1-Node Transient Models

Abstract

A basic philosophy of this book is to develop a modeling approach that considers “the big picture” first, and then more detail only if needed. Since heat transfer rates are driven by temperature differences, the first step in analyzing heat transfer problems is usually to solve for the temperature field, which depends on spatial position and time. Numerical analyses involve the discretization of space (breaking the physical region into discrete subelements) and time (taking finite steps forward in time). This chapter considers “1-node” models, where the region of space in question is lumped into a single node. Time dependence often results in easily solved ordinary differential equations, or ones that can be easily solved numerically. By treating a defined system as a single object with average properties, that is, as a single node, a first approximation can often be made that is either sufficient to make design decisions, or directs attention to intelligently develop a more detailed approach. Furthermore, key functional relationships and associated insights are obtained with relatively little mathematical complexity.

Appendices

Appendix 1. Detailed Calculation of Individual Coffee Mug Thermal Resistances

In this appendix, the formulas used to calculate the individual resistance values are detailed. Figures 4.1 and 4.9 are repeated here for reference. Input parameter values were listed in Figs. 4.4 and 4.10. The rationale for choosing values is described, much of the detail of which is developed in later chapters.

1.1 Top Channel Resistors

1.1.1 Free Surface Convective Resistance, Liquid Side: R_ct

For this resistance, the coffee/air interface at the free surface is treated like a convective resistance. On the liquid side, the resistance is:

$$ {R}_{ct}=\frac{1}{h_{coffee/ surface}{A}_{surface}}=\frac{1}{h_{coffee/ surface}\left(\pi {D}^2/4\right)} $$

The value chosen for convection coefficient (300 W/m²/K) is representative of a natural convection for liquids.

1.1.2 Free Surface Convective Resistance, Air Side: R_at

There are three resistances in series: convective, radiative, and evaporative. Each of them is modeled as:

$$ {R}_{at}=\frac{1}{h_{effective}{A}_{surface}}=\frac{1}{h_{effective}\left(\pi {D}^2/4\right)} $$

The values chosen (6.8, 6.6, and 10 W/m²/K for convective, radiative, and evaporative) were chosen based on detailed Nusselt number and radiation coefficient analyses (methods developed in later chapters) for a coffee temperature of 70 °C.

1.2 Side Channel Resistors

1.2.1 Inner Side Wall Convective Resistance: R_cs

The heat transfer between the coffee and the side wall is by the convective mode, and radiation is not important for two reasons: the inner surface does NOT fall into the model for a resistance model because the inner wall “sees” itself. Also, because the liquid convection coefficient is over an order of magnitude larger than typical air values, convection dominates radiation. The thermal resistance is therefore:

$$ {R}_{cs}=\frac{1}{h_{coffee/ side}{A}_{InnerSurface}}=\frac{1}{h_{coffee/ side}\left(\pi DL\right)} $$

The value for h_coffee/side (470 W/m²/K) was obtained from a Nusselt number calculation for a coffee temperature of 70 °C.

1.2.2 Side Wall Conductive Resistance: R_ws

This resistance represents a conductive resistance across the side wall. Following the general outline, two surfaces are identified, and average distance between the surfaces and average areas calculated. For the side wall, the surface 1 is defined as the inner submerged surface. Its area is:

$$ {A}_1=\pi D\left(H-L\right)=0.0198\kern0.5em {\mathrm{m}}^2 $$

Surface 2 is defined as the total surface area of the outer wall exposed to air, including the top rim and the inner surface area above the coffee surface. This 2D conduction problem is discussed in some detail in a future chapter. The area is:

$$ {A}_2=\pi \left(D+2w\right){H}_0+\pi DL+\frac{\pi }{4}\left({\left(D+2w\right)}^2-{D}^2\right)=0.0372\;{\mathrm{m}}^2 $$

The outer area is 88 % greater than the inner one. The geometric average area is:

$$ {A}_{12}=\left({A}_1+{A}_2\right)/2=0.0285\;{\mathrm{m}}^2 $$

The average distance between surfaces 1 and 2 (Δx₁₂) can be taken to be simply the wall thickness, w, although that underestimates the true distance. The thermal resistance is:

$$ {R}_{ws}=\frac{\Delta {x}_{12}}{k_{12}{A}_{12}}=\frac{w}{k_{12}\left({A}_1+{A}_2\right)/2} $$

