Question

1)Based on the latest restriction of the MCO regulations due to COVID-19, you may need to stay at home to save your life from any diseases. Based on this; you may need to isolate your room using suitable insulation form the appendices of Fundamental of Heat and Mass Transfer textbook that could help you to reject the heat to the outdoor environment to save the energy consumption.

(a) Select the suitable insulation to save the energy, providing the reason of this selection.

(b) Draw the circuit resistance of your design.

(c)By using the properties of Malaysia for the outdoor, assume the temperature of the outdoor that could raise the indoor temperature to 27°C. Determine the heat transfer rate (use the area of your room that you are sitting inside right now, using the wall that exposed to the sun), while the heat transfer coefficient and the emissivity for the outdoor and the indoor are 8 W/m² .K, 4 W/m² .K, 0.8 and 0.6, respectively, however, you may take the thermal conductivity of your wall to be 0.5 W/m.K. Also determine the surface temperature after applying the insulation on the wall.

*Malaysia outdoor temperature would be around 30degree celcius.

*wood has been chosen as the insulation material. k=0.087W/m.K, density = 350kg/m3 , specific heat = 1590 J/kg.K

Answer 1

(a). As we all know that currently we all are in lockdown so we cannot get insulation materials from market. So, i think if i want to insulate my room then wood is the best choice(like people in Malaysia did) but i like to do this in a different way. First i make a wooden wall of my wall size and place it ouside the wall while giving a small(but not too small) gap between wall and wood so that total thermal resistance will increase and heat transfer from outside to inside decreases.

Air Gap Date / / Wood wall where Total Thermal Resistance = l + L + 1 + 1 heA Kwood A hA kwall A hi A he External heat transfer coefficient [convective] kwood & Conductive heat transfer coefficient of w ood ih & Convective heat transfer coefficient of air gap kwall & Conductive heat transfer wefficient of wall hi Internal convective heat transfer coefficient A Cross section Area of wall- | Nood | Wall The Vos Too= 30°C Ti, hi he=&W/m²K Kwall = 0.5 wlink Kwood kwall kwall Kwood Ei=0.6 kwood = 0.087 w/mk hi - 4 wlm²k Ti = 27°C T =5.67 X100 Let, 4 = 20cm = 0. 2m 2 L₂ = 50cm = 0.5m & A= 10m2 Q = Heat transfer through (Conduction & Convection) + Heat transfer through Radiation a = (Too - Ti) + 4T A Toy + 47+1-E:)ATA" - I hi ha het kwood A kwall A hiA Here (lei) is taken because Ei is the emmisivity of wall which is used when caning from left to right &currently we don't have suitace temperature . we use Cl-Ei

Date / / 30-27 +HTTX0-8X10x (273+30)" +49 (1-946)X10[273+27)" 0.2 0.5 8810 0.087x10 0.5X10 4X10 +5.64X10-9 [8.417x10'+4.07x10"} 3 -0.0125 +0.229 + 0.1 +0.025 Q = 3 5 .67X10-9 12-54x 10 0.3665 a = 8.185 + 7/101.8 Q = 71109.985 W TQ = 71.109g85 kW or 71.11 kw l For calculating interface we interface surface all the temperature i.e. temperature of wood & wall same above equation but only up to the a = 4T E e ATM + 4HT (1-E:JATH] + To-Ts It L hot kwood A =[40X0-8X 10X(303)" + 40 XC1-0.6)# ox(300) "Jx5.67X108 ti 30- Ts 002 8810 0.087 X10 0.2 - = 71101.8 t 30 - Ts 0.0125 &0.229 71110 - 71101.8 - 30-TS 0.0125to.229 30-Ts = 8.2 X 0.2415 . 30-TS = 1.9803 TS = 30-109803 Ts= 28.0197°c)
As we can see that after insulation surface temperature of wall comes close to room temperature. Therefore, with the help of insulation we can decrease the heat transfer rate.