Answer:
Hello your question is incomplete attached below is the complete question
Answer : Factor of safety for point A :
i) using MSS
(Fos)MSS = 3.22
ii) using DE
(Fos)DE = 3.27
Factor of safety for point B
i) using MSS
(Fos)MSS = 3.04
ii) using DE
(Fos)DE = 3.604
Explanation:
Factor of safety for point A :
i) using MSS
(Fos)MSS = 3.22
ii) using DE
(Fos)DE = 3.27
Factor of safety for point B
i) using MSS
(Fos)MSS = 3.04
ii) using DE
(Fos)DE = 3.604
Attached below is the detailed solution
Answer:
a) 84.034°C
b) 92.56°C
c) ≈ 88 watts
Explanation:
Thickness of aluminum alloy fin = 12 mm
width = 10 mm
length = 50 mm
Ambient air temperature = 22°C
Temperature of aluminum alloy is maintained at 120°C
<u>a) Determine temperature at end of fin</u>
m = √ hp/Ka
= √( 140*2 ) / ( 12 * 10^-3 * 55 )
= √ 280 / 0.66 = 20.60
Attached below is the remaining answers
Answer:
Heat gain of 142 kJ
Explanation:
We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.
Δ ⇒
Δ
Therefore, the gas gained heat by an amount of 142 kJ.