The thermal conductivity k₁₂ is something you “look up” for the material in question, unless there is more specific information available for a material in question. In this case, an Internet search (January 30, 2011) of “thermal conductivity of ceramic mug” yielded:

http://www.askmehelpdesk.com/physics/why-ceramic-mug-cools-liquid-faster-98327.html

$$ {k}_{12}=15\;\mathrm{W}/\mathrm{m}/\mathrm{K} $$

http://www.madsci.org/posts/archives/2004-12/1102383405.Eg.r.html

$$ {k}_{12}=0.5\hbox{--} 20\;\mathrm{W}/\mathrm{m}/\mathrm{K} $$

A good website for material property values is “Engineering Toolbox” http://www.engineeringtoolbox.com/.

Searching in material properties tabs for thermal conductivity… http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html.

The thermal conductivity for porcelain, and the value used here, is published as:

$$ {k}_{12}=1.5\;\mathrm{W}/\mathrm{m}/\mathrm{K} $$

Such a wide range of uncertainty in a material property is not uncommon. Engineering judgment often comes into play.

1.2.3 Outer Side Wall Convective Resistance: R_as

The exposed outer surface of the side wall, including the inner area above the coffee level and the top of the rim (A₂ from the conduction analysis), exchanges heat by convection in parallel with radiation with ambient air. Technically, the inner portion “sees” the coffee and itself in addition to ambient, but this is a small part of the total exchange so this effect is not considered. The resistances are:

$$ {R}_{as, conv}=\frac{1}{h_{air/ sides}{A}_2}\kern1em \mathrm{and}\kern1em {R}_{as, rad}=\frac{1}{h_{radiation}{A}_2} $$

The values for h_air/sides (6.8) and h_radiation (6.6) are obtained from a Nusselt and Kirchhoff analysis for a coffee temperature of 70 °C.

1.3 Bottom Channel Resistors

1.3.1 Inner Bottom Wall Convective Resistance: R_cb

The only difference between this resistance and the inner side wall is the exposed area:

$$ {R}_{cm}=\frac{1}{h_{cm}\left(\pi {D}^2\right)/4} $$

where h_cm is taken as the same natural convection value as between the coffee and the sides.

1.3.2 Bottom Wall Conductive Resistance: R_fb

For the conductive resistance through the mug bottom, consider the shape to be a truncated cone, with surface 1 as the inside surface of the mug, and surface 2 to be the surface area in contact with the floor. The area is taken to be the geometric average:

$$ {A}_{12}=\frac{1}{2}\left[\frac{\pi {D}^2}{4}+\frac{\pi {\left(D+2w\right)}^2}{4}\right] $$

The thermal resistance is therefore:

$$ {R}_{wb}=\frac{\Delta {x}_{12}}{k_{12}{A}_{12}}=\frac{w}{k_{12}{A}_{12}} $$

1.3.3 Conductive Resistance Across the Floor: R_fb

Heat transfer between the floor of the mug and its surroundings is difficult to precisely define without more information as to the nature of the table on which the mug rests. There is an air gap trapped beneath the bottom rim. If (a big “if”) natural convection effects are sufficiently suppressed so that the air is stationary, then a conductive resistance between the mug bottom and the table surface can be estimated. A radiative channel between flat plates exists between the floor and the bottom.

Then there is a conductive series resistance from the table surface toward the interior of the table. To estimate this resistance, assume the table is large and take a long range view so that the heat transfer emanates outward spherically. The inner surface (surface 1) is taken to be a hemisphere with a radius such that the surface area of the hemisphere equals the surface area of the floor underneath the mug. That is:

$$ 2\pi {r}_1^2=\pi {\left(D+2w\right)}^2/4 $$

or

$$ {r}_1=\left(D+2w\right)/\sqrt{8} $$

The rate of heat transfer across a hemispherical shell of inner radius r₁ and outer radius r₂ (from 1 to 2) is

$$ {q}_{12}=-k2\pi {r}^2\frac{dT}{dr} $$

Separating variables and integrating:

