Boiling Heat Transfer

”The formation of steam bubbles along a heat transfer surface has a significant effect on the overall heat transfer rate.”

Boiling

In an industrial facility, convective heat transfer is used to remove heat from a heat transfer surface. The liquid used for cooling is usually in a compressed state, (that is, a subcooled fluid) at pressures higher than the normal saturation pressure for the given temperature. Under certainconditions, some type of boiling (usually nucleate boiling) can take place.

More than one type of boiling can take place within an industrial facility. A discussion of the boiling processes, specifically local and bulk boiling, will help the student understand these processes and provide a clearer picture of why bulk boiling (specifically film boiling) is to be avoided.

Nucleate Boiling

The most common type of local boiling encountered in industrial facilities is nucleate boiling. In nucleate boiling, steam bubbles form at the heat transfer surface and then break away and are carried into the main stream of the fluid. Such movement enhances heat transfer because the heat generated at the surface is carried directly into the fluid stream. Once in the main fluid stream, the bubbles collapse because the bulk temperature of the fluid is not as high as the heat transfer surface temperature where the bubbles were created. This heat transfer process is sometimes desirable because the energy created at the heat transfer surface is quickly and efficiently “carried” away.

Bulk Boiling

As system temperature increases or system pressure drops, the bulk fluid can reach saturation conditions. At this point, the bubbles entering the coolant channel will not collapse. The bubbles will tend to join together and form bigger steam bubbles. This phenomenon is referred to as bulk boiling. Bulk boiling can provide adequate heat transfer provided that the steam bubbles are carried away from the heat transfer surface and the surface is continually wetted with liquid water. When this cannot occur film boiling results.

Film Boiling

When the pressure of a system drops or the flow decreases, the bubbles cannot escape as quickly from the heat transfer surface. Likewise, if the temperature of the heat transfer surface is increased, more bubbles are created. As the temperature continues to increase, more bubbles are formed than can be efficiently carried away. The bubbles grow and group together, covering small areas of the heat transfer surface with a film of steam. This is known as ”partial film boiling”. Since steam has a lower convective heat transfer coefficient than water, the steam patches on the heat transfer surface act to insulate the surface making heat transfer more difficult. As the area of the heat transfer surface covered with steam increases, the temperature of the surface increases dramatically, while the heat flux from the surface decreases. This unstable situation continues until the affected surface is covered by a stable blanket of steam, preventingcontact between the heat transfer surface and the liquid in the center of the flow channel. The condition after the stable steam blanket has formed is referred to as ”film boiling”.

The process of going from nucleate boiling to film boiling is graphically represented in Figure 1. The figure illustrates the effect of boiling on the relationship between the heat flux and the temperature difference between the heat transfer surface and the fluid passing it.

Figure 1 Boiling Heat Transfer Curve
Figure 1 Boiling Heat Transfer Curve

Four regions are represented in Figure 1. The first and second regions show that as heat flux increases, the temperature difference (surface to fluid) does not change very much. Better heat transfer occurs during nucleate boiling than during natural convection. As the heat flux increases, the bubbles become numerous enough that partial film boiling (part of the surface being blanketed with bubbles) occurs. This region is characterized by an increase in temperature difference and a decrease in heat flux. The increase in temperature difference thus causes total film boiling, in which steam completely blankets the heat transfer surface.

Departure from Nucleate Boiling and Critical Heat Flux

In practice, if the heat flux is increased, the transition from nucleate boiling to film boiling occurs suddenly, and the temperature difference increases rapidly, as shown by the dashed line in the figure. The point of transition from nucleate boiling to film boiling is called the point of departure from nucleate boiling, commonly written as DNB. The heat flux associated with DNB is commonly called the critical heat flux (CHF). In many applications, CHF is an important parameter. For example, in a boiler, if the critical heat flux is exceeded and DNB occurs at any location, the temperature difference required to transfer the heat being produced increases greatly.

The amount of heat transfer by convection can only be determined after the local heat transfer coefficient is determined. Such determination must be based on available experimental data. After experimental data has been correlated by dimensional analysis, it is a general practice to write an equation for the curve that has been drawn through the data and to compare experimental results with those obtained by analytical means. In the application of any empirical equation for forced convection to practical problems, it is important for the student to bear in mind that the predicted values of heat transfer coefficient are not exact. The values of heat transfer coefficients used by students may differ considerably from one student to another, depending on what source “book” the student has used to obtain the information. In turbulent and laminar flow, the accuracy of a heat transfer coefficient predicted from any available equation or graph may be no better than 30%.

Conduction Heat Transfer

Conduction heat transfer is the transfer of thermal energy by interactions between adjacent atoms and molecules of a solid. Conduction Conduction involves the transfer of heat by the interaction between adjacent molecules of a material. Heat transfer by conduction is dependent upon the driving “force” of temperature difference and the resistance to heat transfer. The resistance to heat transfer is dependent upon the nature and dimensions of the heat transfer medium. All heat transfer problems involve the temperature difference, the geometry, and the physical properties of the object being studied.

In conduction heat transfer problems, the object being studied is usually a solid. Convection problems involve a fluid medium. Radiation heat transfer problems involve either solid or fluid surfaces, separated by a gas, vapor, or vacuum. There are several ways to correlate the geometry, physical properties, and temperature difference of an object with the rate of heat transfer through the object. In conduction heat transfer, the most common means of correlation is through Fouriers Law of Conduction. The law, in its equation form, is used most often in its rectangular or cylindrical form (pipes and cylinders), both of which are presented below.


