Question 7.8.4: Consider the experiment with a cooling cup of water describe......
Consider the experiment with a cooling cup of water described in Example 1.5.1 in Chapter 1. Water of volume 250 ml in a glass measuring cup was allowed to cool after being heated to 204°F. The surrounding air temperature was 70°F. The measured water temperature at various times is given in the table in Example 1.5.1. From that data we derived the following model of the water temperature as a function of time.
T=129e^{-0.0007t}+70 (1)
where T is in °F and time t is in seconds. Estimate the thermal resistance of this system.
We model the cup and water as the object shown in Figure 7.8.6. We assume that convection has mixed the water well so that the water has the same temperature throughout. Let R be the aggregated thermal resistance due to the combined effects of (1) conduction through the sides and bottom of the cup, (2) convection from the water surface and from the sides of the cup into the air, and (3) radiation from the water to the surroundings. Assume that the air temperature {T}_{o} is constant and select it as the reference temperature. The heat energy in the water is
E=\rho V c_{p}(T-T_{o})From conservation of heat energy
{\frac{d E}{d t}}=-{\frac{1}{R}}(T-T_{o})or, since \rho,V,c_{p},\mathrm{and}\,T_{o} are constant,
\rho V c_{p}{\frac{d T}{d t}}=-{\frac{1}{R}}(T-T_{o})The water’s thermal capacitance is C=\rho V c_{p} and the model can be expressed as
R C{\frac{d T}{d t}}+T=T_{o} (2)
The model’s complete response is
T(t)=T(0)e^{-t/R C}+\left(1-e^{-t/R C}\right)T_{o}=[T(0)-T_{o}]e^{-t/R C}+T_{o}Comparing this with equation (1), we see that
R C={\frac{1}{0.0007}}=1429\,\mathrm{sec}or
R={\frac{1429}{C}}{\frac{\mathrm{°F }}{\mathrm{ft-lb }}}where C=\rho V c_{p}. Because the temperature data was given in °F, we must convert the volume of 250 ml to ft³. Note that V = 250 ml = 2.5 × 10^{-4} m³, so that V = 2.5 × 10^{-4}(3.28 ft/m)³m³ = 8.82 × 10^{-3} ft³. Using the values at room temperature for water, we have ρ = 1.942 slug/ft³, c_{p} = 25,000 ft-lb/slug-°F, and thus C = 423 ft-lb-sec/°F. Therefore the aggregated thermal resistance is
R={\frac{1429}{C}}=3.37{\frac{\textrm{°F }}{\mathrm{ft-lb }}}The usefulness of this result is that this value of R can be used to predict the temperature dynamics of the water/cup system under somewhat different conditions. For example, we can use it to estimate the temperature of a different amount of water if we also know how much the surface area changes. This is because the thermal resistance is inversely proportional to area, as can be seen from (7.7.2) and (7.7.4). Suppose we double the water volume to 500 ml, so that the new value of the thermal capacitance becomes C = 2(423) = 846 ft-lb-sec/°F. Suppose the surface area of the new volume is 5/3 that of the smaller volume. Then we estimate the new value of R to be R = (3/5)(3.37) = 2.02°F/ft-lb. Thus the model of the new water mass is given by equation (2) with RC = (2.02)(846) = 1709 sec.
R={\frac{L}{k A}} (7.7.2)
R={\frac{1}{h A}} (7.7.4)