$$ {\displaystyle \underset{T_1}{\overset{T_2}{\int }}dT}=\frac{-{q}_{12}}{2\pi k}{\displaystyle \underset{r_1}{\overset{r_2}{\int }}\frac{dr}{r^2}} $$

$$ {T}_2-{T}_1=\frac{-{q}_{12}}{2\pi k}{\left[\frac{r^{-1}}{-1}\right]}_{r_1}^{r_2}=\frac{q_{12}}{2\pi k}\left(\frac{1}{r_2}-\frac{1}{r_1}\right) $$

$$ {q}_{12}=\frac{\left({T}_1-{T}_2\right)}{\frac{1}{2\pi k}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)} $$

In this case, surface 1 represents the floor surface (subscript “b” for bottom), which is modeled as a hemisphere of radius r₁ (that gives the same area as that of the floor under the mug), and surface 2 represents infinity (so 1/r₂ → 0 and T₂ = T_∞ is ambient). The heat transfer rate, defining the conductive resistance, is:

$$ {q}_{1\infty }=\frac{T_1-{T}_{\infty }}{R_{1\infty }}=2\pi k{r}_1\left({T}_1-{T}_{\infty}\right) $$

$$ {R}_{b\infty }=\frac{1}{2\pi k{r}_1}=\frac{\sqrt{2}}{\pi k\left(D+2w\right)} $$

The following table details results for the floor resistance estimate. This analysis could use a couple of schematics, but the text in the opening paragraph of this section contains a word picture. In the end, the result suggests that the total resistance through the bottom of the mug (40.2 °C/W) is much higher than between the free surface and ambient (8.51 °C/W), which is less than that between the side wall and ambient (2.21 °C/W). That means the bottom surface behaves as a nearly insulated surface.

Detail of floor resistance
D	0.083	m	Inner diameter of mug
hrim	0.004	m	Height of air gap
wrim	0.001	m	Width of rim in contact with floor
kai r	0.023	W/m/K	Thermal conductivity of air
kfloor	0.17	W/m/K	Thermal conductivity of table (Oak value)
kmug	1.5	W/m/K	Thermal conductivity of mug
Aair gap	0.00541	m2	Area of air gap
Arim	0.000264	m2	Area of rim contact
emiss	0.95		Emissivity of both surfaces
hrad	6.8		Radiation coefficient (between 70 and 25 °C plates)
Rbc,air	32.1	°C/W	Across air gap
Rbc,gap radiation	27.2	°C/W	Across air gap
Rrim	10.1		Through rim
Rbinf	31.2	°C/W	Infinite floor
Rfloor	37.1	°C/W	Effective floor contact resistance
UAfloor	0.027	W/°C	Effective floor contact conductance
Ufloor	4.7	W/m2/K	Effective floor contact coefficient

Appendix 2. Parabolic Fits for Plotting Temperature Profiles

This appendix demonstrates a common technique for approximate methods for boundary layer analysis. If three conditions are specified at two locations (the wall and the free stream), then the shape of a function can be fit to a parabola, as detailed here. The three conditions here are that the temperature at the wall is fixed at the contact temperature, and at the edge of a boundary layer, the temperature is fixed at the initial value, and the temperature gradient is set to zero.

For this case, the heat transfer rate between the two solids will not be constrained, and therefore the heat transfer rate between them will not match. A fourth condition would be required to match the heat transfer rate between the two solids, and the temperature could be fit to a cubic. A good exercise (with a fair degree of additional algebra) would be to develop that cubic fit.

For the purposes of demonstrating the temperature profiles (snapshots of temperature plotted against distance from the wall surface at fixed times), a simple parabolic fit that fixes the temperature as the wall surface, the temperature at the edge of the boundary layer and forces the slope to be zero at the edge of the boundary layer. That is, with “x” the distance from the wall, an assumed parabola is:

$$ {T}_{(x)}=a+bx+c{x}^2 $$

where the coefficients (a, b, c) are determined from constraints imposed on the fit. At the wall surface (x = 0), the temperature is fixed at T_c:

$$ {T}_{(0)}={T}_c=a+b(0)+c{(0)}^2 $$

or simply, a = T_c.

