Few-Node Transient Models

Abstract

Try this experiment. Fill a ceramic mug with boiling water and immediately grab its side. Close your eyes and concentrate on what it feels like. You’ve just experienced the mug heating phase, the topic of this chapter. In order to observe the separate behavior of the coffee and the mug, the capacitance of the coffee and the mug must be treated separately, creating a 2-node model. The goal of this chapter is to develop a general scheme for doing so. The term “Few-Node Model” is loosely defined as one in which space is discretized into clearly identifiable regions that can be given clear names (as opposed to numbers). In the next chapter, called “multi-node,” space will be discretized into finer elements in a way that there is no clear sense of how many nodes there should be beforehand.

Appendices

Appendix 1. Nodal Response Time

Figure 5.3, reproduced here, shows a node with a capacitance, two neighbors and heat generation. The inherent response time of this node is established by determining its response assuming its neighbors are fixed, not variable. In this case, T₁ will approach an equilibrium value that will depend on its two neighbors and the heat generation.

Repeating the governing nodal equation:

$$ {C}_1\frac{d{T}_1}{dt}={\dot{Q}}_1+\frac{T_2-{T}_1}{R_{12}}+\frac{T_{\infty }-{T}_1}{R_{1\infty }} $$

A dimensionless temperature (ψ ₁) that starts at a value of unity (1) and approaches a value of zero at equilibrium, and a dimensionless time defining a response time for node 1 (τ₁ = RC_1equiv, to be determined) are defined:

$$ {\psi}_1=\frac{T_1-{T}_{1 eq}}{T_0-{T}_{1 eq}} $$

$$ \widehat{t}=\frac{t}{\tau_1}=\frac{t}{R{C}_{1 equiv}} $$

Inserting into the nodal equation:

$$ \frac{C_1}{\tau_1}\frac{d\left[{T}_{1 eq}+{\psi}_1\left({T}_0-{T}_{1 eq}\right)\right]}{d\widehat{t}}={\dot{Q}}_1+\frac{T_{2 fixed}-\left[{T}_{1 eq}+{\psi}_1\left({T}_0-{T}_{1 eq}\right)\right]}{R_{12}}+\frac{T_{\infty }-\left[{T}_{1 eq}+{\psi}_2\left({T}_0-{T}_{1 eq}\right)\right]}{R_{1\infty }} $$

Note that T₂ is given a “fixed” superscript. In the system, T₂ is a moving target, so this calculation is a hypothetical “what if.” With T₂ and therefore T_1eq fixed, the first term under the time derivative is zero. Rearranging:

$$ \frac{C_1\left({T}_0-{T}_{1 eq}\right)}{\tau_1}\frac{d{\psi}_1}{d\widehat{t}}=\left[{\dot{Q}}_1+\frac{T_{2 fixed}-{T}_{1 eq}}{R_{12}}+\frac{T_{\infty }-{T}_{1 eq}}{R_{1\infty }}\right]-\left({T}_0-{T}_{1 eq}\right){\psi}_1\left(\frac{1}{R_{12}}+\frac{1}{R_{1\infty }}\right) $$

The term in brackets [ ] is identically zero because the equilibrium value for T₁ is defined by setting the time derivative of the nodal equation to zero:

$$ 0={\dot{Q}}_1+\frac{{T_2}_{fixed}-{T}_{1 eq}}{R_{12}}+\frac{T_{\infty }-{T}_{1 eq}}{R_{1\infty }} $$

Then the factor (T₀ − T_1eq) cancels and the dimensionless equation is:

$$ \frac{d{\psi}_1}{d\widehat{t}}=-\frac{\tau_1}{C_1}\left(\frac{1}{R_{12}}+\frac{1}{R_{1\infty }}\right){\psi}_1 $$

The time constant can now be defined as:

$$ {\tau}_1=\frac{C_1}{\left(\frac{1}{R_{12}}+\frac{1}{R_{1\infty }}\right)}={R}_{1 equiv}{C}_1 $$

and the governing equation becomes the first-order model:

$$ \frac{d{\psi}_1}{dt}=-{\psi}_1 $$

with solution:

$$ {\psi}_1={e}^{-\widehat{t}} $$

It is reemphasized that this temperature response is a hypothetical one, with T₂ fixed. The aim of this derivation is to determine the inherent response of a given node to its immediate environment. In a real simulation, T₁ would not approach this equilibrium value because T₂ is changing with time, unless the system itself has reached equilibrium.

