Side milling cutter is an example of ______ milling cutter.
1 answer:
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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
Answer with Explanation:
Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is
Solving for 'U' we get
Since our assumption is correct thus we conclude that degree of consolidation is 50.46%
The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2
i) = 71% consolidation
ii) = 45% consolidation
iii) = 30% consolidation
Part b)
The degree of consolidation is given by
Thus a settlement of 50.46 centimeters has occurred
For time factor 0.7, U is given by
thus consolidation of 85.59 % has occured if time factor is 0.7
The degree of consolidation is given by
Answer:
Your question has some missing information below is the missing information
Given that ( specific heat of fluid A = 1 kJ/kg K and specific heat of fluid B = 4 kJ/kg k )
answer : 300 kW , 95°c
Explanation:
Given data:
Fluid A ;
Temperature of Fluid ( Th1 ) = 420° C
mass flow rate (mh) = 1 kg/s
Fluid B :
Temperature ( Tc1) = 20° C
mass flow rate ( mc ) = 1 kg/s
effectiveness of heat exchanger = 75% = 0.75
<u>Determine the heat transfer rate and exit temperature of fluid</u> <u>B</u>
Cph = 1000 J/kgk
Cpc = 4000 J/Kgk
Given that the exit temperatures of both fluids are not given we will apply the NTU will be used to determine the heat transfer rate and exit temperature of fluid B
exit temp of fluid B = 95°C
heat transfer = 300 kW
attached below is a the detailed solution
