An energy system can be approximated to simply show the interactions with its environment including cold air in and warm air out
1 answer:
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
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The original length of the specimen
Explanation:
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Final diameter = 30.04 mm
Change in diameter Δd = 0.04 mm
Final length = 105.20 mm
Elastic modulus E = 65.5 G pa = 65.5 × M pa
Shear modulus G = 25.4 G pa = 25.4 × M pa
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We know that the possion's ratio is given by
0.00476
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