May 22

Convection Study Guide

Heat Transfer Type: Convection

The second type of heat transfer is convecvtion. Convection is a heat transfer mode that uses the movement of a fluid (liquid or gas) across a boundry surface. This movement can either be forced or natural. The general equation for this is Netwon’s Law of Convection which is shown in equation 1

q = hA\Delta T

Equation 1

Generally you will know the area and the two different temperatures. The only variable you will not usually know is the film coefficient (h). Solving for the film transfer coefficient can be done depending on how the convection is taking place.

Forced Convection

Forced convection is when the fluid is moved by an external source. Like when you blow on your soup to cool it down because the soup is to hot. Sometimes this flow will be inside a pipe. Other times the flow will be around an object. In order to determine the film transfer coefficient you will need to decide on which equation to use, which is dependent on how the fluid is flowing.

Turbulent Flow In Pipes

When the fluid is flowing inside a pipe, we can use the Nusselt Number (equation 2) to solve for the film coefficient. The Nusselt Number (Nu) is the ratio of convective heat transfer to  conductive heat transfer. When solving for fluid inside a pipe use the diameter of the pipe instead of the characteristic length (L)

Nu = \frac {hL}{k}

Equation 2

If the temperatures across the fluid is not too great the Dittus-Boelter equation (equation 3) can be used to solve for the film coefficient. You will also need to use equations 4 and 5 to find some missing information, the Reynolds Number (Re) and the Prandtl Number (Pr).

Nu = \frac {hL}{k} = 0.0225(Re^{0.8})(Pr^n)

Equation 3

Re = \frac {Dv}{\upsilon }

Equation 4

Pr = \frac {c_p\mu}{k}

Equation 5

The last variable that is not defined in equation 3 is n. This variable depends on if the heat flow is out of the pipe (n = 0.3) or if the flow of heat is into the pipe (n = 0.4)

In order to use equation 3 there is still some rules that need to be followed:
The Prandtl Number (Pr) must be between 0.7 and 16,700
The Reynolds Number (Re) must be greater than 10,000
And you must have more than 10 times the diameter of the pipe in length (Length/Diameter > 10)

Reynolds Number

The Reynolds Number (Re) is a dimensionless number that is the ratio of inertia forces to viscous forces. Building upon equation 4, we can rewrite the Reynolds Number equation in 3 different ways as shown in equation 6

Re = \frac {\rho v D_h}{\mu} = \frac {v D_h}{\nu } = \frac {Q D_h}{\nu A}

Equation 6

Where v is the velocity, Q is the volumetric flow rate, A is the area, ρ is the density,  Dh is the hydraulic diameter ,  μ is the dynamic viscosity, and v is the kinematic viscosity.

Hydraulic diameter

For a circular pipe the diameter equals the hydraulic diameter, for a channel the hydraulic diameter is 4 times the cross section area that is wetted divided by the perimeter. This is shown in equation 7

D_h = D_{(circular)} = \frac {4A}{P}

Equation 7

Prandtl Number

The Prandtl Number (Pr) is the ratio of kinematic viscosity to thermal diffusivity. Equation 8 shows how the Prandtl Number is calculated

Pr = \frac {c_p\mu}{k} = \frac{\nu}{\alpha}

Equation 8

Where Cp is the specific heat, k is the thermal conductivity, and α is the thermal diffusivity.

May 20

Finned Heat Radiators Study Guide

Finned Heat Radiators

Sometimes an object requires help to be cooled. A good example is the cpu chip in your computer. In order to help with cooling process a heat sink is used. The heat sink uses fins to create a larger surface area to transfer heat. The fins on the heat sink act like a finned heat radiators.

Fin Efficiency

In order to find out if a fin will work we need to calculate the  fin efficiency. The fin efficiency is the ratio of actual heat loss to the ideal heat loss. This is assuming the heat loss across the tip (or face) of the fin is uniform. Equations 1 and 2 show this. It should be noted that in equation 2 the hyperbolic function requires the use of radians.

\eta = \frac {q}{hA\Delta T}

Equation 1

\eta = \frac {tanh(Lm)}{Lm}

Equation 2

m = \sqrt{\frac {hP}{kA}}

Equation 3

In equation 3 the variable P is for the fins perimeter (or circumference). Figure 1 gives a great visual depiction of a finite rod.

Finite Rod
Figure 1: An example of a finned radiator

Infinite Rod

If the rod you are modeling is considered to be infinite then equation 4 and equation 5 can be used.

T_x - T_{ambient}=(T_b - T_{ambient})e^{-mx}

Equation 4

q = \sqrt{hPkA(T_b-T_{ambient})}

Equation 5

Where T_b is the temperature at the base of the fin and x is the distance from the base of the fin to the point being examined

Finite Rod with Adiabatic Tip

If the rod being examined is adiabatic and has a finite length then equations 6 can be used to find the temperature and equation 7 can be used to find the heat transferred from the rod. Where L is the length of the rod.

