An ideal gas undergoes two processes: one frictionless and the other not. In both the cases, the gas is initially at 200 ℉ and 1
2 answers:
You might be interested in
Answer:
a) the amount of energy produced in kJ/K is 0.73145 kJ/K
b) the amount of energy produced in kJ/K is 0.68975 kJ/K
The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.
Explanation:
Draw the T-s diagram.
a)
= 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280
R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar
Δs =
Substitute all parameters in the equation
Δs =
Δs = 5 kg × 0.14629 kJ/kg.K
= 0.73145 kJ/K
b)
Δs =
Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K
T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K
R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar
Δs =
= 5 kg (0.13795 kJ/kg.K)
= 0.68975 kJ/K
The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.
Answer:
The required size of column is length = 15 ft and diameter = 4.04 inches
Explanation:
Given;
Length of the column, L = 15 ft
Applied load, P = 10 kips = 10 × 10³ Psi
End condition as fixed at the base and free at the top
thus,
Effective length of the column, = 2L = 30 ft = 360 inches
now, for aluminium
Elastic modulus, E = 1.0 × 10⁷ Psi
Now, from the Euler's critical load, we have
where, I is the moment of inertia
on substituting the respective values, we get
or
I = 13.13 in⁴
also for circular cross-section
I =
thus,
13.13 =
or
d = 4.04 inches
The required size of column is length = 15 ft and diameter = 4.04 inches
Answer:
8.24μm
Explanation:
The theory of brittle fracture was used to solve this problem.
And if you follow through with the attachment made a the subject of the formula
Such that,
a = 2x(69x10⁹)x0.3/pi(40x10⁶)²
= 4.14x10¹⁰/5.024x10¹⁵
= 8.24x10^-06
= 8.24μm
This is the the maximum length of the surface flaw
