Heat transfer part, short Flashcards
What are the three main modes og heat transfer and how can you distinguish them from each other? Write down the governing equations for each.
Radiation, heat conduction, and heat convection.
Conduction:
Heat conductions takes place within a substance. The particles vibrate more the warmer they are, and when they touch particles that are less warm and vibrate less, they transfer those vibrations which transfers the heat. For gases, there are more space between the particles, so the heat transfer rate is lower. Fourier’s law of conduction:
(Q_cond ) ̇=kA (T_1-T_2)/Δx [W]
Convection:
Heat convection occurs between a solid surface and a fluid, which is in motion. The faster the movement, the more heat transfer. Newton’s law of cooling:
Q ̇_conv=hA(T_1-T_2 ) [W]
Radiation:
Radiation comes from the energy in electromagnetic waves. It is emitted by everything above 0 K. It is the only mode of heat transfer that does not require a medium.
Q ̇_rad=εσA_s (T_s^4-T_surr^4 ) [W]
σ=5,6710^(-8) W/(m^2K^4 )
Explain the concept of thermal resistance.
Like resistance in electric current flow.
Happens for all three modes: conduction resistance, convection resistance, and radiation resistance. It is a body’s ability to withstand heat transfer. The higher the thermal resistance, the lower the ability to transfer heat, or thermal conductivity.
R_cond=L/kA [K/W]
R_conv=1/(hA_s ) [K/W]
R_rad=1/(εσA_s (T_s^2+T_surr^2)(T_s+T_surr)) [K/W]
For multiple resistances, it is also the same as for electric current flow:
R_(tot,series)=∑_i▒R_i
R_(tot,parallel)=∑_i▒1/R_i
For steady state, 1D, heat transfer across a multilayered wall, if the heat transfer rate Q is known, explain how you would determine the temperature drop across each layer.
Figure of wall with three different sections.
Since it is through a wall, it is conduction. Using this formula:
Q ̇=ΔT/R_cond
To calculate the resistance, we have to know the thermal conductivity k of each of the materials in the wall, as well as the lengths L and areas A:
R_cond=L/kA [K/W]
Then, if the heat transfer rate Q is known, the temperature drop through each layer can be calculated.
You want to calculate the temperature as a function of time of a hot steel ball in a cold fluid. In what medium is the lumped system analysis more likely to be applicable: in water or in air? Explain.
Under some conditions, a body will behave like a “lump” meaning that the temperature within the body is close to uniform. The body transfer heat from its surface at the same rate as from the inside to the outside of itself.
The lumped system analysis is applicable when Biot number is smaller than or equal to 0.1. The Biot number is a dimensionless factor which is the ratio between convection at the surface of the body, and conduction within the body:
Bi=hΔT/(kΔT/L_c )=(Convection at the surface of the body)/(Conduction within the body)
Since water has a higher heat transfer coefficient h, it will give a bigger convection. We want the convection as small as possible to get a smaller Biot number, which means that air is a better medium for a lumped system analysis, since its heat transfer coefficient is smaller.
Consider laminar natural convection from a hot vertical plate. Will the heat flux be higher at the top or the bottom of the plate? Explain,
Heat flux is q, which is the same as the heat transfer rate, but without the area included.
The heat flux will be higher at the bottom of the vertical plate, because the boundary layer here is the thinnest. The boundary layer becomes thicker the higher up the plate we move. This is because the convection will move the air upwards along the plate due to buoyancy, and create a natural laminar flow, where the boundary layer increases along the flow. The boundary layer lowers the heat transfer ability. At the bottom of the plate, cooler air, which is heavier, will be sucked in towards the plate because of the vacuum that the buoyancy creates.
Figure of heat flow.
Explain the concepts of forced and natural convection. How do they differ from each other?
Figure of natural, forced and mixed convection.
Natural convection happens due to buoyancy forces. Warmer air is less dense than colder air, so it will rise as it becomes warmer from touching the hot body. This creates a vacuum, which draws in colder air below.
Forced convection comes from an external force, such as a fan, pump, or wind. The heat transfer rate here is higher than with natural convection.
There can also be mixed convection where both modes are included. Then it is important to remember that they can not be added together to find the total convection. For mixed convection, the buoyancy forces are negligible if:
(Gr_(L_c ))/(Re_(L_c)^2 )≪1
(forced convection dominates)
Grashof number plays the same role in natural convection as Re number does in forced convection.
What is the physical significance of the Nusselt number? How is it defined?
Nusselt number is a dimensionless factor used in forced convection, external flow. It represents the fluid’s ability to transfer heat by convection and conduction. It is defined as the heat transfer rate for convection over conduction:
Q ̇_conv/Q ̇_cond =hL/k_solid =Nu
If Nu number = 1, no increase in heat transfer by using convection over conduction.
If Nu number > 1, higher increase in heat transfer by using convection.
If Nu number < 1, decrease in heat transfer by using convection. The fluid is better at conduction than convection.
Adding more insulation to the wall of a house always results in a lower heat transfer rate across the wall. Does this also apply when adding more insulation to a cylindrical pipe (e.g., hot water pipe or electrical cable)? Explain.
Figure of critical radius.
No, it does not always apply. There is a critical radius of insulation for an insulated pipe exposed to convection.