Appendix 2. Detailed Thermal Resistance Formulas for Coffee Mug Problem

Figures 5.6 and 5.7 (repeated side by side) shows the input parameters (left) and derived quantities (right) from the “params” worksheet from “Ch5_2Node_Transient.xls.” The formulas used in the tables on the right side (that reference the cells in the input quantities block) are detailed in this appendix.

The basic nomenclature of the geometry is repeated here, introduced previously in Chap. 2.

The coffee depth is:

$$H - L = \frac{{\rlap{--} V}}{{\pi {D^2}/4}}$$

The mug and coffee capacitances are:

$${C_{mug}} = {m_{mug}}{c_{mug}}\quad {\text{and}}\quad {C_{coffee}} = {\rho _{coffee}}{c_{coffee}}\rlap{--} V$$

The area of the coffee-free surface is:

$$ {A}_{free\kern0.5em surface}=\pi {D}^2/4 $$

The area of contact between the coffee and the mug is the sum of the bottom and sides:

$$ {A}_{inside}=\pi {D}^2/4+\pi D\left(H-L\right) $$

The total outside surface area of the mug (exposed to ambient) includes the outside, bottom (neglecting the bottom ridge area), the top rim surface, and the inside mug surface above the coffee level:

$$ {A}_{outside}= Outside+ Bottom+ Rim+ Inside $$

$$ {A}_{outside}=\pi \left(D+2w\right){H}_o+\pi {\left(D+2w\right)}^2/4+\pi \left[{\left(D+2w\right)}^2-{D}^2\right]+\pi DL $$

The average mug surface area (used in the mug-conductive resistance) is the geometric average of the inside and outside areas:

$$ {A}_{average}=\left({A}_{inside}+{A}_{outside}\right)/2 $$

2.1 Individual Resistances

The RC network is repeated here for reference.

The convective resistance between the coffee and the mug is:

$$ {R}_{cs}=\frac{1}{h_{coffee/ surface}{A}_{top}} $$

The value chosen for the equivalent heat transfer coefficient (h_{coffee/surface} = 470 W/m²/K) is representative of a natural convection between a liquid with water properties and a solid surface.

The equivalent convective/radiative/evaporative resistance between the coffee surface and ambient is:

$$ {R}_{sa}=\frac{1}{h_{air/ top}{A}_{top}} $$

The value chosen for the equivalent heat transfer coefficient is built on values of 6.8, 6.6, and 10 W/m²/K for convection, radiation, and evaporation, respectively.

The convective resistance between the coffee and the mug inner surface is:

$$ {R}_{cm}=\frac{1}{h_{cm}{A}_{inside}} $$

The value chosen for the convection coefficient (470 W/m²/K, with radiation considered negligible in comparison) is representative of natural convection between a liquid with water properties and a vertical surface.

The conductive resistance across the mug wall is:

$$ {R}_m=\frac{w}{k_{mug}{A}_{average}} $$

This resistance is split into two equal parts between the capacity-carrying mug node and the mug surfaces. This model is considered reasonable for nearly full mugs. A detailed analysis of nearly empty mugs is deferred to a later chapter.

Heat transfer from the outer surface of the mug to ambient is split into two parallel channels: side and bottom. For the side, a convective/radiative equivalent resistance for those portions exposed to air is:

$$ {R}_{ma, sides}=\frac{1}{h_{air/ sides}\left({A}_{outside}-\pi {\left(D+2w\right)}^2/4\right)} $$

The area is the outside area subtracted by the surface area of the bottom. The equivalent heat transfer coefficient is the sum of the convective and radiative portions (8.8 and 8.6 W/m²/K, respectively). For the heat transfer through the bottom:

$$ {R}_{ma, floor}=\frac{1}{h_{floor}\pi {\left(D+2w\right)}^2/4} $$

The effective heat transfer coefficient (h_floor = 4.7 W/m²/K) is detailed in Chap. 4, and reflects heat that flows from the bottom of the mug to the floor it is placed on, and then conducted outward into the floor.

