contestada

To maintain a dwelling steadily at 68°F on a day when the outside temperature is 32°F, heating must be provided at an average rate of 740 Btu/min. Compare the electrical power required, in kW, to deliver the heating using (a) electrical-resistance heating, (b) a heat pump whose coefficient of performance is 3.0, (c) a reversible heat pump operating between hot and cold reservoirs at 68°F and 32°F,

Respuesta :

akindeleot

Answer:

a) 13.01 kw

b) 4.336 kw

c) 0.8875 kw

Explanation:

Given:

Heating rate [tex]Q_H[/tex] = 740 Btu/min

[tex]T_H[/tex] = 60°F = 293.15 K

[tex]T_L[/tex] = 32° = 273.15 K

Performance-Coeff. = 3.0

a)

Electrical-Resistance Heating

[tex]W_{elect}[/tex] =  [tex]Q_H[/tex]

[tex]W_{elect}[/tex] = 740 x (60 min / 1hr) x (1 kw / 341 Btu/min)

[tex]W_{elect}[/tex] = 13.01 kw

b)

a Heat Pump

[tex]P_{HP}[/tex] = [tex]\frac{Q_{H}}{Performance-Coeff.}[/tex]

[tex]P_{HP}[/tex] = 13.01 kw / 3.0

[tex]P_{HP}[/tex] = 4.336 kw

c)

Performance coefficient of reversible Heat Pump

Since; [tex]Q_H[/tex] = 13.01 kw

and [tex](Performance-Coeff)_{R.HP} = \frac{T_H}{T_H - T_L}[/tex] ==> [tex]\frac{293.15}{293.15 - 273.15}[/tex] ==> 14.66

Performance-Coeff = [tex]Q_H[/tex] / [tex]W_{elect}[/tex]

∴ [tex](W_{elec})_{R.Heat Pump} = \frac{Q_H}{Perf. Coeff.}[/tex]

[tex](W_{elec})_{R.Heat Pump} = \frac{13.01}{14.66}[/tex]

[tex](W_{elec})_{R.Heat Pump}[/tex] = 0.8875 kw