Answer: The exit temperature of the gas in deg C is .
Explanation:
The given data is as follows.
= 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)
= 100 kPa,
We know that for an ideal gas the mass flow rate will be calculated as follows.
or, m =
=
= 10 kg/s
Now, according to the steady flow energy equation:
= 5 K
= 5 K + 300 K
= 305 K
= (305 K - 273 K)
=
Therefore, we can conclude that the exit temperature of the gas in deg C is .
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Explanation:
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Explanation:
Answer: ε₁+ε₂+ε₃ = 0
Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-
l₀l₀l₀=l₁l₂l₃
taking natural log on both sides
Considering the logarithmic Laws of division and multiplication :
ln(AB) = ln(A)+ln(B)
ln(A/B) = ln(A)-ln(B)
Use the image attached to see the definition of true strain defined as
ln(l1/1o)= ε₁
which then proves that ε₁+ε₂+ε₃ = 0
Answer:
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