Appendix 3. Joulie Property Estimation

Resistance and capacitance value estimation requires assumptions about the construction of the Joulie, detailed in this appendix, and calculated in a separate worksheet (Fig. 5.26). The mass and volume of Joulies are easily measured (34.0 g and 20.6 mL). Thermal properties are obtained online (accessed 1 February 2014), with convection coefficients inside and outside entered as representative values.

Fig. 5.26

Input spreadsheet focused on property values for a single Joulie used as inputs to the simulation input sheet

The Joulies are modeled as equivalent spheres. That is, the equivalent outer diameter is calculated from the measured volume as:

$$ d={\left(\frac{6V}{\pi}\right)}^{1/3} $$

The diameter of the wax (and thereby the wall thickness) is determined by summing the total mass from wax core plus shell:

$$ \begin{array}{c}{m}_{Total}={\rho}_{shell}{V}_{shell}+{\rho}_{wax}{V}_{wax}={\rho}_{shell}\left({V}_{Total}-{V}_{wax}\right)+{\rho}_{wax}{V}_{wax}\\ {}\kern-7.8em ={\rho}_{shell}{V}_{Total}-{V}_{wax}\left({\rho}_{shell}-{\rho}_{wax}\right)\end{array} $$

Solving for the wax volume:

$$ {V}_{wax}=\frac{\rho_{shell}{V}_{total}-{m}_{total}}{\rho_{shell}-{\rho}_{wax}} $$

$$ {d}_{wax}={\left(\frac{6{V}_{wax}}{\pi}\right)}^{1/3} $$

The total specific heat (during single phase stages) is:

$$ {C}_{Joulie}={(mc)}_{Total}={\rho}_{shell}{V}_{shell}{c}_{shell}+{\rho}_{wax}{V}_{wax}{c}_{wax} $$

Yielding effective specific heat:

$$ {c}_{ave}=\frac{\rho_{shell}{V}_{shell}{c}_{shell}+{\rho}_{wax}{V}_{wax}{c}_{wax}}{\rho_{shell}{V}_{shell}+{\rho}_{wax}{V}_{wax}} $$

For the resistances, there are two separate entries for R_wax, representing the thermal resistance between the wax and the inner shell wall. The first is a pure conduction value and the second is assuming a convective mode. The conduction value is much larger than the convective one and also large compared to the outer convection plus conduction across the shell wall (which is negligible). That means that the Biot number of the wax is large, and during heating, the center of the wax will respond much more slowly than its outer layers. Use of the convection resistance is considered a better way to estimate the rate at which heat is absorbed by the Joulie. A more detailed spatial model would be the way to address this issue (and a great assignment!).

These calculations are for a single Joulie. Note that when multiple Joulies are used, the total capacitance is proportional to the number of Joulies, while the thermal resistance is inversely proportional. Adding more Joulies increases the total surface area, reducing the resistance of heat flow. That is:

$$ {C}_{3\kern0.5em Joulie s}=3{C}_{single\kern0.5em Joulie}\kern1em \mathrm{while}\kern1em {R}_{3\kern0.5em Joulie s}=\frac{1}{3}{R}_{single\kern0.5em Joulie} $$

I’d bet that lots of people would not think that through and assume R _{3 Joulies}  = 3R _{1 Joulie} , and in so doing confuse a parallel heat transfer concept with a series effect.

