Answer:
we use
A cylinder in space is of uniform temperature and dissipates 100 Watts. The cylinder diameter is 3" and its height is 12". Assum
e the external surface of the cylinder is covered with Chemglaze Z306 black paint. What temperature does it reach
1 answer:
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Answer:
(a). Entropy change of refrigerant is = 0.7077
(b). Entropy change of cooled space
(c). Total entropy change is dS = 0.0232
Explanation:
Given data
Saturation pressure = 140 K pa
Saturation temperature from property table
= - 18.77 °c = - 18.77 + 273 = 254.23 K
(a). Entropy change of refrigerant is given by
Since heat absorbed by refrigerant Q = 180 KJ
dS = 0.7077
(b). Entropy change of cooled space
= - 10 °c = 263 K
(c). Total entropy change is given by
dS = 0.7077 - 0.6844
dS = 0.0232
This is the value of total entropy change.
Answer:
b). The same for all pipes independent of the diameter
Explanation:
We know,
From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.
We also know,
Factors on which thermal resistance of insulation depends are :
1. Thickness of the insulation
2. Thermal conductivity of the insulating material.
Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.
Answer:
i) S–N plot is attached
ii) fatigue strength = 100 MPa
iii) fatigue life = 5.62 x 10^(5) cycles
Explanation:
i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)
ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.
So let's find the log of this and trace it on the graph attached.
Log(4 × 10^(6)) = 6.6
From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa
iii) The question says we should find the fatigue life for 120 MPa.
Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.
Thus,the fatigue life will be the inverse log of 5.75.
Thus, fatigue life = 10^(5.75)
Fatigue life = 5.62 x 10^(5)
