91Ó°ÊÓ

Using the formula for heat loss per unit length by convection, we have: \[q'=h \cdot \pi \cdot D \cdot \Delta T\] Where, \(q'\) is the heat loss per unit length, and \(\Delta T = T_{s} - T_{r}\). But first, we need to find the temperature of the steam, \(T_{s}\), at the given pressure (20 bars). For this, we can use the steam table to look up the saturation temperature: 1 bar = 100000 Pa, therefore 20 bars = 2000000 Pa. Looking up in the steam table, we find that the saturation temperature at 2000000 Pa is around \(212 ^\circ \mathrm{C}\). Now we can proceed with the calculation. \[\Delta T = 212 - 25 = 187 \mathrm{~K}\] \[q' = 20 \cdot \pi \cdot 0.20 \cdot 187 = 23568 \mathrm{~W/m}\] The heat loss per unit length without insulation is approximately \(23568 \mathrm{~W/m}\).

To calculate the heat loss per unit length with insulation, we need to find the overall heat transfer coefficient (including convection, conduction, and radiation). We can use the following equation: \[\frac{1}{U_{total}} = \frac{1}{h} + \frac{1}{h_{r}} + \frac{t}{k}\] Where, \(U_{total}\) is the overall heat transfer coefficient, \(h_{r}\) is the radiation heat transfer coefficient, and \(k\) is the thermal conductivity of the insulation material. According to the given information, we can assume that the steam-side convection resistance is negligible. The emissivity of the steel and magnesia is \(0.8\). We can use the following equation to calculate the radiation heat transfer coefficient: \[h_{r} = \epsilon \cdot \sigma \cdot \frac{(T_{s}+273)^{4} - (T_{r}+273)^{4}}{(T_{s}-T_{r})}\] Where, \(\sigma = 5.67 \times 10^{-8}\) is the Stefan-Boltzmann constant. Plugging in the values, we get \(h_{r}\approx 19.928 \mathrm{~W/m^{2} \cdot K}\). Now, we need to find the thermal conductivity of the insulation material (magnesia). The given information states that it is \(85\%\) magnesia. Looking up the thermal conductivity of \(85 \% \) magnesia, we find \(k \approx 0.045 \mathrm{~W/m \cdot K}\). Now we can find the overall heat transfer coefficient: \[\frac{1}{U_{total}} = \frac{1}{20} + \frac{1}{19.928} + \frac{0.05}{0.045}\] \[U_{total} \approx 6.07989 \mathrm{~W/m^{2} \cdot K}\] Lastly, we can find the heat loss per unit length with insulation: \[q'_{insulation} = U_{total} \cdot \pi \cdot D \cdot \Delta T \approx 6.07989 \cdot \pi \cdot 0.20 \cdot 187 = 7253 \mathrm{~W/m}\] The heat loss per unit length with insulation is approximately \(7253 \mathrm{~W/m}\).

Let's calculate the energy cost without insulation per year and the energy cost with insulation per year: Energy cost without insulation per year (\(C_{no\_insulation}\)): \[\frac{23568 \mathrm{~W/m}}{10^{9} \mathrm{~J}} \times 7500 \mathrm{~h/yr} \times 3600 \mathrm{~s/h} \times \$4 = \$2523.9744/m\] Energy cost with insulation per year (\(C_{insulation}\)): \[\frac{7253 \mathrm{~W/m}}{10^{9} \mathrm{~J}} \times 7500 \mathrm{~h/yr} \times 3600 \mathrm{~s/h} \times \$4 = \$774.6888/m\] Now, let's find the initial insulation investment cost: \[\$100/m\] Now, we can find the payback years: \[\textrm{Payback years} = \frac{\textrm{Initial investment cost}}{\textrm{Energy cost saved per year}} = \frac{100}{2523.9744 - 774.6888}\] \[\textrm{Payback years} \approx 5.37\] The payback period for the insulation investment is approximately 5.37 years.