Workshop 5.1. A Batch Mode Solar Water Heater

An alternative to burning fossil fuels to produce hot water for domestic needs is to heat water using solar energy. It is a common experience that placing a garden hose in direct sunlight can produce hot water. It is also a common experience that sunlight can be focused using a magnifying glass. Alternatively, a reflecting mirror can be used to focus sunlight. This workshop applies these principles to a specific arrangement in which a “solar pipe” is filled with initially cold water and placed in focused sunlight. The water is heated (without a flow of water) until the water reaches a target temperature, at which point a pump is activated (say by a solar photovoltaic cell) and the hot water is pumped out of the pipe into a separate tank, and the pipe is filled with a fresh batch of cold water, completing the batch cycle. The time it takes to complete a cycle depends on the angle and intensity of sunlight which varies throughout the course of a day.

An alternative mode, called continuous mode, is to use the same arrangement but with a continuous flow of water through the solar pipe. That problem will be studied much later, after the effects of advective flows are developed.

Problem statement: In a solar thermal application (Fig. 5.27), black polypropylene tubing (k_p = 0.15 W/m/°C, ρ_p = 1,400 kg/m³, c_p = 1,600 J/kg/°C) of length L = 3 m, outer diameter (D₀) of 0.040 m, and wall thickness t = 0.0028 m is subjected to focused sunlight distributed evenly on the outer surface (concentrated with a parabolic mirror). The intensity of sunlight that enters the top plane (the solar insolation) has a maximum value of I_0,max = 800 W/m² (on a clear day at noon) and a solar collection width to rod diameter ratio (n) to be determined as a design parameter. An optical efficiency η_opt = 0.9 (defined as the ratio of the solar radiation that strikes the pipe to the collected radiation) accounts for collected solar energy not absorbed by the pipe. In the end view, the angle of the sun remains perpendicular to the collecting plane through proper orientation of the mirror (and changes gradually through the course of a year). From the side view, the angle between the incoming solar rays and the perpendicular to the pipe varies through the course of a day, from θ°= 90° at sunrise to 0° at noon to 90° at sunset. The absorption coefficient of incident radiation is α = 0.85 (85 % of the radiation that strikes the pipe is absorbed, 15 % is reflected). At the start of a cycle, the tube is filled with liquid water (ρ_w = 1,000 kg/m³, c_w = 4,200 J/kg/K) at a temperature T₀ = 15 °C. The tube is exposed to ambient at T_∞ = 25 °C. The objective is to raise the temperature of the water to a target of T_final = 50 °C before pumping it into a separate collection tank and starting a new cycle.

Fig. 5.27

Schematic of a solar pipe placed at the focal point of a parabolic mirror

Write an Excel spreadsheet or program code that implements an explicit numerical formulation of a 2-node model (tube and water). Assume the inner surface is exposed to the water with a constant convection coefficient of h_i = 300 W/m²/°C, and the outer surface is exposed to ambient with a constant convection coefficient of h_o = 15 W/m²/°C. Use the program to determine the size of the solar collection area (nDL) that would cause the outer surface of the pipe to reach its melting point (T_melt = 160 °C) at the same time the water reaches its target of 50 °C when the solar flux equals its maximum value at noon. Then determine the time it takes to heat a fresh batch of water to its target as a function of solar angle (which can be related to time of day) for fixed values of solar flux. Report a minimum intensity required to heat the water to the target temperature, the maximum intensity before the outer surface of the pipe reaches its melting point (T_melt = 160 °C), and the size of the solar collection area (nD), assuming the maximum intensity of direct sunlight is 800 W/m² (a typical maximum value on a clear day at noon). Prepare a table of relevant quantities and plots of

• required heating time as a function of intensity (above the minimum required)

• all nodal temperatures (including non-capacity-carrying nodes) vs. time with maximum intensity

• temperature vs. radial position at fixed times with maximum intensity

Note: A specific 2-node model is developed on the next page. It is advised that you develop your own model first before consulting that model. Then decide which model to implement.

Model Development

The 2-node thermal RC network shown in Fig. 5.28 treats node 1 as the water and node 2 as the pipe. The pipe node (2) is placed in the middle of the pipe wall. The solar absorption is modeled as a heat source absorbed on the surface (not directly on the capacity-carrying node). The surface node is in thermal contact with ambient (through a parallel convective/radiative channel with an assumed constant effective heat transfer coefficient) and the pipe center (through a conductive channel). The pipe is in thermal contact with the water (through a series conductive/convective resistance).

