Generally, frictional losses are more predominant for the machines being not 100% efficient. This friction leads to the loss of energy in the form of heat, into the surroundings. Some of the supplied energy may be utilised to change the entropy (measure of randomness of the particles) of the system.
Answer: The energy system related to your question is missing attached below is the energy system.
answer:
a) Work done = Net heat transfer
Q1 - Q2 + Q + W = 0
b) rate of work input ( W ) = 6.88 kW
Explanation:
Assuming CPair = 1.005 KJ/Kg/K
<u>Write the First law balance around the system and rate of work input to the system</u>
First law balance ( thermodynamics ) :
Work done = Net heat transfer
Q1 - Q2 + Q + W = 0 ---- ( 1 )
rate of work input into the system
W = Q2 - Q1 - Q -------- ( 2 )
where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw
Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw
Q = 18 Kw
Insert values into equation 2 above
W = 6.88 Kw
Answer:
the restoring force is = 3/4NKT
Explanation:
check the attached files for answer.
Answer:
1. 
2. 
Explanation:
1.
Given:
- height of the window pane,

- width of the window pane,

- thickness of the pane,

- thermal conductivity of the glass pane,

- temperature of the inner surface,

- temperature of the outer surface,

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

here:
A = area through which the heat transfer occurs = 
dT = temperature difference across the thickness of the surface = 
dx = t = thickness normal to the surface = 


2.
- air spacing between two glass panes,

- area of each glass pane,

- thermal conductivity of air,

- temperature difference between the surfaces,

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>


