]^{2}[/itex] to ºF[itex]^{2}[/itex], 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.
 

Related Threads on Dimensional analysis, conversion of units in an equation

Replies
0
Views
2K
Replies
1
Views
678
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
6
Views
935
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
23
Views
4K
Replies
2
Views
19K
  • Last Post
Replies
4
Views
731
Top