Fig. 5.28

2-Node RC network for a solar pipe in batch mode

The nodal equations are (water = 1, pipe = 2, pipe outer surface = abs):

$$ {C}_1\frac{d{T}_1}{dt}=\frac{T_2-{T}_1}{R_{1i}+{R}_{i2}} $$

$$ {C}_2\frac{d{T}_2}{dt}=\frac{T_1-{T}_2}{R_{1i}+{R}_{i2}}+\frac{T_{abs}-{T}_2}{R_{2a}} $$

The solar flux is absorbed at the outer surface, not at the center, and a nodal balance on the surface node is:

$$ 0={\dot{Q}}_{abs}+\frac{T_2-{T}_{abs}}{R_{2a}}+\frac{T_{\infty }-{T}_{abs}}{R_{a\infty }} $$

Expressing the time derivatives with a forward difference approximation and rearranging for the recursion relations, the future values (superscript “p + 1”, after a time step Δt) are explicitly expressed in terms of present values (superscript “p”):

$$ {T}_1^{\left(p+1\right)}={T}_1^{(p)}+\frac{\Delta t}{C_1}\left(\frac{T_2^{(p)}-{T}_1^{(p)}}{R_{1i}+{R}_{i2}}\right) $$

$$ {T}_2^{\left(p+1\right)}={T}_2^{(p)}+\frac{\Delta t}{C_2}\left[\left(\frac{T_1^{(p)}-{T}_2^{(p)}}{R_{1i}+{R}_{i2}}\right)+\left(\frac{T_{abs}^{(p)}-{T}_2^{(p)}}{R_{2a}}\right)\right] $$

After rearrangement, the temperature at the absorbing surface is expressed as:

$$ {T}_{abs}^{(p)}=\left({\dot{Q}}_{abs}+\frac{T_2^{(p)}}{R_{2a}}+\frac{T_{\infty }}{R_{a\infty }}\right)/\left(\frac{1}{R_{2a}}+\frac{1}{R_{a\infty }}\right) $$

The time constants for the two nodes are given by:

$$ {\tau}_1=\left({R}_{1i}+{R}_{i2}\right){C}_1 $$

and

$$ {\tau}_2=\frac{C_2}{\frac{1}{R_{1i}+{R}_{i2}}+\frac{1}{R_{2a}+{R}_{a\infty }}} $$

The time step can be chosen to be a fraction of the smallest of these.

The rate of heat absorption is defined by:

$$ {\dot{Q}}_{abs}=\left(\begin{array}{l}\mathrm{fraction}\ \mathrm{of}\\ {}\mathrm{collected}\ \mathrm{radiation}\\ {}\ \mathrm{that}\ \mathrm{strikes}\ \mathrm{pipe}\end{array}\right)\left(\begin{array}{l}\mathrm{fraction}\ \mathrm{of}\\ {}\mathrm{radiation}\ \mathrm{that}\ \mathrm{strikes}\\ {}\mathrm{pipe}\ \mathrm{that}\ \mathrm{is}\ \mathrm{absorbed}\ \end{array}\right)\left(\begin{array}{l}\mathrm{Rate}\ \mathrm{of}\ \mathrm{Solar}\\ {}\mathrm{Radiation}\\ {}\ \mathrm{Collected}\end{array}\right) $$

$$ {\dot{Q}}_{abs}=\left({\eta}_{opt}\right)\left(\alpha \right)\left(nDL{I}_0 \cos \theta \right) $$

