WFT calculations

1) What Wft would need to be applied to give a Dft of 45 um using a paint of 56% VS?
Dft
Wft = ×100
%VS
45
Wft = ×100 → Wft = 80.3 um
56
2) What WFT would need to be applied to give a DFT of 60 um using a paint of 40% VS?
Dft
Wft = ×100
%VS
60
Wft = ×100 → Wft = 150 um
40
3) A paint of 38% vs was used to give a DFT of 45 um what would be the WFT?
Dft
Wft = ×100
%VS
45
W ft = ∗100 → Wft = 118.421 um
38
4) A Dft of 55 um was obtained from a paint of 55% vs, what was the Wft applied?
Dft
Wft = ×100
%VS
55
Wft = ×100 → Wft = 100 um
55
5) What WFT would be applied to leave a DFT of 65 um using a paint of 49% VS?
Dft
Wft = ×100
%VS
65
Wft = ×100 → Wft = 132.6 um
49

DFT calculations
1) What would be the DFT if 20 litres of paint, VS. 45% covered an area of 9m x 12m?
Area × Wft
V=
1000
9 ×12× Wft
20= → Wft=185 um
1000
Wft × VS
Dft =
100
45 ×185
Dft = → Dft = 83.25 um
100
2) 25 litres of paint, vs. 65% was used to cover a circular area of 10m diameter. What would
be the resulting DFT?
Area × Wft
V=
1000
Area = π r 2 → Area=π ×52 = 78.5 m
 78.5× Wft 25 ×1000
25= → Wft= =318 um
1000 78.5
Wft × VS
Dft =
100
65× 318
Dft = → Dft = 206.7 um
100
3) What DFT would be obtained if a paint VS content 42% was applied at a WFT of 84 um?
Wft × VS
Dft =
100
42 ×84
Dft = → Dft = 35.28 um
100
4) With a Wft of 130 um, using a paint containing 83% VS, what would be the resulting DFT?
Wft × VS
Dft =
100
130 × 83
Dft = → Dft = 107.9 um
100
5) A paint, VS 65% was applied at a WFT of 130 um, what would be the resulting DFT?
Wft × VS
Dft =
100
130 × 65
Dft = → Dft = 84.5 um
100
VS calculations

1)A Dft of 53 um was obtained from a Wft of 110 um, what was the VS% of the paint?
Dft Dft
Wft = ×100 → %VS= ×100
%VS Wft
53
VS= ×100 → VS=48 %
110
2) A paint was applied at 120 um Wft. The resulting Dft was 65 um, what was the VS%?
Dft Dft
Wft = ×100 → %VS= ×100
%VS Wft
65
VS= ×100 → VS=54 %
120
3) What would be the VS% of a paint if it was applied with a Wft of 120 um and a Dft of
68 um was obtained?
Dft Dft
Wft = ×100 → %VS= ×100
%VS Wft
68
VS= ×100 → VS=56 %
120
4) What was the VS% of a paint with a Dft of 36 um, when the Wft was 108 um?
Dft Dft
Wft = ×100 → %VS= ×100
%VS Wft
 36
VS= ×100 → VS=33 %
108
5) A Dft of 62 um was measured, from a Wft application of 100 um, what would be the
VS% of the paint used?
Dft Dft
Wft = ×100 → %VS= ×100
%VS Wft
36
VS= ×100 → VS=33 %
108

Volume calculations

1 What volume of paint would be required to cover an area of 300 square meters, to a
specified Dft of 65 um, using a paint of 45% VS?
Dft
Wft = ×100
%VS
65
Wft = ×100 → Wft = 144 um
45
V= Area× Wft
V= 300 ×144 × 10−6 =¿ 0.0432 m3 = 43.2 Liters

2) How much paint would be required to coat a tank, roof and side sheets to a Dft of 100 um?
The tank is 5 meters diameter and 6 meters high. The paint to be used is solvent free.
Area = Side wall+ Roof
2
Area = πDh+
πD D=5 m

4
Area = π × 5 ×6 +
π ×5 2 = 94.2+ 19.6 = 113.8 m2
4 h= 6 m
Paint is solvent free it means that its VS is equal to 100%
In this paint, Dft = Wft
V= Area ×Wft
V= 113.8 ×100 ×10−6=0.01138 m3→ V= 11.38 Liters

