HW6 Text Problem NE2.4, using method from class for state transition matrix estimated with the Taylor series

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5 EGR 326 HW6 Text Problem NE2.4, using method from class for state transition matrix estimated with the Taylor series >> A >> x0 2 1 With 3 terms from the Taylor Series: >> MatrxExpCalc e At with three terms = x(t) with three terms = With 4 terms from the Taylor Series: e At with four terms = x(t) with four terms =

6 EGR 326 HW 6 Thermal System Solution March 2017 Model of Thermal Dynamics of a Small Cabin The model is developed from the dynamic thermal equation Q stored = Q flowin +Q generated Q flowout + ( dw dt ) " dt C in % h $ ' = Q solar +Q heater Q convection:conduction # dt & " C p ρv dt % in $ ' = (yw / m 2 )(windowarea)+vi (T T ) in out # dt & define the state variable of this first order system: x = T in T in = 1 " T in ( + Q solar +Q heater + T + % out C $ * - h # R th ) R ' where the [] indicate the input, u, to the system Th,& y = [1]x +[0]u R th Note that in this model, the term yw/m 2 is to indicate that you select the numerical value for y in the units of Watts/m 2 and multiply this value by the area of your windows (in m 2 ). Also, the term (dw/dt) is not needed in this cabin model, unless you are in fact doing work on your house, such as compressing it, or sliding it around. There are many options for how to represent this system, and also for the assumptions you make. You are unlikely to have made all the same assumptions as I have made in these solutions. Matlab Script % Cabin_script.m --> with Thermal_HW6.mdl % J Cardell, March 2017 % Script for Simulink model of the small cabin thermal dynamics % System Parameters Cp = 1005; % Air specific heat J/kg K rho = 1.2; % Air density at 25 C; kg/m^3 vol = 256; % Volume of cabin, m^3 Qsolar = 600*6.0*60; % Solar input: 600W/m^2 % assuming window area = 6.0m^2; *60sec/min Qheater = 1500*60; Rth = 0.1/60; % 1500W heater; * 60sec/min % Thermal resistance of cabin = 0.1 K/W %(flat resp if R=10); /60sec/min Tout = 273; % Outside temperature is 0C = 273K Tout_Rth = Tout/Rth; % For 'Q input' due to convection along outside walls % Run the simulink model % Note that in the model, the initial condition for Tin is set to 20 C sim('thermal_hw6'); % Define data vectors written to Matlab workspace by Simulink 'To % Workspace' block.

7 EGR 326 HW 6 Thermal System Solution March 2017 t = Qin.time; u = Qin.signals.values; y1 = Tinside.signals.values(:,1); % plot the data figure hold on subplot(2, 1, 1) plot(t, u, 'r-', 'LineWidth', 2) title('sum of All Qin') ylabel('heat Transfer (J/min)') grid subplot(2, 1, 2) plot(t, y1, 'b-', 'LineWidth', 2) title('response of Inside Temperature to Q Dynamics') ylabel('inside Temperature (K)') xlabel('t (min)') grid hold off % Run the second model - with the (Tin - Tout)/Rth term explicitly % modeled in this Simulink.mdl file sim('thermal_hw6_v2'); % Define data vectors written to Matlab workspace by Simulink 'To % Workspace' block. t2 = Qin2.time; u2 = Qin2.signals.values; y2 = Tinside2.signals.values(:,1); % plot the data figure (2) hold on subplot(2, 1, 1) plot(t2, u2, 'r-', 'LineWidth', 2) title('sum of All Qin') ylabel('heat Transfer (J/min)') grid subplot(2, 1, 2) plot(t2, y2, 'b-', 'LineWidth', 2) title('response of Inside Temperature to Q Dynamics, Version 2') ylabel('inside Temperature (K)') xlabel('t (min)') grid hold off

8 EGR 326 HW 6 Thermal System Solution March 2017 Version 1 of Cabin Model Model with T in as the dynamic variable, separated from the (T out /R Th ) term in the Simulink diagram. Note that T in is fedback, and divided by R Th in the dynamic loop, and (T out /R Th ) is a separate input source. Version 2 of Cabin Model Model with T in as the dynamic variable as above, yet the (T in T out )/R Th term in this diagram is modeled explicitly as that mathematical term. Note that these models are identical in behavior and graphs produced.

9 EGR 326 HW 6 Thermal System Solution March 2017 Results of running the model: Behavior of the inside temperature 5 x 105 Sum of All Qin Heat Transfer (J/min) Inside Temperature (K) Response of Inside Temperature to Q Dynamics t (min) Thermal System Model Discussion: Response of Inside Temperature Our system is a first order system, with a single energy storage element the interior of the cabin, and a single dynamic, or state, variable the temperature of the interior of the cabin. We could develop a more complex, higher order system, if we included other thermal storage elements such as the walls, the roof, the floor (which could be a huge, concrete, thermal mass) The input plot shows the aggregate input from the solar flux, Qsolar, electric heater, Qheater, and the conduction and convection heat transfer due to the temperature difference between inside and outside temperatures and the thermal resistance of the cabin. For the results plotted above, the sun is shining into the cabin for 10 minutes and the heater is on for 5 minutes. The convectionconduction effect is active for all time. Note that many of you had a more expected use of the inputs, in that while the sun was shining, the heater was not on, and visa-versa. This is probably a better representation of the inputs than I have above! Without Q solar or Q heater, the indoor temperature falls slowly, as seen from 0 to 10 minutes, and after 20 minutes. With the input from the sun and the heater, the indoor temperature rises from around 293 to 301K, or 20 to 28 C.

10 EGR 326 HW 6 Thermal System Solution March 2017 The response time of the indoor temperature is quantified in the time constant of this system, with τ = R Th C h (lumped thermal resistance times lumped thermal capacitance). This is the same expression as for the time constant in an RC circuit! For this system, τ = (0.1 K/W)(1005 J/kg K)(1.2 kg/m 3 )(256 m 3 ) 30,874s 8 hours, 35 minutes. This is the time it would take for the system to reach 63% of its maximum value, which clearly would be way to hot for anyone to survive. Discussion: Dynamics in Thermal Systems The oscillations in electrical and mechanical systems can be interpreted as the exchange of energy between different types of dynamic elements. For electrical systems, energy is exchanged between electrical and magnetic fields in capacitors and inductors. In mechanical systems kinetic and potential energy is exchanged between masses and springs. Thermal systems have only one type of dynamic element, a thermal capacitance that stores energy, and one type of energy (thermal or heat energy). Without two types of dynamic elements to be exchanging energy there can be no oscillations heat always flows from higher to lower temperatures with the dynamic thermal elements that we encounter in this universe. From this perspective, we can see that there is no source for oscillations within thermal systems, only exponential growth and decay, along with dissipation of thermal energy. Also, from thermodynamics we know that the natural response of heat transfer is from high temperature to low temperature, indicating that there will not be overshoot or oscillation in a thermal dynamic system.