Answer:
so heat loss = 4312 W
cost of heat loss daily is $8.28 per day
Explanation:
given data
slab length L = 11 m
slab wide W = 8 m
thickness t = 0.20 m
temperature top T1 = 17°C
temperature bottom T2 = 10°C
thermal conductivity k = 1.4 W/m-K
efficiency ηf = 0.90
priced Cg = $0.02 / MJ
to find out
rate of heat loss and daily cost of the heat loss
solution
we calculate here heat loss by heat transfer
so apply here formula that is
q = (thermal conductivity × area × temperature difference) / thickness
put here all these value we get heat loss
q = [tex]\frac{k*L*W*(T2-T1)}{t}[/tex]
q = [tex]\frac{1.4*11*8*(17-10)}{0.2}[/tex]
q = 4312 W
so heat loss = 4312 W
and
cost of the heat loss is express as
cost of heat loss = [tex]\frac{q*Cg}{efficiency}[/tex]
put here all these value
cost of heat loss is = [tex]\frac{4312*0.02*10^6}{0.9}[/tex] × 24 hr/day × 3600 s/hr
cost of heat loss = 8.279
so cost of heat loss daily is $8.28 per day
The heat loss be "4312 W" and cost will be "$8.28 per day".
According to the question,
We know the relation,
→ [tex]q = \frac{Thermal \ conductivity\times Area\times Temperature \ difference}{Thickness}[/tex]
or,
[tex]= \frac{k\times L\times W(T_2-T_1)}{t}[/tex]
By substituting the values,
[tex]= \frac{1.4\times 11\times 8(17-10)}{0.2}[/tex]
[tex]= 4312 \ W[/tex]
now,
The cost of heat loss be:
= [tex]\frac{q\times C_g}{Efficiency}[/tex]
= [tex]\frac{4312\times 0.02\times 10^6}{0.9}\times 24\times 3600[/tex]
= [tex]8.28 \ per \ day[/tex]
Thus the above response is correct.
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