T_x - T_{ambient}=(T_b - T_{ambient})\frac{cosh(ML-Mx)}{cosh(mL}

Equation 6

q = \sqrt{hPkA(T_b-T_{ambient})tanh(mL)}

Equation 7

Rectangular Fins

A rectangular fin is treated the same as a finite rod. The only difference is the perimeter is calculated as P = 2(base +thickness) as shown in equation 8 and equation 9

T_x - T_{ambient}=(T_b - T_{ambient})\frac{cosh(ML-Mx)}{cosh(mL}

Equation 8

q = \sqrt{h(2(b+t))kA(T_b-T_{ambient})tanh(mL)}

Equation 9

May 15

Transient Heat Flow

Transient Heat Flow

Sometimes you will need to calculate the amount of time an item will take to either cool down or heat up. Since the temperature of the item is no longer constant there is a transient heat flow. What this basically means is the rate of heat transferred (q) will change as  the difference of the two temperatures becomes greater or lesser.
To solve these type of problems we will look at three different ways, Newton’s Law of Cooling, Lumped System Analysis, and using Heisler Charts.

Newton’s Law of Cooling

If the heat transfer coefficient (U) is independent of the temperature then Newton’s Law of Cooling can be used. Newton’s Law of Cooling states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings
To solve a problem using Newton’s Law of Cooling you will need to know at least one set of initial conditions to use with equation 1

.
T_{time} = T_{ambient} + (T_{initial\ condition}-T_{ambient})e^{-rt}
Equation 1

 Where t is the time it takes to reach the initial condition (Ttime) and r is a constant that is dependent to the problem. To find the value for r, equation 1 can be rewritten into equation 2.

r = -\frac {1}{t}*ln\frac {T_{time}-T_{ambient}}{T_{initial\ condition}-T_{ambient}}

Equation 2

Once you have solved for r, you can then use equation 2 to solve for the new time/temperature by replacing t with r.

Lumped System Analysis

If the temperature throughout the object is homogeneous during the cooling/heating process a lumped system analysis may be used. A lumped system analysis is when the object is treated as if the temperature on the edge of the object was the same as the center of the object. In order for this to happen the Biot number must be below 0.10.

The Biot number is a dimensionless number that is the ratio of resistance to internal heat flow (h) to resistance to external heat flow (k). The formula for the Biot Number (Bi) is shown in equation 3 where L is the characteristic length which is shown in equation 4.

Bi = \frac {hL}{k}

Equation 3

L = \frac {Volume}{Surface\ Area}

Equation 4

The next number that needs to be calculated is the Fourier number. The Fourier number (Fo) is a non-dimensional time parameter shown in equation 5. Where t is the time, Cp is the specific heat, ρ is density, α is emissivity.

Fo = \frac {\alpha T}{L^2} = \frac {kt}{\rho c_p L^2}

Equation 5

Now that all the variables have been defined the full equation for a lumped sum analysis is shown in equation 6

T_{time}-T_{ambient} = (T_{initial}-T_{ambient})e^{-BiFo}

Equation 6

If you want to solve for the time it will take to cool/heat an object, equation 6 can be rewritten into equation 7

time = ln(\frac {(T_{time}-T_{ambient})}{(T_{initial}-T_{ambient})} )*\frac {\rho c_pV}{hA}

Equation 7

Heisler Chart

If the Biot number is greater than 0.1 then you will need to use a Heisler Chart, which are graphical representations of the equations that have been solved. A good place to find a copy of the charts is here courtesy of Wikipedia. The Heisler Chart is used to solve for the Fourier Number by using \frac {T_t-T-{ambient}}{T_i-T-{ambient}} and the inverse of the Biot  Number (Bi^{-1}). Once the Fourier Number is known, you can then solve for time (t)

May 8

Conduction Study Guide (Part 2)

Conduction (Cont)

In the first installment of this study guide we looked at Fourier’s law and how heat was transferred inside a solid object by conduction. Thermal conductivity and resistance was also discussed as well as heat transfer through a solid composite slab. In part 2 we will look at the equations that are used when the surface is curved.

Conduction Through a Cylindrical System

When looking at a the heat transfer through a flat object the area is easy to calculate. That area is the width times the height of the object. When the object is curved the radius on the inside and the radius on the outside of the curved surface needs to be taken into account. This is shown in equation 1.8 by dividing the natural logarithm of the outer surface by the inner surface. Figure 1.3 has been provided to give a visual representation.  Though it should be noted that sometimes the wall of the pipe is so thin it is not taken into account.q = 2\pi KL\frac{\Delta T}{ln\frac{r_2}{r_1}}

Equation 1.8

Cylindrical Heat Flow
Figure 3. Pipe with Insulation

When dealing with a curved surface that  has more than one layer equation 9 should be used. Equation 1.9 takes into account the film coefficient as well thermal conductivity of the different layers. q = 2\pi L\frac{\Delta T}{\frac {1}{hr_1} + \frac {1}{k}ln\frac{r_2}{r_1} + \frac {1}{k}ln\frac{r_3}{r_2} + \frac {1}{hr_3}}

