weight of 1000 pounds is suspended from two cables. The allowable stress in the cables is 1500 psi. Find the minimum diameter fo
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Answer:
a) 358.8K
b) 181.1 kJ/kg.K
c) 0.0068 kJ/kg.K
Explanation:
Given:
P1 = 100kPa
P2= 800kPa
T1 = 22°C = 22+273 = 295K
q_out = 120 kJ/kg
∆S_air = 0.40 kJ/kg.k
T2 =??
a) Using the formula for change in entropy of air, we have:
∆S_air =
Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K
Solving, we have:
[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]
Solving for T2 we have:
Taking the exponential on the equation (both sides), we have:
[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]
b) Work input to compressor:
= 184.1 kJ/kg
c) Entropy genered during this process, we use the expression;
Egen = ∆Eair + ∆Es
Where; Egen = generated entropy
∆Eair = Entropy change of air in compressor
∆Es = Entropy change in surrounding.
We need to first find ∆Es, since it is unknown.
Therefore ∆Es =
∆Es = 0.4068kJ/kg.k
Hence, entropy generated, Egen will be calculated as:
= -0.40 kJ/kg.K + 0.40608kJ/kg.K
= 0.0068kJ/kg.k
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