Math  /  Geometry

QuestionQ2: It is proposed to illuminate a class room of dimensions 6×8×2.85 m6 \times 8 \times 2.85 \mathrm{~m} to an illuminance (E) of 400 lx at the bench level. The specification calls for luminaires having one 1050 mm 40 W fluorescent natural tube with an initial output of 3200 lumens with white metal base an prismatic plastic diffuser (its UF is given in Table -2). Determine the number of luminaires required for this installation when the MF is 0.7 , respectively. The reflection coefficients are: ( C=\mathrm{C}= 0.70, W=0.3, F=0.20.70, \mathrm{~W}=0.3, \mathrm{~F}=0.2 ) Assuming the working table height is 0.85 m \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \multicolumn{6}{|l|}{Room Reflection} & \multicolumn{6}{|l|}{Room Index} \\ \hline c & W & F & 75 & 1.00 & 1.25 & 1.50 & 2.00 & 2.50 & 3.00 & 4.00 & 5.00 \\ \hline \multicolumn{3}{|l|}{\multirow[t]{3}{*}{.70 .50 .30 .10}} & 44 & . 50 & .56 & . 60 & 65 & .69 & .72 & .7i5 & .777 \\ \hline & & & . 38 & . 44 & .50 & .54 & . 60 & . 64 & 68 & . 72 & .74 \\ \hline & & & . 33 & 40 & 46 & .50 & . 56 & 61 & . 64 & . 69 & . 72 \\ \hline \multirow[t]{3}{*}{. 50} & . 50 & 20 & 39 & 44 & 49 & .5. & . 57 & 60 & 62 & .65 & . 67 \\ \hline & .30 & & 34 & 40 & 44 & 48 & 53 & . 56 & . 59 & 62 & . 64 \\ \hline & . 10 & & .30 & .36 & 41 & 44 & . 50 & .53 & . 56 & . 60 & . 62 \\ \hline \multirow[t]{3}{*}{} & . 50 & 20 & 34 & 39 & 42 & 45 & 49 & 51 & . 53 & .55 & . 57 \\ \hline & .30 & & 30 & . 35 & . 39 & 42 & 46 & 49 & .51 & . 53 & . 55 \\ \hline & 10 & & 27 & .32 & . 36 & .39 & 44 & 47 & 49 & 52 & 54 \\ \hline \multirow[t]{2}{*}{} & 0.00 & . 0 & 22 & 26 & 29 & . 31 & 34 & . 36 & 38 & 40 & . 41 \\ \hline & & & & & & - & \rightarrow UF & 4 & & & \\ \hline \end{tabular}
Note: The reflection coefficients for a standard room are: (0.7,0.5,0.2)(0.7,0.5,0.2)

Studdy Solution
Calculate the number of luminaires needed using the formula:
Number of Luminaires=Total LumensLumens per Luminaire×UF×MF\text{Number of Luminaires} = \frac{\text{Total Lumens}}{\text{Lumens per Luminaire} \times \text{UF} \times \text{MF}}
=192003200×0.50×0.7=19200112017.14= \frac{19200}{3200 \times 0.50 \times 0.7} = \frac{19200}{1120} \approx 17.14
Since the number of luminaires must be a whole number, round up to the nearest whole number:
Number of Luminaires=18\text{Number of Luminaires} = 18
The number of luminaires required is:
18 \boxed{18}

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