At the edge of the boundary layer, x = δ, the temperature is set to ambient:

$$ {T}_{\left(\delta \right)}={T}_{\infty }={T}_c+b\delta +c{\delta}^2 $$

where the result for coefficient “a” has been inserted. Also, the slope at the edge of the boundary layer is set to zero:

$$ {\left.\frac{dT}{dx}\right|}_{x=\delta }=0=b+2c\kern.1em \delta $$

Multiplying this result by d and subtracting from the previous:

$$ {T}_{\left(\delta \right)}={T}_{\infty }={T}_c-c{\delta}^2 $$

$$ c=\frac{T_c-{T}_{\infty }}{\delta^2} $$

Inserting this into the third constraint:

$$ b=-2c\delta =\frac{-2\left({T}_c-{T}_{\infty}\right)}{\delta } $$

A curve fit that yields an approximate shape for the temperature profile therefore is:

$$ {T}_{(x)}={T}_c-2\left({T}_c-{T}_{\infty}\right)\frac{x}{\delta }+\left({T}_c-{T}_{\infty}\right){\left(\frac{x}{\delta}\right)}^2 $$

A somewhat more compact and dimensionless form of this expression is:

$$ \frac{T_{(x)}-{T}_{\infty }}{T_c-{T}_{\infty }}=\left(\frac{x}{\delta}\right)\left(\frac{x}{\delta }-2\right) $$

This formula does not give the exact shape of the temperature profile, but an approximate fit. The time dependency of the profile is contained in the boundary layer thickness, δ, which grows with time.

The inner wall of the mug is at the contact temperature, and the temperature falls within the thermal boundary layer, eventually reaching ambient. There is no single mug temperature. However, an average mug temperature can be obtained by integration (whose value would yield the correct total energy stored, assuming constant specific heat):

$$ {T}_{ave}w={\displaystyle \underset{x=0}{\overset{w}{\int }}Tdx} $$

where the coordinate x is measured from the inner wall of the mug outward. Within the boundary layer (x < δ_m), the temperature profile was fit to a parabola to generate the plot of Fig. 4.36, and outside the boundary layer (x > δ_m), the temperature is ambient. Therefore, the integration can be broken into two pieces:

$$ {T}_{ave}w={\displaystyle \underset{x=0}{\overset{\delta_m}{\int }}{T}_{(x)}dx}+{\displaystyle \underset{\delta_m}{\overset{w}{\int }}{T}_{\infty }dx} $$

The parabolic fit for temperature is: ${T}_{(x)}={T}_c-2\left({T}_c-{T}_{\infty}\right)\frac{x}{\delta_m}+\left({T}_c-{T}_{\infty}\right){\left(\frac{x}{\delta_m}\right)}^2$

Performing the integrations:

$$ {T}_{ave}w={\left.\left[{T}_c-2\left({T}_c-{T}_{\infty}\right)\frac{x^2}{2{\delta}_m}+\left({T}_c-{T}_{\infty}\right)\frac{x^3}{3{\delta}_m^2}\right]\right|}_0^{\delta_m}+{T}_{\infty }{\left.\left[x\right]\right|}_{\delta_m}^w $$

Doing the algebra: ${T}_{ave}={T}_{\infty }+\frac{\delta_m}{3w}\left({T}_c-{T}_{\infty}\right)$

Inserting the expression for the boundary layer thickness as a function of time:

$$ {T}_{ave}={T}_{\infty }+\frac{4\sqrt{\alpha_mt}}{3w\sqrt{\pi }}\left({T}_c-{T}_{\infty}\right) $$

Similarly, the average temperature of the coffee can be calculated during this time:

$$ {T}_{ave, coffee}\pi {R}^2={\displaystyle \underset{r=0}{\overset{R}{\int }}T2\pi rdr} $$

where in this case, the cylindrical shape is considered, so the integration is conducted over a cylindrical shell from the centerline of the coffee (r = 0) to the inner wall of the mug (r = R). The coordinates r and x are related simply by r = R + x. Breaking the integral into two parts, inside and outside the thermal boundary layer:

$$ {T}_{ave, coffee}\pi {R}^2={\displaystyle \underset{r=0}{\overset{R-{\delta}_c}{\int }}{T}_{coffee, init}2\pi rdr}+{\displaystyle \underset{R-{\delta}_c}{\overset{R}{\int }}T2\pi rdr} $$

Since the radius (R = 32 mm) is much larger than the maximum thermal boundary layer thickness (d_c,max = 2.7 mm), the average coffee temperature changes very little during the initial heating, while the mug has increased from ambient (25 °C) to 39.4 °C.

