# EMEC 303

PROJECT 4

EMEC303

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Abstract

Theproject involves creating a comfortable cabin in terms of temperatureand balancing the heat from within. The project is aimed at solvingthe issue of having a cold night caused by temperature loss throughwalls and the roof of a cabin. The problem is solved mathematicallyusing ODE equations, specifically the fourth order Runge-Kuttaequations. MATLAB is used for the modelling of the solution. The endproduct is a comfortable cabin whose temperature is maintained at20^{0}C.

Thetemperatures in the getaway wilderness are usually quite low at thisseason, thus most revelers tend to rent out cabins. These cabins havefire places that help heat up the house. However, it is usually aproblem to maintain a comfortable temperature due to the fact that alot of heat is lost through the walls and roof. Moreover, ensuringthat the fire is not too much is also a problem. A door linksdownstairs to upstairs that can be closed or opened to controltemperature. It is also warmer upstairs by 5^{0}C due to warm air rising and cold air sinking as a result of densitydifferences. The cabin measures 5m x 6m x 3m with a roof height of3m.

Figure1: Schematic of the cabin

Qinis heat flux of the fireplace, and the maximum it can get to is 2500W/m^{2}.Q1 is the heat transfer through the downstairs walls, Q2 the heattransfer from downstairs to upstairs and Q3 from upstairs through theroof. They are governed by the following equations:

Q1= k1 (T1 – Tout)……………………………………………………………………(1)

Q3= k3 (T2 – Tout)…………………………………………………………………….(2)

Q2= k2 (T1 – T2 + 5)………………………………………………………………….(3)

Wherek1, k3 and k2 are the coefficients of thermal resistance given by 0.2W/ (m^{2}K),0.5 W/ (m^{2}K)and k2 varies from 1 to 10 W/ (m^{2}K)respectively. The initial temperatures for downstairs and upstairsare 5^{0}C and 7^{0}C respectively. The energy required to change the cabin’stemperature depends on the cabin’s size, heat capacity of air (Cair= 1005 J/ (KgK)) and density of air (ρ _{air}= 1.2kg/m^{3}).The outside temperature is modelled using the equation below:

Tout(t) = -10sin (2πt/86400)…………………………………………………………..(4)

NumericalMethods

Tosolve this thermodynamic problem and maintain a constant temperatureof 20^{0}C without repeated adjusting of the door, fourth order Runge-Kuttaequations were used in addition to MATLAB modelling. These twotechniques were preferred because of the equations provided and theirease of application in modelling applications of ODE. The initialconditions are also provided and therefore approximation ispractical. Errors are also minimal for fourth order equations. TheODE equations were solved and the error minimized to get the point ofthe required constant temperature (Lewis, 1961).

Resultsand Discussion

Amean temperature of 20^{0}Cis maintained for both downstairs and upstairs by balancing thefireplace heat and the level to which the door linking downstairs toupstairs is kept open. The best way to keep the cabin comfortable wasto ensure that the door was kept in a fixed position and the heatflux at the fireplace controlled such that the temperature downstairsvaried between 15^{0}Cand 36^{0}C.The variation upstairs was between 18^{0}Cand 37^{0}C,giving the same mean of 20^{0}C.Below is the illustration in Fig. 2:

Figure2: Plot of various cabin temperatures

FromFigure 2, it is clear that the temperature upstairs is always higherthan the one downstairs. It is attributed to the rising of warm airand descending of cool air due to density differences.

References

Lewis,G., & Randall, M. (1961). *Thermodynamics*(2nd Ed.). New York: McGraw-Hill.