Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure
of 140 kPa. The refrigerant absorbs 180 kJ of heat from the cooled space, which is maintained at −10°C, and leaves as saturated vapor at the same pressure. Determine (a) the entropy change of the refrigerant, (b) the entropy change of the cooled space, and (c) the total entropy change for this process.
1 answer:
Answer:
A) ΔS_refrigerant = 0.70754 Kj/K
B) ΔS_space = -0.68441 Kj/K
C) ΔS_total = 0.02313 Kj/K
Explanation:
A) From he table attached, at Pressure of 140 KPa, and by interpolation, we get, Temperature of T = -18.77°C
Converting to degree kelvin yields;
T = -18.77 + 273 = 255.23 K
Formula for entropy change of refrigerant is given as;
ΔS_refrigerant = Q_in/T_refrigerant
We are given Q = 180 KJ
Thus, ΔS_refrigerant = 180/255.23 = 0.70754 Kj/K
B) Formula for entropy change of cooled space is given as;
ΔS_space = Q_out/T_s pace
T_space = -10°C = 273 - 10 = 263K
Thus, ΔS_space = -180/263 = -0.68441 Kj/K
C) the total entropy change would be;
ΔS_total = ΔS_refrigerant + ΔS_space
Thus,
ΔS_total = 0.70754 - 0.68441 = 0.02313 Kj/K
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Answer:
critical clearing angle = 70.3°
Explanation:
Generator operating at = 50 Hz
power delivered = 1 pu
power transferable when there is a fault = 0.5 pu
power transferable before there is a fault = 2.0 pu
power transferable after fault clearance = 1.5 pu
using equal area criterion to determine the critical clearing angle
Attached is the power angle curve diagram and the remaining part of the solution.
The power angle curve is given as
= Pmax sinβ
therefore : 2sinβo = Pm
2sinβo = 1
sinβo = 0.5 pu
βo = ⁰
also ; 1.5sinβ1 = 1
sinβ1 = 1/1.5
β1 = = 41.81⁰
∴ βmax = 180 - 41.81 = 138.19⁰
attached is the remaining solution
The critical clearing angle = ≈ 70.3⁰
Answer:
the required diameter is 0.344 m
Explanation:
given data:
flow is laminar
flow of carbon dioxide Q = 0.005 Kg/s
for flow to be laminar, Reynold's number must be less than 2300 for pipe flow and it is given as
arrange above equation for diameter
\frac{\rho Q D}{\mu A }<2300
dynamic density of carbon dioxide = 1.47× Pa sec
density of carbon dioxide is 1.83 kg/m³
D = 0.344 m