Workshop 4.1. Cooking 1-Node Hot Dogs

A hot dog of diameter D (=11/16″), length L is to be cooked by several methods (boiling baking, barbequing, microwaving). The hot dog is initially at 5 °C and is placed into a cooking environment with a fluid temperature T _∞. It is considered to be cooked when the average temperature exceeds 70 °C. Estimate the cooking times and the surface and center temperatures at that time for each method. Develop a 1-node extended model, where the total thermal capacitance of the hot dog is lumped into a single node and a thermal channel with a conductive resistance from the interior to the surface is in series with a convective resistance between the outer surface and ambient. Use input parameters from the input parameter tables shown.

For the convection coefficient, the value for boiling water (800 W/m²/K) is on the high end of natural convection in liquid water. There is expected to be significant fluid motion induced by the boiling process. For baking (at 350 °F = 177 °C) and microwaving, a value of 15 W/m²/K is typical of natural convection in air, in parallel with a radiative coefficient. The value for barbequing (25 W/m²/K) is taken to be somewhat higher due to the strong updraft caused by the fire.

For the barbequing and microwaving cases, some of these input parameters were chosen after the fact, to give cooling times consistent with experience. For barbequing, the fluid temperature is taken to be 500 °C. In practice, raising the distance between the coals and the food lowers the fluid temperature in contact with the food. There is an important radiation effect from the coals that is not considered here as it does not fit into either radiation model. For the microwaving case, the value of the rate of heat generation is given in Watts absorbed per unit volume, and the value (1(10)⁷ W/m³) is chosen to be one which gives a cooking time consistent with empirical experience. That’s cheating, a little… Or, rather, it is using experimental results and a theoretical model to evaluate an approximation for the effective rate of absorption of microwave energy, for which no attempt to model from first principles is made here.

Note: A detailed model development follows. Conduct your own analysis before referring to it.

Model Development

A general 1-node model with a heat source (for the microwave case) and a thermal channel between the capacity-carrying node and ambient with a conductive and convective/radiative channel in series is shown in Fig. 4.39.

Fig. 4.39

General 1-node model applicable to hot dog model

The energy balance applied to this general RC network is:

$$ C\frac{dT}{dt}=\dot{Q}+\frac{T_{\infty }-T}{R_k+{R}_h} $$

The solution (with constant parameter values) has been given in Chap. 1:

$$ \frac{T-{T}_{SS}}{T_0-{T}_{SS}}={e}^{\frac{-t}{C\left({R}_k+{R}_h\right)}} $$

where ${T}_{ss}={T}_{\infty }+\dot{Q}\left({R}_k+{R}_h\right)$ is the steady-state temperature and T₀ is the initial temperature. Note that the microwave case is the only one for which there is internal heat generation, and there is a difference between the steady-state and ambient fluid temperatures.

The surface temperature is directly related to the average hot dog temperature through the voltage divider channel between the capacity-carrying node and the outer hot dog surface:

$$ \frac{T-{T}_{\infty }}{R_k+{R}_h}=\frac{T_s-{T}_{\infty }}{R_h}=\frac{T-{T}_s}{R_k} $$

$$ {T}_s={T}_{\infty }+\frac{R_h\left(T-{T}_{\infty}\right)}{R_k+{R}_h}=T-\frac{R_k\left(T-{T}_{\infty}\right)}{R_k+{R}_h} $$

