Answer:
a. 8.228 kW
b. 28.8 kW
Explanation:
a) Carnot COP = T_high/(T_high - T_low) = (15+273)/((15+273)-(-20+273)) = 8.228
Heating provided / Work input = COP
Heating provided = 8.228*1 = 8.228 kW
b)
Carnot COP = T_high/(T_high - T_low) = (15+273)/((15+273)-(5+273)) = 28.8
Heating provided / Work input = COP
Heating provided = 28.8*1 = 28.8 kW
Answer:
a
The heat provided is [tex]E_t = 0.1480 kW[/tex]
b
The heat provided is [tex]E_T = 0.5656 kW[/tex]
Explanation:
From the question we are told that
The temperature of interior is [tex]T_h = 158 ^oC = 158 + 273 = 431\ K[/tex]
The coefficient of performance of the heater is
[tex]COP= \frac{T_h}{T_h +T_c}[/tex]
Where [tex]T_c[/tex] is the temperature of the cold reservoir with a value [tex]T_c = 2208 ^oC = 2208+273 =2481 K[/tex]
So
[tex]COP= \frac{431}{431 + 2481 }[/tex]
[tex]COP= 0.1480[/tex]
The coefficient of performance is also represented mathematically as
[tex]COP = \frac{Heat \ provide \ to\ the \ research }{Workdone}[/tex]
Let assume the workdone is = 1 N m
[tex]Heat \ provide \ to\ the \ research (E_t) = 1 * 0.1480[/tex]
[tex]E_t = 0.1480 kW[/tex]
The coefficient of performance of the heater is
[tex]COP= \frac{T_h}{T_h +T_c}[/tex]
Where [tex]T_c[/tex] is the temperature of the cold reservoir with a value [tex]T_c = 58 ^oC = 58+273 =331 K[/tex]
So
[tex]COP= \frac{431}{431 + 331 }[/tex]
[tex]COP= 0.5656[/tex]
The coefficient of performance is also represented mathematically as
[tex]COP = \frac{Heat \ provide \ to\ the \ research }{Workdone}[/tex]
Let assume the workdone is = 1 N m
[tex]Heat \ provide \ to\ the \ research (E_T) = 1 * 0.5656[/tex]
[tex]E_T = 0.5656 kW[/tex]
