I have added the answer as a pic due to difficulties pasting the text here.
Answer:
=4/5 because I'm not going to go back in a year meaning that they are you are
Answer:
The correct answer is "1341.288 W/m".
Explanation:
Given that:
T₁ = 300 K
T₂ = 500 K
Diameter,
d = 0.2 m
Length,
l = 1 m
As we know,
The shape factor will be:
⇒ ![SF=\frac{2 \pi l}{ln[\frac{1.08 b }{d} ]}](https://tex.z-dn.net/?f=SF%3D%5Cfrac%7B2%20%5Cpi%20l%7D%7Bln%5B%5Cfrac%7B1.08%20b%20%7D%7Bd%7D%20%5D%7D)
By putting the value, we get
⇒ ![=\frac{2 \pi l}{ln[\frac{1.08\times 1}{0.2} ]}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%20%5Cpi%20l%7D%7Bln%5B%5Cfrac%7B1.08%5Ctimes%201%7D%7B0.2%7D%20%5D%7D)
⇒ 
hence,
The heat loss will be:
⇒ 



Answer:
Recognize that there is a moral dilemma.
Determine the actor. ...
Gather the relevant facts. ...
Test for right versus wrong issues. ...
Test for right versus right paradigms. ...
Apply the resolution principles. ...
Investigate the trilemma options. ...
Make the decision.
Answer:
The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C
Explanation:
The properties of water at 100°C and 1 atm are:
pL = 957.9 kg/m³
pV = 0.596 kg/m³
ΔHL = 2257 kJ/kg
CpL = 4.217 kJ/kg K
uL = 279x10⁻⁶Ns/m²
KL = 0.68 W/m K
σ = 58.9x10³N/m
When the water boils on the surface its heat flux is:

For copper-water, the properties are:
Cfg = 0.0128
The heat flux is:
qn = 0.9 * 18703.42 = 16833.078 W/m²

The tube surface temperature immediately after installation is:
Tinst = 100 + 20.4 = 120.4°C
For rough surfaces, Cfg = 0.0068. Using the same equation:
ΔT = 10.8°C
The tube surface temperature after prolonged service is:
Tprolo = 100 + 10.8 = 110.8°C