I believe number 4 I could be wrong but I think it’s 4
Answer:
When water is surrounding T_s = 34.17 degree C
When air surrounding T_S = 1434.7 degree C
from above calculation we can conclude that air is less effective than water as heat transfer agent
Explanation:
Given data:
length = 300 mm
Outer diameter = 30 mm
Dissipated energy = 2 kw = 2000 w
Heat transfer coefficient IN WATER = 5000 W/m^2 K
Heat transfer coefficient in air = 50 W/m^2 K
we know that 
From newton law of coding we have

is surface temp.
T - temperature at surrounding
![P = hA(T_s - T_{\infity})[tex]\frac{P}{\pi hDL} = (T_s - T_{\infity})](https://tex.z-dn.net/?f=P%20%3D%20hA%28T_s%20-%20%20T_%7B%5Cinfity%7D%29%3C%2Fp%3E%3Cp%3E%5Btex%5D%5Cfrac%7BP%7D%7B%5Cpi%20hDL%7D%20%3D%20%20%28T_s%20-%20%20T_%7B%5Cinfity%7D%29)
solving for[/tex] T_s [/tex] w have



When air is surrounding we have



from above calculation we can conclude that air is less effective than water as heat transfer agent
Answer:
YES
Explanation:
Entropy is an extensive property of the system entropy change that value of entropy change can be determined for any process between the states whether reversible or not. i have attached the formula to calculate entropy change which is independent of whether the system is reversible or not and can be determined for any process.
Answer:
Hello, I'm good. Thank you for asking
Answer:
(a) Calculate the field, armature, and load currents versus load = 306A
(b) Determine the terminal voltage at no-load and at rated load conditions = 300V
(c) Calculate the voltage regulation of the generator. Use the no-load voltage as the base value = 6.67%
(d) Plot the terminal voltage as a function of the load. Determine the load that corresponds to a 5% voltage drop using the no-load voltage as the base = 294.74V
Explanation:
CHECK THE ATTACHED FILES FOR DETAILED EXPLANATION.