3) How much paint would be needed to cover a circular area of 10 meters diameter, using a
paint 65% vs to a DFT of 60 um?
Dft
Wft = ×100
%VS
60
Wft = ×100 → Wft = 92 um
65
Area ¿ π r 2 → Area = π × 52= 78.5 m2
V= Area ×Wft
V= 78.5× 92× 10−6= 0.0072 m3→ V= 7.2 liters
4) A circular area of 7 meters radius is to be coated to a Dft of 45 um. What volume of
paint would be required if the VS content was 48%?
 Dft
Wft = ×100
%VS
45
Wft = ×100 → Wft = 94 um
48
Area ¿ π r 2 → Area = π × 72 = 153.86 m2
V= Area ×Wft
V= 153.86 × 94 ×10−6= 0.01446 m3→ V= 14.46 liters
5) How much paint would be needed, at 55% VS, to coat an area of 250 square meters to a Dft of 60 um?
Dft
Wft = ×100
%VS
60
Wft = ×100 → Wft = 109 um
55
Area ¿ 250 m2
V= Area ×Wft
V= 250 ×109 ×10−6= 0.02725 m3→ V= 27.25 liters

Density and SG exercise

1) What would be the weight of 16.5 litres of paint with a SG of 1.45?

 gr
SG= 1.45 → Density ¿ ρ) = 1.45 3
Cm
Weight
Density ¿ ρ) = → Weight = ρ×Volume
Volume
Weight = 1.45×16.5 ×1000=¿23925 gr = 23.92 Kg

2) What is the density of a paint if 7.5 litres weighs 9.75 kg?

Weight
Density ¿ ρ) =
Volume
9.75 ×1000 gr
Density ¿ ρ) = = 1.3 3
7.5× 1000 Cm

3) What would be the relative density of paint in question two?

( Density of X)∨ρ
SG or RD =
Density of Water
gr
1.3 3
Cm
SG or RD = = 1.3
gr
1
Cm3

4) If the weight of 25 litres of paint is 37.5 Kg, what would be the SG?
Weight
Density ¿ ρ) =
Volume
37.5× 1000 gr
Density ¿ ρ) = = 1.5 3
25× 1000 Cm
( Density of X)∨ρ
SG or RD =
Density of Water
gr
1.5
Cm3
SG or RD = = 1.5
gr
1 3
Cm
5) A 2 pack epoxy should be mixed at one part base to one part activator, the base has a density of
1.4gm/cc and the activator 0.9 gm/cc. What would be the density of the mixed components?
A × Density of Base+ B × Density of Activator
Density of Mix =
A+B
A=1, B=1
gr
Density of Base = 1.4
Cm3
gr
Density of Activator = 0.9 3
Cm
1× 1.4+1 ×0.9 gr
Density of Mix = =1.15
2 Cm3
6) A 2 pack paint is mixed at a ratio of six parts pack A (density 1.3gm/cc) to one part pack B
(density 0.9gm/cc). What would be the density of the combined parts?
 A × Density of Base+ B × Density of Activator
Density of Mix =
A+B
A=6, B=1
gr
Density of Base = 1.3
Cm3
gr
Density of Activator = 0.9 3
Cm
6 ×1.3+1 ×0.9 gr
Density of Mix = =1.24
7 Cm3

7) A mixed 2 pack paint has a density of 1.35gm/cc. The density of the base was 1.5gm/cc and the
activator 0.9gm/cc. The mixing ratio was 3:1. Has the paint been mixed correctly?
A × Density of Base+ B × Density of Activator
Density of Mix =
A+B
A=3, B=1
gr
Density of Base = 1.5
Cm3
gr
Density of Activator = 0.9 3
Cm
3× 1.5+1× 0.9 gr
Density of Mix = =1.35 3
4 Cm
Calculation of density of mix and measurement by Density cup shows same result and it means that mixing is OK.
8) A mixed 2 pack paint has a density of 1.35gm/cc. Mixed at a ratio of 6:1, base density
1.45gm/cc, activator density 0.95gm/cc. Has the paint been mixed correctly?

A × Density of Base+ B × Density of Activator

Density of Mix =
A+B
A=6, B=1
gr
Density of Base = 1.5
Cm3
gr
Density of Activator = 0.9 3
Cm
6 ×1.45+1 ×0.95 gr
Density of Mix = =1.378
7 Cm3
Density of mix by calculation is greater than density measurement by density cup. It means that mixing is not
correct and because it is less than calculation it means that activator is more than enough.