While not essential, it is useful to define Biot and Fourier numbers:

$$ Bi={R}_k/{R}_h\kern0.5em \mathrm{Gives}\kern0.5em \mathrm{SPATIAL}\kern0.5em \mathrm{INFO}\kern1em Fo=t/C{R}_k\kern0.5em \mathrm{Gives}\kern0.5em \mathrm{TEMPORAL}\kern0.5em \mathrm{INFO} $$

$$ \frac{T-{T}_{SS}}{T_0-{T}_{SS}}={e}^{\frac{-t}{C{R}_k\left(1+\raisebox{1ex}{${R}_h$}\!\left/ \!\raisebox{-1ex}{${R}_k$}\right.\right)}}={e}^{-\left(\frac{Bi}{1+ Bi}\right)Fo} $$

$$ {T}_s={T}_{\infty }+\frac{\left(T-{T}_{\infty}\right)}{Bi+1}=T-\frac{Bi\left(T-{T}_{\infty}\right)}{Bi+1} $$

Consider the Biot number limits:

For Bi ≫ 1, there is good convection and Ts → T_∞, that is, the hot dog cooks on the edges first, and the center lags behind.

For Bi ≪ 1, there is good conduction, and Ts → T_ave, that is, the hot dog cooks with a uniform temperature.

Evaluating the capacitance is unambiguous:

$$ C=\rho cL\pi {D}^2/4 $$

Similarly, evaluating the convective resistance is unambiguous because convection occurs at a single, specific surface:

$$ {R}_h=1/\left( h\pi DL\right) $$

The convection coefficient is considered to be an average value. For the barbequing case, it is known from experience that the hot dog cooks more on the bottom side. That effect is probably due to radiation heat coming from the hot coals, an effect that could be accounted for by a parallel radiation channel acting on the bottom surface (but from a different temperature than the hot gases flowing past it, requiring a new RC network be drawn). But think of the hot dog as being turned frequently during the heating, and the higher convection coefficient value takes the nonuniform heating effect into account.

Evaluation of the conductive resistance requires some more thought because conduction occurs between two surfaces that have different surface area. As suggested by the sketch in Fig. 4.40, the cylindrical hot dog is modeled as an equivalent wall, with one “surface” (of zero surface area) being the centerline, and the other surface being the outer surface of the hot dog, with surface area πDL (with L the length of the hot dog), a distance D/2 away. The capacity-carrying node is placed at some radius between the centerline and the surface.

Fig. 4.40

Conceptual model for estimating hot dog conductive resistance

The general form of a conductive resistance is:

$$ {R}_k=\frac{1}{k}{\displaystyle \underset{x_1}{\overset{x_2}{\int }}\frac{dx}{A}} $$

which has the general form of an average distance between surfaces (Δx) divided by the thermal conductivity, and an average area. A few models (summarized in Table 4.4) are considered before obtaining results for the base case hot dog.

Table 4.4 Summary of conduction models for 1-node hot dog problem

MODEL 1: The most straightforward conduction model (for conductive resistance between the hot dog surface and the capacity-carrying node) is to “place” the capacity-carrying node at the centerline of the hot dog (at r = 0), and use the basic estimate for R _k as the distance between surfaces dividing by the thermal conductivity times average surface area.

MODEL 2: The centerline in a cylindrical geometry acts like an insulated boundary because the surface area approaches zero with the radius. Conceptually, heat conducted from the surface into the interior does not require it to conduct all the way to the centerline. Therefore, placement of the capacity-carrying node at the centerline will overestimate the conductive resistance. This model places the capacity-carrying node at the midpoint between the surface and the centerline (at r = D/4). The conductive resistance is based on the distance between the surface and node, and on the average areas. When executed, the functional dependencies are the same as Model 1, but the Biot number is reduced by a factor of 2/3.

MODEL 3: This model places the capacity-carrying node at the centroid of a wedge of the hot dog (at r = D/3), rather than at the midpoint between the centerline and center. The centroid is more representative of the solid than the geometric center. The Biot number is 2/5 of the value of Model 1.

MODEL 4: The model places the capacity-carrying node at the centroid of a wedge, but uses the exact formula for conduction between two surfaces at different radial positions, that is, across a cylindrical shell. The Biot number is reduced by a factor of 4.93 compared to Model 1.

Model 4 is considered to be conceptually the most accurate for this case. However, the important take-home message is that all models give the same functional dependency. Moving forward, methods for breaking the hot dog into multiple nodes will be developed, and the differences between models for conduction will be less important.

