analyze the heat balance equation in the context of the race conditions
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Mathematically, it can be expressed as: \[ Q_{in} - Q_{out} = \Delta U \] where \( Q_{in} \) is the heat added to the system, \( Q_{out} \) is the heat lost from the system, and \( \Delta U \) is the change in internal energy. Show more…
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According to the example which is discussed in class, we can establish the governing equation for the change of CPU temperature versus time, with I.C. as t=0, T=TH. ̑c_pV dT/dt = -hA(T - T∞) + J.H. Now we can nondimensionalize this governing equation and its I.C. We can define dimensionless temperature θ = (T - T∞) / (Tss - T∞) , where Tss is the steady state temperature, which has been discussed in class, Tss = T∞ + J.H./hA. We also need to find a reference time scale tc to nondimensionalize the variable t to t* (t*=t/tc). (1) Please substitute all the introduced dimensionless variables into the governing equation, and show the finalized dimensionless governing equation and dimensionless IC. (2) From Step (1), you should have a combination of parameters together with tc in front of dθ/dt*, we can find the actual tc at this step if we make the coefficient be equal to 1. Please show the expression of tc. (3) After Step (2), you should have an ODE as dθ/dt* = -θ + 1, please solve this ODE together with the dimensionless IC. (4) Once you find the final solution, please plot it out using Matlab or Excel, and show the value of θ when t* is equal to 5, and you may discuss the physical meaning of this result.
Adi S.
Jogging in the Heat of the Day. You have probably seen people jogging in extremely hot weather and wondered Why? As we shall see, there are good reasons not to do this! When jogging strenuously, an average runner of mass 68 $\mathrm{kg}$ and surface area 1.85 $\mathrm{m}^{2}$ produces energy at a rate of up to $1300 \mathrm{W}, 80 \%$ of which is converted to heat. The jogger radiates heat, but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around $33^{\circ} \mathrm{C}$ instead of the usual $30^{\circ} \mathrm{C}$ . (We shall neglect conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is 40.0 $^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right)$ ? (Remember that he radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must the jogger's body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is $2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}$. must get rid of per second? (d) How much water must the jogger's body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is $2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}.$
The conservation of heat can be used to develop a heat balance for a long, thin rod. If the rod is not insulated along its length and the system is at a steady state, the equation that represents the system is: d^2T/dx^2 + k(T_a - T) = 0 with boundary conditions: T(0) = 40 and T(10) = 200. The term k is a heat transfer coefficient (m^-2) that parameterizes the rate of dissipation to the surrounding air and T_a is the temperature of the surrounding air (°C). Given k = 0.01m^-2, T_a = 20 and the step size of h = 2. Find the approximate solution of the differential equation for a 10m rod by using i. Shooting method with the first z_0 = 10 and the second initial guess z_0 = 20. ii. Finite difference method.
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