The use of Equations 2-4 and 2-5 in determining the amount of heat transferred by conduction is demonstrated in the following examples.

Conduction-Rectangular Coordinates

Example:

    1000 Btu/hr is conducted through a section of insulating material shown in Figure 2 that measures 1 ft2 in cross-sectional area. The thickness is 1 in. and the thermal conductivity is 0.12 Btu/hr-ft-° F. Compute the temperature difference across the material.
Figure 2 Conduction Through a Slab
Figure 2 Conduction Through a Slab

Example:

    A concrete floor with a conductivity of 0.8 Btu/hr-ft-° F measures 30 ft by 40 ft with a thickness of 4 inches. The floor has a surface temperature of 70° F and the temperature beneath it is 60° F. What is the heat flux and the heat transfer rate through the floor?

Solution:

Using Equations 2-1 and 2-4:

Using Equation 2-3:

Equivalent Resistance Method

It is possible to compare heat transfer to current flow in electrical circuits. The heat transfer rate may be considered as a current flow and the combination of thermal conductivity, thickness of material, and area as a resistance to this flow. The temperature difference is the potential ordriving function for the heat flow, resulting in the Fourier equation being written in a form similar to Ohms Law of Electrical Circuit Theory. If the thermal resistance term Δx/k is writtenas a resistance term where the resistance is the reciprocal of the thermal conductivity divided by the thickness of the material, the result is the conduction equation being analogous to electrical systems or networks. The electrical analogy may be used to solve complex problems involving both series and parallel thermal resistances. The student is referred to Figure 3, showing the equivalent resistance circuit. A typical conduction problem in its analogous electrical form is given in the following example, where the “electrical” Fourier equation may be written as follows.

Figure 3 Equivalent Resistance
Figure 3 Equivalent Resistance

Electrical Analogy

Example:

A composite protective wall is formed of a 1 in. copper plate, a 1/8 in. layer of asbestos, and a 2 in. layer of fiberglass. The thermal conductivities of the materials in units of Btu/hr-ft-oF are as follows: kCu = 240, kasb = 0.048, and kfib = 0.022. The overall temperature difference across the wall is 500XF. Calculate the thermal resistance of each layer of the wall and the heat transfer rate per unit area (heat flux) through the composite structure.


Conduction-Cylindrical Coordinates

Heat transfer across a rectangular solid is the most direct application of Fouriers law. Heat transfer across a pipe or heat exchanger tube wall is more complicated to evaluate. Across a cylindrical wall, the heat transfer surface area is continually increasing or decreasing. Figure 4 is a cross-sectional view of a pipe constructed of a homogeneous material.

Figure 4 Cross-sectional Surface Area of a Cylindrical Pipe
Figure 4 Cross-sectional Surface Area of a Cylindrical Pipe

The surface area (A) for transferring heat through the pipe (neglecting the pipe ends) is directly proportional to the radius (r) of the pipe and the length (L) of the pipe.

A = 2πrL

As the radius increases from the inner wall to the outer wall, the heat transfer area increases.

The development of an equation evaluating heat transfer through an object with cylindrical geometry begins with Fouriers law Equation 2-5.

From the discussion above, it is seen that no simple expression for area is accurate. Neither the area of the inner surface nor the area of the outer surface alone can be used in the equation. For a problem involving cylindrical geometry, it is necessary to define a log mean cross-sectional area (Alm).

Substituting the expression 2πrL for area in Equation 2-7 allows the log mean area to be calculated from the inner and outer radius without first calculating the inner and outer area.

This expression for log mean area can be inserted into Equation 2-5, allowing us to calculate the heat transfer rate for cylindrical geometries.

Example:

A stainless steel pipe with a length of 35 ft has an inner diameter of 0.92 ft and an outer diameter of 1.08 ft. The temperature of the inner surface of the pipe is 122° F and the temperature of the outer surface is 118° F. The thermal conductivity of the stainless steel is 108 Btu/hr-ft-° F.

Calculate the heat transfer rate through the pipe.

Calculate the heat flux at the outer surface of the pipe.


Example:

A 10 ft length of pipe with an inner radius of 1 in and an outer radius of 1.25 in has an outer surface temperature of 250° F. The heat transfer rate is 30,000 Btu/hr. Find the interior surface temperature. Assume k = 25 Btu/hr-ft-° F.

The evaluation of heat transfer through a cylindrical wall can be extended to include a composite body composed of several concentric, cylindrical layers, as shown in Figure 5.

Figure 5 Composite Cylindrical Layers
Figure 5 Composite Cylindrical Layers

Example:

A thick-walled coolant pipe (ks = 12.5 Btu/hr-ft-° F) with 10 in. inside diameter (ID) and 12 in. outside diameter (OD) is covered with a 3 in. layer of asbestos insulation (ka = 0.14 Btu/hr-ft-° F) as shown in Figure 6. If the inside wall temperature of the pipe is maintained at 550° F, calculate the heat loss per foot of length. The outside temperature is 100° F.

Figure 6 Pipe Insulation Problem
Figure 6 Pipe Insulation Problem