Pomodoro
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Difficulty
Name the dominant mode in each situation:
Heat flowing through a stationary brick wall
Air removing heat from a hot engine block
Sunlight warming a roof
Solution
The dominant modes are:
Through a stationary wall: conduction
From the engine block to moving air: convection
From the sun to the roof: radiation
A plane wall has $k = 0.8\ \text{W/(m K)}$, area $A = 3\ \text{m}^2$, thickness $L = 0.05\ \text{m}$, and surface temperatures $T_1 = 60^\circ\text{C}$ and $T_2 = 20^\circ\text{C}$.
What is the heat transfer rate through the wall?
Solution
Use Fourier's law for a plane wall:
Substitute the values:
A flat layer has thickness $L = 0.10\ \text{m}$, conductivity $k = 2\ \text{W/(m K)}$, and area $A = 4\ \text{m}^2$.
What is its conduction resistance?
Solution
For a plane wall,
So
Equivalently,
A surface has area $A_s = 1.5\ \text{m}^2$, convection coefficient $h = 12\ \text{W/(m}^2\text{ K)}$, surface temperature $T_s = 85^\circ\text{C}$, and surrounding fluid temperature $T_\infty = 25^\circ\text{C}$.
What is the convective heat transfer rate?
Solution
Use Newton's law of cooling:
Substitute:
A surface has $h = 15\ \text{W/(m}^2\text{ K)}$ and area $A_s = 0.4\ \text{m}^2$.
What is the convection resistance?
Solution
Use
So
That is approximately
A hot surface has area $A = 2.5\ \text{m}^2$ and linearized radiation coefficient $h_r = 6\ \text{W/(m}^2\text{ K)}$.
If the surface is $30\ \text{K}$ warmer than the surroundings, what is the radiative heat transfer rate?
Solution
Using the linearized radiation model,
Substitute:
A solid has volumetric heat generation rate $\dot{q}''' = 2.5\times 10^5\ \text{W/m}^3$ and volume $V = 0.004\ \text{m}^3$.
What is the total heat generation rate?
Solution
Use
So
Three thermal resistances are connected in series:
What is the equivalent resistance?
Solution
For resistances in series,
So
Two surfaces have areas $A_1 = 2\ \text{m}^2$ and $A_2 = 3\ \text{m}^2$.
If the view factor from surface 1 to surface 2 is $F_{1\to 2} = 0.30$, what is $F_{2\to 1}$?
Solution
Use reciprocity:
Solve for $F_{2\to 1}$:
A solid has $h = 20\ \text{W/(m}^2\text{ K)}$, characteristic length $L_c = 0.01\ \text{m}$, and conductivity $k = 200\ \text{W/(m K)}$.
What is the Biot number, and is the lumped-capacitance assumption reasonable?
Solution
Use
So
Since $Bi \ll 0.1$, the lumped-capacitance approximation is reasonable.
Difficulty
A wall separates indoor air at $100^\circ\text{C}$ from outdoor air at $20^\circ\text{C}$.
The wall has area $A = 2\ \text{m}^2$, inner convection coefficient $h_1 = 10\ \text{W/(m}^2\text{ K)}$, outer convection coefficient $h_2 = 20\ \text{W/(m}^2\text{ K)}$, thickness $L = 0.10\ \text{m}$, and conductivity $k = 0.5\ \text{W/(m K)}$.
What is the heat transfer rate?
Solution
Build the resistance network:
Compute each term:
Now use
so
A pipe has inner radius $r_1 = 0.02\ \text{m}$ and outer radius $r_2 = 0.04\ \text{m}$.
The insulation has conductivity $k = 0.04\ \text{W/(m K)}$ and length $L = 2\ \text{m}$.
What is the cylindrical conduction resistance of the insulation?
Solution
For radial conduction through a cylinder,
Substitute:
A hot surface at $80^\circ\text{C}$ is connected to a cold surface at $20^\circ\text{C}$ through three resistances in series:
What is the heat transfer rate, and what is the temperature immediately after the contact resistance?