MODEL 4 results:

Spreadsheet results using conduction Model 4 are summarized in Fig. 4.41 and plots of the average hot dog temperature and the surface temperature are shown in Figs. 4.42, 4.43, 4.44, and 4.45. The conductive resistance and capacitance are independent of the method of cooking and are calculated in the top section (derived parameters) of a worksheet. This block refers to the input parameter block of the problem statement. The convective resistance depends on the method of cooking. Cooking in boiling water results in a large (but not effectively infinite) Biot number (4.73) event, which means conduction through the hot dog is slow compared to the rate at which heat is transferred by convection from the water to the surface. There is a large spatial gradient, with the surface temperature much closer to the fluid temperature than the core temperature. The other three cases are low Biot number cases (but not effectively zero), which means that conduction is effective relative to convection and the hot dog cooks nearly uniformly. Comparing the boiling and baking cases, the cooking time is faster in boiling water, despite the fact that the fluid temperature (100 °C) is lower than that in an oven (177 °C). But at the time the hot dog is cooked, the surface temperature in water is 95 °C (25 °C above the hot dog average, and only 5 °C below the fluid temperature) while the surface temperature is 79 °C for baking (9 °C above the hot dog, and almost 100 °C below the fluid temperature).

Fig. 4.41

Results of 1-node hot dog workshop for conduction Model 4

Fig. 4.42

Boiling: Biot number = 4.73, T_inf = 100 °C

Fig. 4.43

Baking: Biot number = 0.089, T_inf = 177 °C

Fig. 4.44

Barbeque, Biot number = 0.148, T_inf = 500 °C

Fig. 4.45

Microwave, Biot number = 0.089, T_inf = 25 °C

For barbequing, the cooking time is shorter than boiling, because the fluid temperature (500 °C) is so high. Meanwhile, the surface temperature is 125 °C. In this case, despite the low Biot number, the hot dog appears to heat nonuniformly, but that is merely because the fluid temperature is so high. The surface temperature is closer to the average hot dog temperature than it is to the fluid temperature. The fluid temperature chosen for this case is not well characterized. Nevertheless, it is well known from experience that it is a function of the vertical distance between the source of the fire and the hot dog. It burns on the outside if placed too close.

Finally, microwaving yields the shortest cooking times (but to me, a hot dog in a microwave is too rubbery…). The equilibrium (steady-state) temperature of 3,200 °C represents the temperature the hot dog would achieve in theory if the microwave were left on indefinitely. Of course, something would happen to the hot dog to change the model long before it reached that point.

The surface temperature at the microwave cooking time is lower than surface temperature. That is, heat flows outward from the core of the hot dog to the surface. Heat is absorbed uniformly throughout the interior, yet the temperature is not uniform. In order for heat to flow from the interior to the surface, and ultimately to the ambient, there must be a temperature gradient driving it. From Fig. 4.45, at the start of the cooking event, the surface temperature is higher than average. That is because the hot dog is taken from the refrigerator, and it is being warmed by the air temperature at the same time it is being heated from within. Once the hot dog average temperature exceeds the air temperature, the core is hotter than the surface.

Workshop 4.2. Beverage Chilling Methods

Beverages are commonly purchased at room temperature, but are consumed cold. This workshop develops a means to predict the time it takes to chill a room temperature beverage to a desirable drinking temperature. A standard beverage chilling time is defined here as the time it takes to chill a beverage from 25 to 5 °C using a specific cooling method. The Cooper Cooler™ is a consumer appliance developed at the Cooper Union that is designed to do so rapidly. The user prepares an ice/water slurry in a reservoir and the Cooper Cooler sprays a cold recycled jet of water on a horizontal beverage while is it rotated along its natural axis. For this workshop, five different chilling methods are considered: three conventional modes and two Cooper Cooler modes.

• 1. Refrigerator: The beverage is placed in a standard household refrigerator.

• 2. Freezer: The beverage is placed in a freezer.

• 3. Ice/water bath: The beverage is placed in a cooler filled with ice, and enough water to completely immerse the beverage.

• 4. Cooper Cooler™ Spray-only Mode: A jet of cold water is sprayed onto a stationary, horizontal beverage.

• 5. Cooper Cooler™ Spray and Spin Mode: A jet of cold water is sprayed onto a rotating horizontal container.