Workshop 5.2. Cooper Cooling with a Finite Reservoir

In Workshop 4.2, various methods of chilling a beverage were considered, the fastest of which involved the process of Cooper Cooling™ in which a horizontal beverage is rotated along its axis while being sprayed with a cold jet of water. In that problem statement, the temperature of the water jet was a fixed input value, a result that would occur if the source of the water was a large (infinite) reservoir of ice and water, and/or the water heated by the beverage was not returned and recycled. However, the first marketed consumer product that incorporated the Cooper Cooling process has a finite reservoir. Jet temperature data and a schematic of the process are shown in Fig. 5.29, taken from Jaime Cachiero’s Master’s Thesis. The cold water jet is drawn from an ice/water bath and is pumped onto the beverage. Heat transferred from the beverage is added to the jet before the jet is returned to the ice water bath. Furthermore, a large portion of the electrical input to the pump is added to the recirculating water. In the reservoir, the water is in contact with ice, which removes heat from the water jet. Since the beverage temperature changes with time, the rate at which heat is added to the jet changes. In practice, during the first four or so beverages, the jet temperature fluctuates around approximately 2 °C (the input value used for Workshop 4.2). The goal of this workshop is to develop a 3-node model (beverage, reservoir, ice) that predicts this behavior and can be used to explore other scenarios.

Fig. 5.29

Schematic of Cooper Cooling with a Finite Reservoir. The data shows the measured temperature of the water jet before it is used to chill a horizontal rotating beverage. A warm beverage is added every 60 s (copyright Wei Dai, reproduced with permission)

Note: A specific 3-node model is developed on the next page. It is advised that you develop your own model first before consulting that model. Then decide which model to implement.

Model Development

It is good practice to draw a thermal resistance network for this problem in which capacitance is assigned to the beverage and reservoir, and a rechargeable battery at the melting point of the reservoir (T_mp) is used to represent the ice. The mass of the ice, but not its temperature, changes as it melts. Notice that if a reservoir fluid other than water is used (like salt water), the melting point can differ from that of water. Also, the capacitance of the reservoir changes with time in a predictable way as ice melts. The composition of the reservoir might change if a reservoir other than water were used.

The governing equation for the beverage is the same as that in Workshop 4.2, namely:

$$ {C}_{bev}\frac{d{T}_{bev}}{dt}=\frac{T_{res}-{T}_{bev}}{R_{bev}} $$

The difference in this model is that the reservoir temperature is not a fixed input, and this differential equation cannot be integrated directly.

The liquid water in the system is considered to be at an average temperature (T_res) that exchanges heat with the beverage (at T_bev), the ice (at T_mp, the melting point of the reservoir), and ambient (at T_∞) and picks up electrical energy that is converted to thermal energy in the pump (VI_pump):

$$ {C}_{res}\frac{d{T}_{res}}{dt}=\frac{T_{bev}-{T}_{res}}{R_{bev}}+\frac{T_{mp}-{T}_{rex}}{R_{ice}}+\frac{T_{\infty }-{T}_{res}}{R_{\infty }}+V{I}_{pump} $$

The energy stored in the ice is in the form of latent (phase) energy, and an energy balance on the ice is given by:

$$ {L}_f\frac{d{M}_{ice}}{dt}=\frac{T_{res}-{T}_{mp}}{R_{ice}} $$

L_f is the heat of fusion of ice (330 kJ/kg) and the thermal resistance between the ice and the reservoir (R_ice) is one of external flow forced convection on a surface area that changes with time.

5.1 Explicit Formulation

The nodal equations written in the form of an explicit formulation are with superscripts used to indicate all values that change with time:

$$ {T}_{bev}^{\left(p+1\right)}={T}_{bev}^{(p)}+\frac{\Delta {t}^{(p)}}{C_{bev}}\left(\frac{T_{res}^{(p)}-{T}_{bev}^{(p)}}{R_{bev}}\right) $$

$$ {T}_{res}^{\left(p+1\right)}={T}_{res}^{(p)}+\frac{\Delta {t}^{(p)}}{C_{res}^{(p)}}\left(\frac{T_{bev}^{(p)}-{T}_{res}^{(p)}}{R_{bev}}+\frac{T_{mp}-{T}_{res}^{(p)}}{R_{ice}^{(p)}}+\frac{T_{\infty }-{T}_{res}^{(p)}}{R_{\infty }}+V{I}_{pump}\right) $$

$$ {M}_{ice}^{\left(p+1\right)}={M}_{ice}^{(p)}+\frac{\Delta {t}^{(p)}}{L_v}\left(\frac{T_{res}^{(p)}-{T}_{mp}}{R_{ice}^{(p)}}\right) $$