Solution
First find the total resistance:
Then the heat rate is
The temperature drop across the first resistance is
The temperature drop across the contact resistance is
So the temperature immediately after the contact resistance is
A body has mass $m = 2\ \text{kg}$ and specific heat $c_p = 500\ \text{J/(kg K)}$.
It cools by convection with $h = 10\ \text{W/(m}^2\text{ K)}$ and area $A_s = 0.5\ \text{m}^2$.
The initial temperature is $100^\circ\text{C}$, the surrounding temperature is $20^\circ\text{C}$, and the time is $100\ \text{s}$.
What is the body temperature after 100 seconds?
Solution
For lumped cooling,
where
Compute the time constant:
Now evaluate the temperature:
A heat exchanger transfers heat from a hot stream to a cold stream.
The hot stream has $\dot{m}_h = 0.5\ \text{kg/s}$, $c_{p,h} = 4000\ \text{J/(kg K)}$, and cools from $90^\circ\text{C}$ to $60^\circ\text{C}$.
The cold stream has $\dot{m}_c = 1.0\ \text{kg/s}$ and $c_{p,c} = 3000\ \text{J/(kg K)}$.
If losses to the surroundings are negligible, what is the cold-stream outlet temperature?
Solution
Use the energy balance:
So
Set this equal to the cold-side heat gain:
Solve:
A counterflow heat exchanger has:
Hot stream: $\dot{m}_h = 1.0\ \text{kg/s}$, $c_{p,h} = 4200\ \text{J/(kg K)}$, $100^\circ\text{C} \to 75^\circ\text{C}$
Cold stream: $\dot{m}_c = 0.7\ \text{kg/s}$, $c_{p,c} = 3000\ \text{J/(kg K)}$, $20^\circ\text{C} \to 70^\circ\text{C}$
Overall heat transfer coefficient $U = 350\ \text{W/(m}^2\text{ K)}$
What area is required?
Solution
First compute the heat transfer rate from the hot stream:
Check the cold side:
Now find the terminal temperature differences:
The log-mean temperature difference is
Use
so
A correlation gives $Nu = 25$ for a flow over a surface with $k = 0.6\ \text{W/(m K)}$ and characteristic length $L_c = 0.02\ \text{m}$.
What convection coefficient does this imply?
Solution
Use the definition of the Nusselt number:
Solve for $h$:
Substitute:
A fluid has $\rho = 1.2\ \text{kg/m}^3$, $V = 10\ \text{m/s}$, $L = 0.05\ \text{m}$, $\mu = 1.8\times 10^{-5}\ \text{Pa s}$, $c_p = 1005\ \text{J/(kg K)}$, and $k = 0.026\ \text{W/(m K)}$.
Compute $Re$ and $Pr$. What does the Reynolds number suggest about the flow?
Solution
First the Reynolds number:
Now the Prandtl number:
A large Reynolds number suggests inertia effects are strong and turbulence is more likely.
Difficulty
Indoor air is at $22^\circ\text{C}$ and outdoor air is at $-8^\circ\text{C}$.
A wall has area $A = 10\ \text{m}^2$, inside convection coefficient $h_1 = 8\ \text{W/(m}^2\text{ K)}$, outside convection coefficient $h_2 = 25\ \text{W/(m}^2\text{ K)}$, drywall thickness $L_1 = 0.013\ \text{m}$ with $k_1 = 0.17\ \text{W/(m K)}$, and insulation thickness $L_2 = 0.09\ \text{m}$ with $k_2 = 0.04\ \text{W/(m K)}$.
What is the heat loss rate?
Solution
Write the total resistance:
Compute each term:
Then
A pipe has outer radius $r_0 = 0.015\ \text{m}$, insulation conductivity $k = 0.06\ \text{W/(m K)}$, and surrounding convection coefficient $h = 3\ \text{W/(m}^2\text{ K)}$.
