A 3floor hospital ward has a gross floor area of 60000 SF De

A 3floor hospital ward has a gross floor area of 60,000 SF. Determine the number of occupants and the minimum number of fixtures required. Assume the occupancy is equally male and female.

– Assume that 75% of the GFA is for sleeping areas and 25% for outpatient areas

Solution

Area= 60000 SF

number of floors= 3

area per floor= 20000 SF

For ground floor out of this only 75% is used means 15000 SF for sleeping.

Consider,

1. one bed is of 6.66*2.95 = 20 SF

2. Gap between two beds= 16 SF

3. Total area for each bed is= 36 SF

4. Middle corrider = 8.2 feet

5. Area for washrooms = 9.84*11.48 = 113 SF

6. Area for untensils= 5.74*6.56 = 37 SF

7. Lights per ward= with gap of 3 feet

8. Fans per ward= with gap of 3 feet

9. one ward area= 2500 SF

10. Area for corrider of one ward= 500 SF

11. Number of toilets per ward= 3

So total space occupied by 3 toilets and 3 utensils area= 150*3= 450 SF

Space remaining in ward= 2500-450 = 2050 SF

So it may be 45*45 in length and width

area lost by middle space is 45*8.2 = 370 SF

So area remaining = 2050-370= 1680 SF

So number of beds = 1680/36 = 46 beds

For ground floor number of rooms= 5

For First and second floor number of rooms =6

So total beds for GF= 46*5 = 230 beds

total beds for FF= 46*6 = 276 beds

Total beds for SF= 46*6 = 276 beds

Total beds= 230+276+276 = 782 beds

Number of toilets on ground floor= 3*5= 15

Number of toilets on first floor= 3*6= 18

Number of toilets on ground floor= 3*6= 18

Total toilets= 51

As per specifications 22 tubes and 22 fans are required for one room

So tubes and fans for GF= 22*5 = 110

tubes and fans for FF= 22*6 = 132

tubes and fans for SF= 22*6 = 132

Total number of tubes and fans required = 374