For a cylinder, the critical radius of insulation is:
r_(cr,cylinder)=k_(insulating material)/h_surroundings
You should always add so much insulation so that you get below the point of the bare heat transfer rate. It increases in the beginning because the surface area increases faster than the insulation, which will give a bigger heat transfer rate. Otherwise, there will still be a bigger heat transfer rate, than without any insulation.
For a sphere, it is:
r_(cr,sphere)=2k/h
How does the local heat transfer coefficient vary along the flow over a flat plate?
Figure of h_x over plate (laminar, transition, turbulent flow).
The local heat transfer coefficient h_x will generally speaking fall along the flow over a flat plate. However, if the flow changes from laminar to turbulent, the heat transfer coefficient will increase a lot in the transitional region. It will start decreasing again as soon as it becomes a turbulent flow.
What is the difference between the fin effectiveness and the fin efficiency?
Figure of A_b when removing fin.
The fin effectiveness is how good it is to have a fin, compared to not having a fin. It is a ratio between the two. Effectiveness should be above 1. If it is below, it will have an insulating effect. If it is equal to 1, it might as well not be there, it has no effect. If it is above 1, the fin is effective.
ε_fin=Q ̇fin/Q ̇(no fin) =(Actual heat transfer rate)/(Heat transfer rate from A_b )
where A_b is the surface area that the fin covers
Figure of heat transfer along a fin.
The fin efficiency is how good the temperature distribution is along the fin. It is a ratio between the actual distribution and the ideal distribution, if the thermal conductivity was very high, which means that it will not lose any heat trough the material. The efficiency will really always be below 1.
η_fin=Q ̇fin/Q ̇(fin,max) =(Actual heat transfer rate)/(Heat transfer rate if entire fin is T_b )
Consider fluid flow in a tube whose surface temperature remains constant. What is the appropriate temperature difference for use in Newton’s law of cooling with an average heat transfer coefficient? What does this specific temperature difference represent?
If the surface of the tube remains a constant temperature, then the fluid flow will increase its temperature along the tube. The temperature will increase at a slower and slower pace, exponentially. The temperature difference, that should be used to calculate the heat transfer rate Q is the log mean temperature:
Q ̇=hA_s ΔT_lm
ΔT_lm=(ΔT_e-ΔT_i)/ln((ΔT_e)/(ΔT_i ))
We need to know the inlet temperature and the desired exit temperature.
Figure of “exponential” fluid temperature.
What is the physical significance of the Grashof number?
Figure of mixed convection.
For mixed convection, the buoyancy forces are negligible if:
(Gr_(L_c ))/(Re_(L_c)^2 )≪1
(forced convection dominates)
Gr_(L_c )=(Buoyancy forces)/(Viscous forces)=(gβ(T_s-T_∞ ) L_c^3)/v^2
If the Grashof number is greater than 1, it means that the buoyancy forces are greater than the viscous forces of the fluid. Grashof number plays the same role in natural convection as Re number does in forced convection.
What does the effective thermal conductivity of an enclosure represent? How is it related to the Nusselt number?
The effective thermal conductivity k_eff is calculated as:
k_eff=kNu
If The Nu number is 1, then the effective thermal conductivity is just equal to the thermal conductivity of the fluid in the enclosure, and there will be no natural convection in the enclosure. The effectivity represents the enclosure’s ability to transfer heat from one side of the enclosure to the other (hot to cold).
Figure of convection in enclosure.
Can the outlet temperature on the cold side become larger than the outlet temperature on the hot side of a parallel flow heat exchanger? What about a counterflow heat exchanger? Explain.
No, the outlet temperature cannot become larger on the cold side, than on the hot side of a parallel flow heat exchanger. Because the two flows have the same direction, they start with the maximal temperature difference. Along the flow, this difference will become smaller as heat transfer happens between the two pipes. However, they can only exchange heat until they have the same temperature, and at that point, no more heat transfer will be happening.
Yes, it can happen in a counter flow heat exchanger. Here, the two flows have opposite directions. The hot side’s hottest flow meets the cold side’s hottest flow. Since they are not the same temperature, heat transfer occurs. Then, as the hot flow moves along, it will continue to be cooled down by the cold side. But the cold side will continue getting colder, since we are moving towards its inlet. Therefore, heat transfer can continue to occur, and if it happens for long enough, the cold outlet might be warmer than the hot outlet.
Figure of parallel and counter flow.
What is fouling in a heat exchanger and how does it affect heat transfer and pressure drop?
Fouling is build-up of unwanted material from scales, eroded pipe, chemical reactions, and minerals in the fluid. Mostly comes when the flow through the heat exchanger is slow. It increases the heat transfer resistance of the pipe as it acts like an insulation, that does not add to the outside diameter of the pipe.
The inside diameter of the pipe is decreased due to fouling. This means that there is a smaller cross-section area for the flow, and the resistance is increased over the surface. This will all make for an increased pressure drop across the heat exchanger, which will reduce the flow rate or require more pumping power.
R_tot=(conv.fluid to fouling)+(cond.in internal fouling)+(cond.in pipe)+(cond.in outside fouling)+(cond.to outside)
Figure of fouling and equation.
