Answer:C 0.12 V
Explanation:
Given
Concentration of
Concentration of
Standard Potential for Ni and Fe are and
Which option identifies the type of device the engineer will develop in the following scenario?
2 answers:
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Answer:
t= 4.5 mm
Explanation:
Given that
P = 520 KPa ( gauge)
Maximum allowable normal stress ,σ= 150
d= 2.6 m
Wall thickness = t
The normal stress for pressure vessel given as
( hoop stress)
We always take maximum stress for safe design.
Now by putting the values
t= 4.5 mm
So the minimum thickness, t, of the wall is 4.5 mm
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