IEEE 1584 Equation 4 Units

[bdfig]

Hi, in IEEE 1584 equation (4) (page 11) it states "En is the incident energy (J/cm^2) normalised..."

Then in equation (6) E = 4.184*Cf*En(t/0.2)(610^x/D^x)
E is the incident energy (J/cm^2)

and in section E.3.1
Equation (E.1) E= Cf*En*(t/0.2)(610^x/D^x) in cal/cm^2

Is En from equation (4) in J/cm^2 as stated or cal/cm^2?

 

[jghrist]

J/cm^2 as stated. cal/J = 4.187 (according to Mathcad - it must be 4.184 according to IEEE ;-))

 

[bdfig]

If En is in J/cm^2 and E is in J/cm^2 then why are we introducing the 4.184 cal->J conversion factor.

 

[isu20cpree]

Although the metric unit of energy is the joule, heat is commonly also measured in units called calories (there are about 4.19 joules in a calorie), or in larger units called Calories (note the capital C). A Calorie is 1000 calories, and should always be called a kilocalorie.

 

[bdfig]

What I am saying is that IEEE 1584 provides the following formula to convert from normalized incident energy to the actual incident energy:

E = 4.184*Cf*En(t/0.2)(610^x/D^x)

And E and En are both in J/cm^2.

Therefore why do we have a 4.184 factor?

 

[jghrist]

Therefore why do we have a 4.184 factor?

So that you can have one equation give the answer in J/cm² and the other equation give the answer in cal/cm². I have no idea why IEEE decided to use one unit in the main standard and another unit in Appendix E, but that's what they did and it's as simple as a conversion factor.

 

[bdfig]

Jghrist, ignoring appendix E and only looking at Section 5.3:

En is in J/cm^2 and E is in J/cm^2