Respuesta :
Answer:
Explanation:
Energy consumed per day = (2.2 x 1.5 + 4 x .075 x 5 + 2.5 x 1 + 2.3 )kWh
= 3.3 + 1.5 + 2.5 + 2.3
= 9.6 kWh
Energy consumed in a month = 9.6 x 30
= 288 kWh
cost = 288 x .105
= $ 30.24
b ) 288 kWh = 288 x 60 kW
= (288 x 60)/ 4.2 kcal
= 4114.28 kcal
coal required per month = ( 4114.28 / 7500) x (1 / .37 )
= 1.48 kg
coal required per month= 1.48 x 12
=17.76 kg
=
Answer:
a) [tex]TC=\$ 26.69625[/tex]
b) [tex]m=945.998\ kg[/tex]
Explanation:
Given:
power usage by the heater, [tex]P_h=2.2\ kW[/tex]
duration of operating the heater, [tex]t_h=1.5\ h.day^{-1}[/tex]
power usage by the light bulb, [tex]P_l=0.075\ kW[/tex]
duration of operating the light bulb, [tex]t_l=5\ h.day^{-1}[/tex]
power usage by the electric stove, [tex]P_s=2.5\ kW[/tex]
duration of operating the stove, [tex]t_s=1\ h.day^{-1}[/tex]
miscellaneous power usage each day, [tex]\dot P_m=2.3\ kWh.day^{-1}[/tex]
a)
Given rate of electricity, [tex]C=\$ $0.105\ kWh^{-1}[/tex]
Now the total power consumption each day:
[tex]P_D=\dot P_m+P_s.t_s+P_l.t_l+P_h.t_h[/tex]
[tex]P_D=2.3+2.5\times 1+0.075\times 5+2.2\times 1.5[/tex]
[tex]P_D=8.475\ kWh.day^{-1}[/tex]
Total power consumption in one month:
[tex]P_M=P_D\times 30[/tex]
[tex]P_M=8.475\times 30[/tex]
[tex]P_M=254.25\ kWh[/tex]
Total monthly cost, [tex]TC=P_M\times C[/tex]
[tex]TC=254.25\times 0.105[/tex]
[tex]TC=\$ 26.69625[/tex]
b)
calorific value of coal, [tex]CV=7500\time 10^3\ cal.kg^{-1}=3.138\times 10^7\ J.kg^{-1}[/tex]
efficiency of the powerplant, [tex]\eta=0.37[/tex]
The energy output of the powerplant:
[tex]E_o=0.37\times CV[/tex]
[tex]E_o=0.37\times 31380000[/tex]
[tex]E_o=1.16106\times 10^7\ J[/tex]
The yearly power consumption at this rate will be:
[tex]P_Y=P_M\times 12[/tex]
[tex]P_Y=254.25\times 12[/tex]
[tex]P_Y=3051\ kWh[/tex]
[tex]P_Y=3051\times 10^3\times3600\ J[/tex]
Now the mass of coal required:
[tex]m=\frac{P_Y}{E_o}[/tex]
[tex]m=\frac{3051000\times 3600}{1.16106\times 10^7}[/tex]
[tex]m=945.998\ kg[/tex]
