Help with Sample PE Power Exam Illumination Question

[wfg42438]

I tried looking at the question below and I can't seem to understand the practice exam solution.



Equations and Variable Definitions provided in NCEES handbook:













When looking at the question below what i gathered was the following:
Givens:
E (Illumination)= 50 fc
Length of Room= 50 Ft & Width of Room =30 ft
CU=0.78

Unknown:
Luminous Flux=?
Based on the NCEES handbook definitions and equations below I figured what the question is after is the luminous flux as the output is in lumens. Here my only unknown is "I" which is in candelas.

So using the equation from 2.2.3.2 i Figured let me solve for "I" given E=50 fc and D=50 ft * 30ft
This returns I=75000 candelas

Of course, if I plug this into the luminous flux equation i get ridiculously large number meaning what I'm plugging in is wrong.

Based on the Solution below they simply find the luminous intensity in candelas and then divide that by the coefficient of utility.

This seems wrong as the luminous intensity is in candelas and the question clearly requires the luminous flux in lumens.

Also, for luminous flux isn't "I" supposed to be in candelas?




Please let me know if someone out there can explain where im making a mistake. Im sure it something simple I may be overlooking.
Thanks in Advance!

 

[akyip]

It's been a bit of a long time since I posted here with solutions, so I am a bit rusty.

Attached is what I have. For lighting illumination problems, I prefer using this formula I first learned from the Complex Imaginary PE Power exam sets:

Luminous Flux = (FC * Area) / (CU * MF)

It may also be written as:

Luminous Flux = (E * Area) / (CU * MF)

or

Luminous Flux = (FC * Area) / (CU * BF * LLF)

They are variations of the same basic luminous flux equation that I have seen from Complex Imaginary.

FC or E is the illumination, typically in footcandles.

I've seen MF (maintenance factor) in the NCEES practice exam and also the Complex Imaginary set (I think). In some other practice exam problem sets, I've seen BF (ballast factor) and LLF (light loss factor) also used.

Again, the main equation I use for luminous flux is:

Luminous Flux = (FC * Area) / (CU * MF)

If MF, BF, or LLF are not given, you can just assume that MF = 1.

 

[wfg42438]

Thanks for the feedback.

The only thing I'm wondering is there any way for me to use the equation below provided?


My main concern is during the exam what i will have are NCEEs equations. So i prefer to be able to use them in the event i blank out and forget the shortcut you have provided.

 

[akyip]

Luminous flux is in lumens (lm). Illuminance (typically called E, although I guess the NCEES reference book uses I) is given in footcandles or lux.

Footcandles (fc) is lumens over square foot (lm/ft^2).
Lux (lx) is lunens over square meters (lm/m^2).

So I guess one way to look at that NCEES equation is:

I = illuminance
A = W * L = area
CU = coefficient of utilization
MF = 1 is assumed for maintenance factor if not given

 

[wfg42438]

Thanks for the feedback

Your method makes sense but the equations given by NCEES for luminous flux puzzle me now since the units dont work out
Based on their variable definitions they Say "I" is the illuminance in candelas
but for the luminous flux to work out does "I" not need to be in foot candelas meaning I should really be "E"??




Link to NCESS handbook where the equations above originate from:
https://account.ncees.org/reference-handbooks/56/download
Also how do we end up with lumens when the numerator returns candelas?