If a thin layer of insulation is added, will the heat loss initially increase or decrease?
Solution
Compute the critical radius for a cylinder:
So
Because the current radius $0.015\ \text{m}$ is smaller than the critical radius, adding a thin layer of insulation initially increases heat loss.
A surface has area $A = 1.2\ \text{m}^2$, surface temperature $T_s = 120^\circ\text{C}$, ambient temperature $T_\infty = 20^\circ\text{C}$, convection coefficient $h = 10\ \text{W/(m}^2\text{ K)}$, and linearized radiation coefficient $h_r = 5\ \text{W/(m}^2\text{ K)}$.
If the surrounding radiation temperature is the same as the ambient temperature, what is the total heat loss rate?
Solution
Treat convection and linearized radiation as parallel heat-transfer paths:
Substitute:
A body has a time constant of $\tau = 60\ \text{s}$, initial temperature $T_i = 20^\circ\text{C}$, and surrounding temperature $T_\infty = 200^\circ\text{C}$.
How long does it take to reach $110^\circ\text{C}$?
Solution
Use the lumped transient relation:
Substitute the target temperature:
The left side is
So
Take the natural log:
Two plates in a bolted joint are connected by conduction resistances $R_1 = 0.20\ \text{K/W}$ and $R_2 = 0.20\ \text{K/W}$, with a contact resistance of $R_{contact} = 0.10\ \text{K/W}$ between them.
The hot side is at $80^\circ\text{C}$ and the cold side is at $20^\circ\text{C}$.
What is the heat transfer rate, and what temperature drop occurs across the contact resistance?
Solution
Add the resistances:
Then
The drop across the contact resistance is
Difficulty
A thermal circuit has a hot-side resistance of $0.15\ \text{K/W}$, then two parallel paths with resistances $0.60\ \text{K/W}$ and $0.30\ \text{K/W}$, and finally a cold-side resistance of $0.05\ \text{K/W}$.
If the hot side is at $60^\circ\text{C}$ and the cold side is at $10^\circ\text{C}$, what is the total heat transfer rate?
Solution
First reduce the parallel part:
So
Now add the series resistances:
Finally,
A small metal part has $h = 18\ \text{W/(m}^2\text{ K)}$, $L_c = 0.005\ \text{m}$, $k = 120\ \text{W/(m K)}$, mass $m = 0.8\ \text{kg}$, specific heat $c_p = 500\ \text{J/(kg K)}$, and area $A_s = 0.4\ \text{m}^2$.
Its initial temperature is $100^\circ\text{C}$, the surroundings are at $25^\circ\text{C}$, and the time of interest is $100\ \text{s}$.
Is the lumped-capacitance model reasonable, and what is the temperature after 100 seconds?
Solution
Check the Biot number:
Since $Bi \ll 0.1$, the lumped model is reasonable.
Now find the time constant:
Use the lumped cooling formula:
So
A counterflow heat exchanger has the following data:
Hot stream: $\dot{m}_h = 1.0\ \text{kg/s}$, $c_{p,h} = 4200\ \text{J/(kg K)}$, $100^\circ\text{C} \to 75^\circ\text{C}$
Cold stream: $\dot{m}_c = 0.7\ \text{kg/s}$, $c_{p,c} = 3000\ \text{J/(kg K)}$, $20^\circ\text{C} \to 70^\circ\text{C}$
Overall heat transfer coefficient $U = 350\ \text{W/(m}^2\text{ K)}$
What area is required?
Solution
The heat transfer rate is
The terminal temperature differences are
So the log-mean temperature difference is
Now solve for area:
A small object in a vacuum cools only by radiation to a large surrounding enclosure. The object is small enough that the lumped-capacitance model is valid.
What governing equation should you write for $T(t)$?
Solution
Because the object is lumped, its temperature is uniform in space and depends only on time. The energy balance is
If you want a linearized network model, you can write
with
So the workflow is: identify radiation as the only mode, confirm lumped behavior, and then write the transient energy balance.
