Study the diagram. Point C identifies the _____ of the wave.
2 answers:
You might be interested in
Answer:
e. TA>T>Tc
Explanation:
a) In this case, we cannot say for sure QA>QB>QC. This is because the magnitude of the heat flow will depend on the specific heat and the mass of each sample. Due to the equation:
if we did an energy balance of the system, we would get that>
QA+QB+QC=0
For this equation to be true, at least one of the heats must be negative. And one of the heats must be positive.
We don't know either of them, so we cannot determine if this statement is true.
b) We can say for sure that QA<0, because when the two samples get to equilibrum, the temperatrue of A must be smaller than its original temperature. Therefore, it must have lost heat. But we cannot say for sure if QB<0 because sample B could have gained or lost heat during the process, this will depend on the equilibrium temperature, which we don't know. So we cannot say for sure this option is correct.
c) In this case we don't know for sure if the equilibrium temperature will be greater or smaller than TB. This will depend on the mass and specific heat of the samples, just line in part a.
d) is not complete
e) We know for sure that A must have lost heat, so its equilibrium temperature must be smaller than it's original temperature. We know that C must have gained heat, therefore it's equilibrium temperature must be greater than it's original temperature, so TA>T>Tc must be true.
Answer:
W(r,out) = 5.81 MW
= 86.1 %
Explanation:
we use here steam table for get value of h1, s1 etc
so use for 6MPa and 600 degree
Enthalphy of steam h1 = 3658.8 kJ/kg
Entropy of steam s 1 is = 7.1693 kJ /kg.K
and
for 50 kPa and 100 degree
Enthalphy of steam h2 = 2682.4 kJ/kg
Entropy of steam s2 is = 7.6953 kJ /kg.K
so we use here energy balance equation that is
..............1
put here value and we get m
m =
solve it we get
m = 5.156 kg/s
so by energy balance equation
m1 = m
2 + W(r,out)
W(r,out) = m(1 -
2)
W(r,out) = h1 - h2 + ΔKE + ΔPE - To(s1-s2)
W(r,out) =
W(r,out) = W(a,out) - m.To.(s1-s2) ........................2
put here value
W(r,out) = 5000 - ( 5.156 × (25 + 273) ×( 7.1693 - 7.6953)
W(r,out) = 5908.19 = 5.81 MW
and
second law deficiency is
=
..............................3
put here value
= \frac{5}{5.81}
= 86.1 %