My Solution Attempt: Visualized

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The discussion revolves around a solution attempt for calculating the average illuminance level from a point luminaire with a luminous intensity of 2000 cd, located 3 m from a plane with an annular cavity. Participants express difficulty in following the calculations presented in images, suggesting that the work should be typed out, preferably using LaTeX, and that variables should be defined rather than using numeric values. There is confusion regarding the term "point luminaire" and the specifics of the annular cavity, leading to frustration over understanding the final illuminance value of 879.55 lux. Suggestions for improvement emphasize clarity and adherence to forum rules regarding the presentation of mathematical work. The conversation highlights the importance of clear communication in technical discussions.
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Homework Statement
A point luminaire with a luminous intensity of 2000 cd in all directions is 3 m away from the center of a plane
exists. In the middle of this distance, it has an annular cavity with an outer diameter of 3 m and an inner diameter of 2.4 m.
The plane has a parallel screen. Accordingly, the average illuminance level in the illuminated part of the plane
calculate.
Relevant Equations
Relevant Equations Luminous intensity [cd] = luminous flux [lm] / solid angle [sr].
Illuminance is calculated with the following formula: Lux [lx] = luminous flux [lm] / area [m2].
My solution attempt can be seen on these pictures.
 

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heisenbergs said:
Homework Statement: A point luminaire with a luminous intensity of 2000 cd in all directions is 3 m away from the center of a plane
exists. In the middle of this distance, it has an annular cavity with an outer diameter of 3 m and an inner diameter of 2.4 m.
The plane has a parallel screen. Accordingly, the average illuminance level in the illuminated part of the plane
calculate.
Relevant Equations: Relevant Equations Luminous intensity [cd] = luminous flux [lm] / solid angle [sr].
Illuminance is calculated with the following formula: Lux [lx] = luminous flux [lm] / area [m2].

My solution attempt can be seen on these pictures.
Welcome to the forum, but…

Your working is too difficult to follow.
According to forum rules, images are for diagrams and printed matter, such as text book extracts. Your working should be typed in, preferably using LaTeX.
It will also help if you avoid use of the numeric values; just define variables for them and work algebraically. That is good style anyway because it has many advantages.

I for one do not know what a "point luminaire" is, and a web search has not enlightened me. Please explain about the cavity.
 
In addition: I spent quite some time trying to decypher what you are doing to end up with 879.55 lux but had to give up, slightly frustrated. (Note: the suggested accuracy is non-existent and the units don't look like lx at all, but I can be mistaken...)

Nevertheless.
:welcome: ##\qquad## !​

(rules/guidelines are here)

##\ ##
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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