Equation 1.9

Conduction Through a Spherical System

The last geometric system we need to look at is a spherical system. At first you would think the equations would be worse than the cylindrical system. But looking at equation 1.10 for a sphere made of just one material shows us different. q = 4 \pi k \frac {\Delta T}{\frac {1}{r_1} - \frac {1}{r_2}}

Equation 1.10

Critical Radius

When adding insulation to a flat surface the insulation will decrease the heat transfer. Now when you add insulation to a curved surface you are reducing the conduction resistance, but at the same time you are increasing the outer surface area which affects the film coefficient (convection resistance) The heat transfer may increase or decrease depending on which effect dominates.  In order to calculate the amount of insulation that is  needed to prevent the adverse effect, we need to calculate the critical radius. The critical radius of insulation is the ratio of the thermal conductivity (k) divided by the film coefficient (h) This can be seen in equation 1.11 which is for cylindrical object and equation 1.12 which is for spherical objects. r_{critical} = \frac {k_{insulation}} {h_{min}}

Equation 1.11

r_{critical} = 2\frac {k_{insulation}} {h_{min}}

Equation 1.12

 

May 3

Conduction Study Guide (Part 1)

Heat Transfer Mode: Conduction

Conduction is one of the three modes of heat transfer. Conduction occurs by molecular vibration in a solid material. An example would be heating one end of a pipe up and after a period of time the other end of the pipe becomes hot.

Fourier’s law

The steady state equation for heat transfer through a flat slab by conduction is called Fourier’s law. (Equation 1.1)
\dot{Q}=-ka\frac{\mathrm{d} T}{\mathrm{d} x} = -ka\frac{T_{2}-T_{1}}{L}

Equation 1.1

Fourier’s law states that the rate of heat transfer (Q) is equal to the area (A) times the thermal conductivity (k) multiplied by the change in temperature (T) divided by the thickness (L). It should be also noted that the rate of heat transfer is negative; this is to indicate that the heat flow is in the opposite direction of the thermal gradient or heat always goes from a warm area to a cold area.

Thermal Conductivity

Thermal conductivity is how well a substance transfers heat. For instance air has a low thermal conductivity which is why when you open an oven you do not get burned when you place your hand inside. On the other hand copper has a high thermal conductivity which is why it is used in cookware. Some typical thermal conductivities are given in table 1.1.

Table 1.1
Typical Thermal Conductivities
Material k (BTU ft/hr ft² R)
Aluminum 137
Copper 231.7
Gold 183.2
316 Stainless Steel 7.8
Titanium 12.7

Conduction Through a Composite Slab

When there is only one material and the transfer of heat is steady then the value for area (A) and thermal conductivity (k) do not change. Only the temperature will change as the thickness changes or vice versa. When more than one material is present the thermal resistance (R) needs to be adjusted for each layer.

Thermal Resistance

Thermal resistance is the measure of how well a material resist the flow of heat. This is the R-value you hear people talk about on insulation. Equation 1.2 shows thermal resistance, and equation 1.3 shows Fourier’s equation modified for a composite wall as show in figure 1.1.

R = \sum \frac{L_{n}}{A_nk_n}

Equation 1.2

Q=\frac{A\Delta T}{\sum \frac{L}{k}}

Equation 1.3

Composite wall slab
Figure 1.1.  Heat transfer through a composite slab

Overall Coefficient of Heat Transfer

Sometimes you will see Equation 3 written using the overall coefficient of heat transfer (U) as shown in equation 1.5. The overall heat transfer equation is defined in equation 1.4.

U=\frac {1}{R}

Equation 1.4

Q=UA\Delta T

Equation 1.5

Film Coefficient

On the inside and the outside of the  material there is also a film coefficient (h) that must be taken into account. The film coefficient (also known as the convection heat transfer coefficient) is a measure of the heat transfer between a solid surface and fluid. The film coefficient can sometimes be relatively low such as 1.65 for still air, or it can be high as in the case of evaporating water (3000). Table 1.2 has a listing of some common film coefficients.

Table 1.2
Typical Film Coefficients
Film Material h (BTU/hr ft² F)
Outside Air 4
Inside Horizontal Surface  1.1
Inside Vertical Surface 1.47
 Wall Air Space 1.15
 Ceiling air Space  1

When you calculate the overall heat coefficient of heat transfer, R (or U) value, you will need to  use equation 1.6. Also equation 1.7 has been provided to show you what the combined equation for heat transfer through a composite slab would look like.

R=\sum \frac{L_n}{k_n}+\sum \frac{1}{h_n}

Equation 1.6

Q=\frac{A \Delta T}{\sum \frac{L_n}{k_n}+\sum \frac{1}{h_n}}

Equation 1.7