Ch9.3 Model for a Hot Water Heater

• In this section we derive a differential equation for temperature of water in hot water tank.
• The problem applies principles similar to those in described in the coffee cup problem, except that in this case heat is added to the system via the heating element.

I just went to an emotional wedding.

Even the cake was in tiers.

• We rely on hot water heaters for all of our hot water needs, and often take their operation for granted.
• Hot water heaters are typically hidden within a utility closet or mechanical room.
• They can operate for years without needing maintenance.

• Consider a cylindrical electric hot water heater.
• The heating element supplies heat at a constant rate of 3.6 kW (kJoules/sec).
  
• The hot water heater shuts off during night to save energy.
  
• How long will it take to reach 60\( ^{\circ} \) C (140 \( ^{\circ} \) F) from 15\( ^{\circ} \) C (59 \( ^{\circ} \) F) in the morning?

• Determine temperature of water in hot water heater as a function of time.
• Once established, use model to predict time required for water to be heated to a specified temperature.

• The water in tank is well stirred so that temperature remains homogeneous throughout.
• Heat lost from surfaces according to Newton's law of cooling.
• Specific heat \( c \) and Newton cooling coefficient \( h \) remain constant.

The compartment diagram for heat content of tank is:

The word equation is obtained from compartment diagram and balance law:

\[ \small{ \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, heat \, in \, tank } \end{Bmatrix} = \begin{Bmatrix} \mathrm{rate \, of \, change } \\ \mathrm{of \, heat \, produced} \\ \mathrm{by\, heating \, element} \end{Bmatrix} - \begin{Bmatrix} \mathrm{rate \, of \, change } \\ \mathrm{of \, heat \, lost \, to} \\ \mathrm{surroundings} \end{Bmatrix} } \]

• \( U(t)= \) temperature in Celcius of the water at time t
• \( u_0= \) initial temperature of the water
• \( u_f= \) final temperature
• \( u_s= \) temperature of surroundings
• \( m= \) mass of water in the heater, in kg.
• \( q= \) rate of heat energy supplied, in W = J/sec.
• \( S= \) surface area m² of the tank
• \( Q = \) rate of change of heat content, in J/s.
• \( c= \) specific heat of water, in J/(kg C).
• \( h= \) Newton cooling coefficient, in W/(m² C).

To derive IVP, we start by looking back at word equation.

\[ \small{ \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, heat \, in \, tank } \end{Bmatrix} = \begin{Bmatrix} \mathrm{rate \, of \, change } \\ \mathrm{of \, heat \, produced} \\ \mathrm{by\, heating \, element} \end{Bmatrix} - \begin{Bmatrix} \mathrm{rate \, of \, change } \\ \mathrm{of \, heat \, lost \, to} \\ \mathrm{surroundings} \end{Bmatrix} } \]

Using results from Ch9.2, we obtain

\[ \small{ \begin{aligned} cm \frac{dU}{dt} &= q - hS \left[ U(t) - u_s \right] \\ \\ \frac{dU}{dt} &= \frac{q}{cm}-\frac{hS}{cm}\left[ U(t) - u_s \right], \,\,\, U(0) = u_0 \end{aligned} } \]

Check that units make sense and balance on both sides.

\[ \small{ \begin{aligned} \left[\frac{dU}{dt}\right] & = \frac{C}{sec} \\ \left[\frac{q}{cm}\right] & = \frac{ \frac{J}{sec} } { \left(\frac{J}{kg ~C}\right)kg } = \frac{C}{sec} \\ \left[\frac{hS}{cm}\left(U - u_s \right)\right] & = \frac{ \left(\, \frac{J}{m^2 sec \, C}\right)\left(m^2~\right)\left(\, C \, \right) } { \left(\frac{J}{kg \, C \,}\right) \,kg \,} = \frac{C}{sec} \end{aligned} } \]

Use amount of change on \( \small{[t, t+\Delta t]} \), rather than rate of change:

\[ \small{ \begin{Bmatrix} \mathrm{change \, in} \\ \mathrm{heat \, energy } \end{Bmatrix} = \begin{Bmatrix} \mathrm{amount \, of } \\ \mathrm{heat \, produced \, by } \\ \mathrm{heating \, element} \end{Bmatrix} - \begin{Bmatrix} \mathrm{amount \, of} \\ \mathrm{ heat \, lost \, to } \\ \mathrm{surroundings} \end{Bmatrix} } \]

For this word equation, use:

• Change in heat energy = \( cm \left[ U(t + \Delta t) - U(t) \right] \)
• Amount heat produced = \( q \Delta t \)
• Amount heat lost to surroundings = \( hS \left[ U(t^{*}) - u_s \right]\Delta t \)
• \( U(t^{*}) \) = temperature at \( t^{*} \in [t, t + \Delta t] \)

The ODE can then be derived as shown below.

\[ \small{ \begin{aligned} cm \left[ U(t + \Delta t) - U(t) \right] &= q \Delta t - hS \left[ U(t^*) - u_s \right] \Delta t \\ \\ \lim_{\Delta t \rightarrow 0} \frac{U(t + \Delta t) - U(t)}{\Delta t} &= \frac{q}{cm} - \frac{hS}{cm} \lim_{\Delta t \rightarrow 0} \frac{\left[ U(t^*) - u_s \right]\Delta t}{\Delta t} \\ \\ \frac{dU}{dt} &= \frac{q}{cm} - \frac{hS}{cm}\left[ U(t) - u_s \right] \end{aligned} } \]

The exact solution will be found and explored in Ch10.2.

\[ \small{ \begin{aligned} \frac{dU}{dt} &= \frac{q}{cm}-\frac{hS}{cm}\left[ U(t) - u_s \right], \,\,\, U(0) = u_0 \\ \frac{dU}{dt} + \frac{hS}{cm}U(t) &= \frac{q+hSu_s}{cm} \end{aligned} } \]

The numerical solution is included in the textbook for Ch9.3, using the parameter specifications below:

• \( q = 3600 W = 3600 \frac{J}{sec} \)
• \( c = 4200 \, \frac{J}{kg \,^\circ C} \)
• \( m = 250 \, kg \)
• \( h = 12 \, \frac{W}{m^2 \,^\circ C}= 12 \frac{J}{m^2 sec \,^\circ C } \)
• \( S = 3.06 m^2 \)
• \( u_0 = u_s = 15^\circ C \)

WaterHeater <- function(T) {
#Ch9.3: Perform Rk4 for hot water heater
#T is time length in hours for [0, T]

#Parameters  
N <- 10000       #N = number of time steps
T1 <- T*36000    #Convert hours to seconds
h <- T1/N        #h = step size
UL <- 60         #UL = target temperature

#System Parameters
t0 <- 0      #seconds (SI units)
q <- 3600    #Joules/sec (SI units)
c <- 4200     
h2 <- 12     #h2 because h is used for step size 

#Slope formula from ODE
f <- function(U){q/(c*m)-((h2*S)/(c*m))*(U-us)}  

#Runge-Kutta Loop (Generate temperature vector U)
for(i in 1:N) {
a1 <- h*f(U[i]);          
b1 <- h*f(U[i]+0.5*a1);   
c1 <- h*f(U[i]+0.5*b1);  #c = specific heat
d1 <- h*f(U[i]+c1);        
U[i+1]<-U[i]+(1/6)*(a1+2*b1+2*c1+d1);
t[i+1] <- t[i] + h  } 

#Plot results 

#Convert [0,T1] in seconds to [0,T] in hours
plot(t/36000,U,               
      main = "Hot Water Heater",
      xlab = "t (hours)",           
      ylab = "Temperature (C)",      
      type="l",lwd=2,col="blue", 
      xlim=c(0,T),                
      ylim=c(0,100)  )

#Place additional graph in plot  
lines(t/36000,z, type="l",lwd=2,col="red") 

• The graph of numerical solution shows target temperature reached after about half an hour.
  
• This graph is different than what is in the textbook, where the time is closer to 5 hours.
  
• Waiting half an hour for a shower is preferable to 5 hours!

• Assuming water in tank is well stirred allows us to model temperature \( U(t) \) as a function of time \( t \) only.
• Otherwise temperature would be function of time and position \( U(x,y,z,t) \), leading to a PDE model.
  
• Further, we would need to account for heat transfer by conduction and convection between different points in tank.

• The book points out that the model no longer applies after the temperature reaches boiling point (\( 100^\circ \) Celcius).
  
• This is because model doesn't account for latent heat, which is an additional form of energy associated with a change in state without a change in temperature; see [1,2,3].

[1] Mathematical Modeling with Case Studies, Barnes and Fulford, CRC Press, 2015

[2] Latent heat, https://en.wikipedia.org/wiki/Latent_heat, retrieved on 9/25/2022.

[3] Latent heat figure, https://www.teachoo.com/12526/3426/Latent-Heat-of-Vaporization-and-Fusion/category/Extra-Questions/, retrieved 9/25/2022.