Answer:
a) 0.39795 kJ/K
b) 79.589.37 kJ
Explanation:
m = Mass of air = 2 kg
Temperature = 200 K
P₁ = Initial pressure = 300 kPa
P₂ = Final pressure = 600 kPa
R = mass-specific gas constant for air = 287.058 J/kgK
a) For isentropic process
∴ Entropy is generated in the process is 0.39795 kJ/K
b)
∴ Amount of lost work is 79.589.37 kJ
Answer:
Yield strength, tensile strength decreases with increasing temperature and modulus of elasticity decreases with increasing in temperature.
Explanation:
The modulus of elasticity of a material is theoretically a function of the shape of curve plotted between the potential energy stored in the material as it is loaded versus the inter atomic distance in the material. The temperature distrots the molecular structure of the metal and hence it has an effect on the modulus of elasticity of a material.
Mathematically we can write,
where,
E(t) is the modulus of elasticity at any temperature 'T'
is the modulus of elasticity at absolute zero.
is the mean melting point of the material
Hence we can see that with increasing temperature modulus of elasticity decreases.
In the case of yield strength and the tensile strength as we know that heating causes softening of a material thus we can physically conclude that in general the strength of the material decreases at elevated temperatures.
Answer:
Explanation:
Hello,
a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:
- and
: extent of the reactions (2).
- ,
,
,
and
: Molar flows at the second stream (5).
On the other hand, we've got the following equations:
- : oxygen mole balance.
- : methane mole balance.
- : water mole balance.
- : formaldehyde mole balance.
- : carbon dioxide mole balance.
Thus, the degrees of freedom are:
It means that we need two additional equations or data to solve the problem.
b. Here, the two missing data are given. For the fractional conversion of methane, we define:
And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:
In such a way, one realizes that the output formaldehyde's molar flow is:
Which is equal to the first reaction extent , therefore, one computes the second one from the fractional conversion of methane as:
Now, one computes the rest of the output flows via:
-
-
-
-
-
The total output molar flow is:
Therefore the output stream composition turns out into:
Best regards.
