Answer:
(a). max possible efficiency = 55.62%
(b). max power output = = 133.5 MW
Explanation:
From the question we were given the Maximum temperature in the system as
Tmax = 500°C
Minimum temperature in the system Tmin = 70°C
the Heat supplied to the boiler Qb = 240000 KJ/s
we use the temperature conversion factor from °C to K
given T(K) = T (°C) + 273
⇒ Tmax = 500 + 273 = 773 K
⇒ Tmin = 70 + 273 = 343 K
(a). we are to determine the maximum possible thermal efficiency;
(Πth)max = 1 - Tmin/Tmax
(Πth)max = 1 - 343/773 = 0.5562
(Πth)max = 55.62%
(b). to determine the maximum possible power output for the plant we have;
(Πth)max = Wmax/Qb
where Wmax rep the maximum power output
(Πth)max = 0.5562
Qb = 240000
∴ Wmax = 0.5562 × 240000 = 133505.6 Kw
Wmax = 133.5 MW
cheers i hope this helps
The problem of the Graduation date about the cover was that we had a new name on the back of the earth lol but
Answer:
Find the attachments sequence wise for complete solution.
