Answer:
Explanation:
given data:
innner radius r_i = = 5cm
outer radius r_0 = 7 cm
temperature at outer surface = 80 degree celcius
temperature at inner surface = 100 degree celcius
thermal resistance per unit length is given as
heat loss rate per unit length
q = 373.474 K
1) for pure COPPER
k = 387 W/m degree celcius
q = 373.474 * 387 = 144534.438 W
2) for pure ALUMINIUM
k = 200 W/m degree celcius
q = 373.474 * 200 = 74694.84 W
3)
1) for pure IRON
k = 62W/m degree celcius
q = 373.474 * 62 = 23155.416 W
Answer:
Explanation:
From the question we are told that:
Initial Pressure
Initial Temperature
Final Pressure
Final Temperature
Work Output
Generally Specific Energy from table is
At initial state
With
Specific Volume
At Final state
Generally the equation for The Process is mathematically given by
Assuming Mass to be Equal
Where
Therefore
Answer:
a.)
US Sieve no. % finer (C₅ )
4 100
10 95.61
20 82.98
40 61.50
60 42.08
100 20.19
200 6.3
Pan 0
b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4
c.) Cu = 3.33
d.) Cc = 1
Explanation:
As given ,
US Sieve no. Mass of soil retained (C₂ )
4 0
10 18.5
20 53.2
40 90.5
60 81.8
100 92.2
200 58.5
Pan 26.5
Now,
Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g
⇒ w = 421.2 g
As we know that ,
% Retained = C₃ = C₂×
∴ we get
US Sieve no. % retained (C₃ ) Cummulative % retained (C₄)
4 0 0
10 4.39 4.39
20 12.63 17.02
40 21.48 38.50
60 19.42 57.92
100 21.89 79.81
200 13.89 93.70
Pan 6.30 100
Now,
% finer = C₅ = 100 - C₄
∴ we get
US Sieve no. Cummulative % retained (C₄) % finer (C₅ )
4 0 100
10 4.39 95.61
20 17.02 82.98
40 38.50 61.50
60 57.92 42.08
100 79.81 20.19
200 93.70 6.3
Pan 100 0
The grain-size distribution is :
b.)
From the diagram , we can see that
D10 = 0.12
D30 = 0.22
D60 = 0.12
c.)
Uniformity Coefficient = Cu =
⇒ Cu =
d.)
Coefficient of Graduation = Cc =
⇒ Cc = = 1
The answer & explanation for this question is given in the attachment below.