The reservoir capacitance equals the specific heat of the reservoir times the mass of the reservoir, which equals the initial mass of the reservoir (M_res,0) plus the mass of ice that has melted (the initial mass of ice less the remaining ice, M_ice,0 – M_ice):

$$ {C}_{res}^{(p)}={c}_{res}\left({M}_{res,0}+{M}_{ice,0}-{M}_{ice}^{(p)}\right) $$

The thermal resistance that models the melting of the ice requires an assumption about the shape of the ice, and how that changes. For example, if ice cubes (actual cubes of dimension “w”) are used, and it is assumed that they remain as cubes (even though in practice the edges become rounded), then the surface area can be related to the mass. That is, if a total of n_cubes are added initially of dimension w₀, the initial mass of ice is given by:

$$ {M}_{ice,0}={n}_{cubes}{w}_0^3 $$

As the ice melts, the number of cubes does not change, but their size does, so that the remaining mass of ice can be related to the ice cube size by:

$$ {M}_{ice}={n}_{cubes}{w}^3 $$

The thermal resistance between the ice and the reservoir is given by:

$$ {R}_{ice}^{(p)}=\frac{1}{h_{ice}{A}_{ice}^{(p)}}=\frac{1}{h_{ice}{n}_{cubes}6{\left({w}^{(p)}\right)}^2} $$

Notice that the thermal resistance increases as the ice cubes become smaller. A value of 1,500 W/m²/K would be reasonable for the convection coefficient in this case.

The values for the beverage capacitance and thermal resistance can be taken from Workshop 4.2, and simulations with different beverages can be conducted. The electrical input to the pump can be varied, but VI_pump = 20 W is a good starting place.

There are a variety of scenarios that can be used to run a simulation. For example, a prechilling period can be conducted in which case a number of ice cubes are placed into an input mass of reservoir at an initial temperature, without actively chilling a beverage, and letting the initial ice/water reservoir come to equilibrium before adding beverages. A decision about how long to chill each individual beverage can be made and that simply involves resetting the temperature of the beverage. Chilling consecutive beverages can be simulated by either fixing the chilling time or allowing the system to operate until the beverage reaches a target temperature. In the former case, the final temperature of each beverage will be different. In the latter case, the total time required to chill each beverage will be different. Eventually, the reservoir will not be able to chill a beverage to that target.

Additional Workshop Ideas

Workshop 5.2: Solve the 2-node mug of coffee problem using a variable step-size algorithm that fixes the maximum change in temperature (absolute value) of either the mug or coffee.

Workshop 5.3: Solve the 2-node mug of coffee problem using a variable step-size algorithm that fixes the maximum fractional change in temperature of either the mug or coffee.

Workshop 5.4: Solve the 2-node mug of coffee problem using the backward implicit method with a fixed time step.

Workshop 5.5: Solve the 2-node mug of coffee problem using a Styrofoam cup.

Workshop 5.6: Solve the 2-node mug of coffee problem using the explicit numerical scheme with variable thermal resistances on the side channel. Assume an emissivity of 0.9 for the radiation coefficient, and use the following formulas (derived in a future chapter) for the convection coefficients. For convection between the outer surface of the mug and ambient:

$$ {h}_{air}=\left(1.5646-0.0015{T}_{film, air}\right)\frac{{\left|\left({T}_{mug}-{T}_{ambient}\right)\right|}^{0.25}}{H_0^{0.25}} $$

where T _film,air is the average of the mug and ambient. For convection between the inner surface of the mug and the coffee:

$$ {h}_{coffee}=\left(32.38{T}_{film, coffee}^{0.441}\right)\frac{{\left|\left({T}_{coffee}-{T}_{mug}\right)\right|}^{0.25}}{H_0^{0.25}} $$

where T _film,coffee is the average of the coffee and mug. Compare this model with the model any of the constant convection coefficient models (and the data).