Model Development

All five chilling methods are modeled with the same RC network which the reader can draw based on the following description. The capacitance of the beverage and container are lumped into a single node, T _bev . For heat to be transferred from the beverage to the coolant (at T _coolant ), it must flow from the beverage to the inner wall of the container (a convective resistance), across the container wall (conductive), and from the outer wall to the coolant (convective/radiative). In other words, there is a single thermal channel from the beverage to the coolant with three resistances in series. The differences in chilling times are due to different values of the various capacitances (for different beverage types), thermal resistances, and coolant temperature.

The governing nodal equation is:

$$ {C}_{bev}\frac{dT}{dt}=\frac{T_{coolant}-{T}_{bev}}{R_{eq}} $$

where R _eq is the equivalent thermal resistance (sum of three series resistances). The solution to this now familiar governing first-order equation is:

$$ \frac{T_{bev}-{T}_{coolant}}{T_{initial}-{T}_{coolant}}={e}^{\frac{-t}{R{C}_{bev}}} $$

This solution can be used, once estimates of R and C are made, to prepare plots of beverage temperature vs. time. To determine the standard chilling time (t_chill), the natural logarithm of both sides is taken and the beverage temperature set to the desired final temperature:

$$ {t}_{chill}=-R{C}_{bev} \ln \left(\frac{5-{T}_{coolant}}{25-{T}_{coolant}}\right) $$

A spreadsheet template for a specific beverage type is shown in Fig. 4.46. Blocks for input parameters, derived parameters, and results are shown in which suggested values for the case of a 12 oz. bottled beverage are listed. Input values considered to be common for different beverage types are listed in the results block. Blank boxes are left for the formulas needed to implement the method. The beverage is considered to have a clearly identified outer diameter and treated as an equivalent cylinder to obtain the length. A thin-wall approximation is suggested in which the inner and outer areas are treated as approximately equal when thermal resistances are calculated.

Fig. 4.46

Spreadsheet templates for beverage chilling workshop

The coolant temperatures are taken to be those typical of a refrigerator, freezer, and the melting point of water for the three conventional cooling methods. For the Cooper Cooling methods, a coolant temperature of 1 °C is suggested for spray only, and 2 °C for spray and spin. In practice, the coolant is recycled from an ice/water bath, and heat added to the coolant from the beverage warms it somewhat, even as the ice in the bath cools it. The modeling of this process involves more than one node and is discussed later. The coolant temperature is warmer for the spray and spin because of the higher heat transfer rates from the beverage.

The value for inner heat transfer coefficient is taken to be a typical value for the natural convection of fluid with nominal water properties (300 W/m²/K) for all stationary beverages. When the beverage is rotated, there is mixing inside which produces a sort of wind chill on the inside and the value used (1,500 W/m²/K) is representative of a mild forced convection with water. The outer convection/radiative coefficient is taken to be representative of natural convection in air, with a parallel radiation channel, for the refrigerator and freezer. For the ice/water bath, the outer coefficient is representative of natural convection in water. A value representative of a forced convection with water (stronger than for the inside) is listed for the two cases, where a jet of cold water is sprayed.

For the three conventional cooling methods (refrigerator, freezer, ice/water bath), the coolant is considered to be in contact with the entire surface area of the beverage (sides, top, and bottom) whereas for the two Cooper Cooling cases, only the side surface is considered to be in good thermal contact with the coolant.

It is suggested that the case of a 12 oz. canned beverage be compared with the 12 oz. bottled beverage, in which case the only difference is the thermal resistance of the container wall. Also, the cases of a 0.5, 1, and 2 L plastic bottles be considered, where both the thermal capacitance and surface area vary. It is also suggested that plots of temperature vs. time be prepared with all five cooling methods on the same set of axes for a given beverage. Include the inner and outer wall temperatures. A final suggestion is to conduct some experiments and compare results. Notice that in an experiment, the natural dependent variable to set is the chilling time, and this analysis can help inform an intelligent choice for that time for a single experiment. Too short a time, and the temperature difference between initial and final temperature will be small and experimental uncertainty will be high. Too long a time, and the beverage will be too close to the final equilibrium to obtain a reasonable dimensionless temperature.