An aggregate blend is composed of 55% aggregate A (Sp. Gr. 2.631), 25% aggregate B (Sp. Gr. 2.331) and 20% sand (Sp. Gr. 2.609).
1 answer:
You might be interested in
Answer:
q = 1.73 W
Explanation:
given data
small end = 5 cm
large end = 10 cm
high = 15 cm
small end is held = 600 K
large end at = 300 K
thermal conductivity of asbestos = 0.173 W/mK
solution
first we will get here side of cross section that is express as
...............1
here x is distance from small end and S1 is side of square at small end
and S2 is side of square of large end and L is length
put here value and we get
S = 5 +
S = m
and
now we get here Area of section at distance x is
area A = S² ...............2
area A = m²
and
now we take here small length dx and temperature difference is dt
so as per fourier law
heat conduction is express as
heat conduction q = ...............3
put here value and we get
heat conduction q =
it will be express as
now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K
solve it and we get
q (30) = (0.173) × (600 - 300)
q = 1.73 W
Answer:
The value of R is 10101
Explanation:
As per the given data
D = 1000100100
G = 100101
Redundant bit = 6-bits - 1-bit = 5-bits
No add fice zero to D
D = 100010010000000
Now calculate R as follow
R = D / G
R = 100010010000000 / 100101
R = 10101
Workings are attached with this question
