A concrete plant uses the following proportions per m3 of concrete: cement 310 kg; water 180 kg; fine aggregate 750 kg and coarse aggregate 1080 kg both in SSD conditions. If the fine aggregate in the field has 4% excess water find the revised quantity of water. cehgg

150 kg

Step-by-step explanation:

Given that

cement = 310 kg  

Fine aggregate = 750 kg

coarse aggregate = 1080 kg

mass of water = 180 kg

given that aggregate in field is 4% excess water

Determine the revised quantity of water

mass of concrete per m^3 = mass of cement + mass of water + mass of fine aggregate + mass of coarse aggregate

therefore mass of concrete = 310 + 180  + 750 + 1080 =  2320 kg

given that fine aggregate in field has 4% excess water this simply means it was void of air in the concrete the new weight = (1.04 * 750) = 780

Hence the revised quantity of water ( x )

= 310 + x + 780 + 1080 = 2320

therefore x = 150 kg

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47