A) The rate of heat loss per unit length
- No insulation = 3747 w/m
- With insulation = 163 w/m
B) The number of years needed to pay back the initial investment is :
Given data:
From table A-6 saturated water at p = 20 bar : Tsat = Ts = 486 k
From Table A-3 Magnesia 85% ( T ≈ 392 K ) : K = 0.058 W/m.k
Diameter = 0.2 m
pressure = 20 bars
Heat transfer coefficient = 20 W/m² k
A) Determine the heat loss rate per unit length
i) No insulation we will apply the formula below
q' = ε * π * D * σ ( Ts⁴ - T sur⁴ ) + b ( π * D )( Ts - T∞ )
= 0.8 * π * 0.2 * 5.67 * 10⁻⁸ ( 486⁴ - 298⁴ ) + 20 ( π * 0.2 ) ( 486 - 298 )
= 3727 w/m
ii) With Insulation
we will apply the formula below
Q cond = Qconv + Q rad
[tex]\frac{Ts,i - Ts,o}{In(\frac{Do}{D1})/ 2\pi k }[/tex] = h* π * Do ( Tso - T∞ ) + ε * π * D * σ ( Ts⁴ - T sur⁴ ) ----- ( 2 )
Insert values into Equation ( 2 )
Applying trial and error method
T,s,o = 305 k
Therefore
q' = [tex]\frac{(486 - 305 ) / In(0.3 / 0.2 )}{2\pi (0.058)}[/tex]
= 162.65 w/m ≈ 163 w/m
B) Determine the payback period of initial investment
First step : calculate the yearly energy savings
yearly energy savings = Energy savings per year * cost / energy
= ( 3727 - 163 ) J/s.m * 3600 s/h * 7500 h/yr * $4 / 10⁹ J
≈ $ 385 / yr.m
Final step : calculate payback period
Payback period = Insulation cost / savings per year
= $100 /m / $385/yr.m
= 0.26 yr
Hence we can conclude that The rate of heat loss per unit length
- No insulation = 3747 w/m
- With insulation = 163 w/m
The number of years needed to pay back the initial investment is :
0.26 yr
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