Workshop 5.7: A hybrid implicit method

For a stiff problem (${\tau}_2\ll {\tau}_1$), a legitimate approach is to use a forward difference approximation for the node with the larger response time (i.e., coffee) and a backward difference for the node with the shorter response time (mug). The future value of T₁ is obtained (from before) as:

$$ {T}_1^{\left(p+1\right)}={T}_1^{(p)}+\frac{\Delta {t}^{(p)}}{{C_1}^{(p)}}\left[{\dot{Q}}_1^{(p)}+\frac{{T_2}^{(p)}-{T_1}^{(p)}}{{R_{12}}^{(p)}}+\frac{{T_{\infty}}^{(p)}-{T_1}^{(p)}}{{R_{1\infty}}^{(p)}}\right] $$

The backward difference equation for T₂ is rearranged as:

$$ {T}_2^{\left(p+1\right)}\left(\frac{{C_2}^{\left(p+1\right)}}{\Delta {t}^{\left(p+1\right)}}+\frac{1}{{R_{12}}^{\left(p+1\right)}}+\frac{1}{{R_{2\infty}}^{\left(p+1\right)}}\right)={\dot{Q}}_2^{\left(p+1\right)}+\frac{C_2^{\left(p+1\right)}{T}_2^{(p)}}{\Delta {t}^{\left(p+1\right)}}+\frac{{T_1}^{\left(p+1\right)}}{{R_{12}}^{\left(p+1\right)}}+\frac{{T_{\infty}}^{\left(p+1\right)}}{{R_{2\infty}}^{\left(p+1\right)}} $$

Since T₁ changes slowly compared with T₂, then for one time step, ${T}_1^{\left(p+1\right)}\approx {T}_1^{(p)}$. Furthermore, when the coefficients vary with time, they will do so slowly, and the R and C values can be evaluated at the present, not the future time. Therefore, a legitimate explicit prediction of the future temperature of T₂ that is inherently stable numerically (let Δt → ∞) is:

$$ {T}_2^{\left(p+1\right)}=\frac{{\dot{Q}}_2^{(p)}+\frac{C_2^{(p)}{T}_2^{(p)}}{\Delta {t}^{(p)}}+\frac{{T_1}^{(p)}}{{R_{12}}^{(p)}}+\frac{{T_{\infty}}^{(p)}}{{R_{2\infty}}^{(p)}}}{\frac{{C_2}^{(p)}}{\Delta {t}^{(p)}}+\frac{1}{{R_{12}}^{(p)}}+\frac{1}{{R_{2\infty}}^{(p)}}} $$

For this scheme, the time step can be chosen to be a value greater than the Second Law restriction on a full explicit scheme.

Workshop 5.8: 2-node Electric Water Heater:

An electrically resisting element made of nichrome (thermal conductivity k_element = 11.0 W/m/K, density r_element = 8,400 kg/m³, specific heat c_element = 450 J/kg/K) of diameter d = 0.015 m with a total length to be determined is designed to be immersed in 1.0 L of liquid water in a cylindrical container of inner diameter D = 0.15 m with a thin wall (negligible thermal capacity and conductive resistance). The water is initially at 15 °C and electrical power at the rate of Q = 1,000 W is delivered to the heating element.

Conduct a 2-node analysis (element and water). Assume that the heat transfer coefficient between the element and the water, and between the water and the container is h_{element,water} = h_{water,container} = 400 W/m²/K, the convective/radiative coefficient between the outer wall and ambient is h_{container,air} = 15 W/m²/K.

Determine:

• 1. The length of the heating element needed so that the water temperature is 90 °C at the point the water starts to boil (i.e., reaches 100 °C) in the thermal boundary layer between the element and the water and the corresponding time required to boil. Note: if heated beyond this point, most of the heat will go toward boiling the water in that boundary layer, and not to raising the temperature of the bulk of the liquid. That is to say, it is difficult to obtain water hotter than 90 °C for this heating element at this heating rate.

• 2. The time to start to boil and the final water temperature obtained for this length heating element, but with a heating power of 500 W.

• 3. The minimum power that would yield 90 °C water and the corresponding time and the temperature of the element at that time.