P4 – Use mathematical techniques to solve construction problems associated with simple perimeters, areas and volumes.
D1 – Independently carry out checks on calculations using relevant alternative mathematical methods and make appropriate judgements on the outcome.
Question 1
Measurements -Cistern 0.5m x 0.3m x 0.5m filled to 0.1m -Pipe work 22mm 30m x 0.02m 15mm 50m x 0.013m -Radiators 8 x (1m x 0.05m x 0.48m) -Boiler Holds 25 litres of water.
Cistern Radiators
V = L x W x D V = 8 x (L x W x D)
V = 0.5m x 0.3m x 0.1m V = 8 x (1m x 0.05m x 0.48m)
V = 0.015m3 V = 8 x (0.024m3)
Converted to Litres 15 l V = 0.192m3 Converted to Litres 192 l
Pipe work
22mm V = πr2h 15mm V = πr2h V = π0.012 x 30 V = π0.00652 x 50 V = 9.424777961x10-3 V = 6.636614481x10-3 V = 0.00942m3 (5dp) V = 0.00664m3 (5dp) Converted to Litres 9.4 l (x 1000) Converted to Litres 6.6 l (x 1000) 9.4 l 6.6 l
A) Determine the volume of water the system contains when cold.
Total Volume = 25 + 15 + 192 + 9.4 + 6.6 = 248 litres
B) When the water in the system is at its working temperature it will expand by 5%. Determine the volume of hot water contained in the system when it is at working temperature
5% = 0.05 x 248 = 12.4
Hot water Volume = 12.4 + 248 = 260.4 l
C) Determine the depth of water in the feed & expansion cistern when the system is at working temperature
12.4 l+15 l = 27.4 l 27.4 / 1000 = 0.0274m3 (extra amount) = Volume = L x W x D = = =0.1826m or 182.6mm
Question 2
a) Surface Area
Cylinder = πr2 + 2πrh
= (π x 0.22) + (2 x π x 0.2 x 0.9)
= (0.1257) + (1.1310) (4dp)
= 1.26m2 (2dp)
Hemisphere = 2 x π x r2 = 2 x π x 0.22 = 0.251m2
Total surface Area = 1.26 + 0.251 = 1.51m2
b) Volume of water
Cylinder = πr2h = π x 0.22 x 0.9 = 0.113m3
Hemisphere = 0.5 x r3 = 0.5 x 0.23 =0.0168m3 (4dp)
Total Volume = 0.113 + 0.0168 = 0.1298m3 (4dp) = 0.1298 x 1000 - (To convert from m3 to litres = x 1000) = 130 l (0dp)
c) Calculate 5% = 0.1298 x 1000
= 130 litres (0dp)
=0.05 x 130
=6.5 l
d) Estimate answers
I would estimate the volume of the cylinder by reclassifying it as a cuboid to simplify the calculation, which allows quick practicality when no calculator is available. = L x W x D = 0.9 x 0.4 x 0.4 = 0.144m3
As detailed the estimate is not greatly accurate but allows you to gauge a region your calculations should be in.
Question 3
1 – Select appropriate methods and calculate the area of the floor and length of skirting.
Area
Area1 – (square) = L x W
= (3.2+0.7) x 4.2 = 3.9 x 4.2 = 16.38m2
Area 2 – (trapezium) = 0.5 (A+B) x H = 0.5 (2.2+1.2) x 0.5 = 0.5 x 3.4 x 0.5 = 0.85m2
Area 3 – (semi-circle) = πr2 x = π12 x