Answer:
Q=0.95 W/m
Explanation:
Given that
Outer diameter = 0.3 m
Thermal conductivity of material
So the mean conductivity
So heat conduction through cylinder
Q=0.95 W/m
Answer:
(a) the velocity ratio of the machine (V.R) = 1
(b) The mechanical advantage of the machine (M.A) = 0.833
(c) The efficiency of the machine (E) = 83.3 %
Explanation:
Given;
load lifted by the pulley, L = 400 N
effort applied in lifting the, E = 480 N
distance moved by the effort, d = 5 m
(a) the velocity ratio of the machine (V.R);
since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.
V.R = distance moved by effort / distance moved by the load
V.R = 5/5 = 1
(b) The mechanical advantage of the machine (M.A);
M.A = L/E
M.A = 400 / 480
M.A = 0.833
(c) The efficiency of the machine (E);
Answer:
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Answer:
A). Dry unit weight = 1657.08Kg/m3
B). Porosity = 0.37
C). Void ratio = 0.593
D). 0.712
Explanation:
Total unit weight, Y = 120pcf =1922.2 Kg/m3
Specific gravity of solids, Gs = 2.64
Water content, w = 16%
A). Dry unit weight
Yd = Y/(1+w)
= 1922.2/(1+0.16) = 1657.08Kg/m3
B). Porosity
However void ratio, e = Gs×Yw/Yd, where Yw = 1000Kg/m3
Void ratio = 2.64×1000/1657.08 = 0.593
And porosity = e/(1+e) =0.593/(1+0.593) = 0.37
C). void ratio, e = 0.593
D). Degree of saturation, S = m×Gs/e where m =water content
S = 0.16×2.64/0.593 = 0.712