Chapter 9 Transport Phenomena

Transcription

1 Chapter 9 Transport Phenomena GOALS After you have mastered the contents of this chapter you will be able to achieve the following goals: Transport Equation Write the quantitative equation for the transport process of a system whose variables are given. Continuity State the continuity equation for a system, and explain the flow properties of a system in terms of that equation. Transport Problems Use algebraic and graphical methods to solve transport problems for one dimensional systems. PREREQUISITES Before beginning this chapter, you should have achieved the goals of Chapter 1, Human Senses, and Chapter 2, Unifying Approaches.

2 Chapter 9 Transport Phenomena OVERVIEW Scientists have long recognized the symmetry and simplicity of nature. As you consider the basic aspects of flow (called Transport Phenomena) please notice the basic uniformity of the expressions utilized for each case. Thus, explaining heat flow, electrical charge flow, water flow, and diffusion are related by a basic model for flow. SUGGESTED STUDY PROCEDURE Before you begin to study this model, be familiar with two of the Chapter Goals: Transport Equation and Transport Process. These two chapter goals are summarized by the equations under Algorithmic Problems in the Chapter Summary. Please note the form of the Transport Equations: flow (9.1 and 9.3) and current density (9.4, 9.15, and 9.16). For a brief explanation of each of these equations and an example of how each might be used, see the following section of this Study Guide. Next, read text sections Carefully consider the Examples and Questions given at the end of the sections. Remember to look in this Study Guide for answers to the questions asked in the text. Now read the Chapter Summary and complete Summary Exercises 1, 2, 3, 6, 7, and 8. Check your answers carefully. Now do Algorithmic Problems 1-4 and complete Exercises and Problems 4, 5, 9, and 11. For additional practice with the concepts presented in this chapter, turn to the Examples section of this Study Guide chapter. Now you should be prepared to attempt the Practice Test on Transport Phenomena provided at the end of this chapter. If you have difficulties with any of the questions, refer to this study procedure for additional assistance. This study procedure is outlined below Chapter Goals Suggested Summary Algorithmic Exercises Text Readings Exercises Problems & Problems Transport 9.1, 9.2, 9.3, 9.4, 1, 2, 3, Equations 9.5, 9.6, 9.7 Transport 9.2, 9.5 6, 7, 8 1, 2, 3, 4 4, 5, 9, 11 Problems Continuity 9.6 4, 5 3

3 ANSWERS TO QUESTIONS FOUND IN THE TEXT SECTION 9.1 Introduction The water flows down a slope. There is a difference in elevation from one place to another. The greater the difference in elevation the faster is the flow of water. The ladle is a different temperature at different locations. This temperature difference may be thought of as the cause of a flow of heat. The cup is less susceptible to the flow of heat than is the silver ladle. We can talk about an electric potential being of different value at different locations in an electric circuit, so we call it a flow of electricity, or electric current. SECTION 9.2 Temperature Differences and Gradients Example - The gradient of the height in the easterly direction is greatest; i.e., most positive, on the west side of the hill. It is least; i.e., most negative, on the east side of the hill. The gradient of the height in the easterly direction is the negative of the gradient of the height on the westerly direction. The gradient of height in the southerly direction is greatest on the north side, least on the south side, and zero on the top of the hill. The water flowing down the hill will go fastest in the steepest place; i.e., where the contour lines are the closest together, which occurs on the southeast portion of the hill. As time goes on the temperatures at various places on the ladle handle will increase until constant values are obtained and the heat flow becomes stable. 1. A negative value of the temperature gradient means that the gradient points from a high temperature region to a low temperature region. 2. The temperature gradient is a larger negative number at 4 cm than at 3 cm. It is a smaller negative number at 5 cm than at 3 cm. 3. The gradient is just the slope of the line tangent to the curve at the point of interest. The relative size of the gradient can be found by looking at the steepness of these tangent lines. 4. The grad x T has its largest, or least negative, value at the distance of 7 cm. Grad x T is smallest, or most negative at 4 cm. It is not zero any place between zero and seven. SECTION 9.4 Heat Flow 5. Heat can escape from the portion of the handle directly into the air. 6. Yes, your hand on the rod will be hotter than a hand in air the same distance from the fire. SECTION 9.5 Water Flow 7. The water flow between 150 and 200 seconds appears to be at a constant rate of about 5 liters/50 sec. or 0.1 liter/second. 8. The fact that the amount of water in the bucket stays the same after 300 seconds can be explained by a condition where the inflow and outflow of water become equal. It can be obtained by (1) turning off the water so both inflow and outflow are zero or (2) filling the bucket then the inflow equals the outflow as the bucket overflows.

4 SECTION 9.7 How to Increase the Flow 9. The units of current density for heat flow are joules per square meter per second. 10. The temperature gradient in SI units is measured in degrees Celsius per meter. 11. The coefficient of thermal conductivity must have the units of heat current divided by temperature gradient, joules/ meter ø C sec. 12. The various values for the temperature gradients as estimated from the graph (See Fig. 9.2) are -3 ø C/cm, -9 ø C/cm, and -7 ø C/cm at the locations of 1 cm, 3 cm, and 5 cm respectively. The current densities can be found by multiplying the temperature gradients by the coefficient of thermal conductivity 420 J/m ø Cs. The current densities are then 1 x 10 5 J/m 2 s, 4 x 10 5 J/m 2 s, and 3 x 10 5 J/m 2 s respectively. SECTION 9.8 Diffusion 13. From Equation 9.16 the diffusion coefficient D must have the same units as the ratio of the mass current density J (moles/m 2 sec) to the concentration gradient Δc/Δx (moles/m 4 ); so D has the units of m 2 /sec. 14. Equation 9.19 has consistent units since the gradient of the concentration is multiplied by the ratio of D 2 /D, it is the same as being multiplied by D as in Equation WORD STATEMENT OF EQUATIONS The rate of heat flow (I) is equal to the change in the quantity of heat (DH) divided by the time required to produce such a change (Δt), I = ΔH/Δt (9.1) The rate of matter flow (I) is equal to the change in the quantity of matter (DQ) divided by the time required to produce such a change, I =ΔQ/Δt (9.3) The current density (J) is the amount of something that flows through a unit area in a unit of time. Hence, if the area is of size A, the time is of duration t, and the amount of something is Q; J = Q/At. (9.4) The change of the mass density of a substance per unit time is equal to the change in current density per unit length, Δp/Δt = ΔJ/l (9.13) For incompressible fluids, the density does not change, so Δp/Δt = 0, and the current density cannot change from one location in the liquid to another one. The thermal energy current through an object Jx is equal to the negative of the coefficient of thermal conductivity of the object times the temperature gradient across the object. J x = -K H ΔT/Δx (9.15) The matter current of diffusive flow (J) is equal to the negative product of the diffusion constant D and the gradient of the mass concentration. J = -D Δc/Δx (9.16)

5 EXAMPLES TRANSPORT PROBLEMS 1. While studying the flow of oxygen in living systems a pathologist quick froze a human blood cell and then used a microprobe to analyze the concentration of oxygen at various locations in the cell at one instant in time. These data are shown below. (a) Where are regions of maximum oxygen concentration gradients? Minimum oxygen concentration gradients? (b) As time goes on, if the cell were not frozen, explain in words what you expect to happen to the oxygen concentration in the blood cell. (c) By using a microprobe around the outside of the warmed cell the scientist measured a maximum oxygen diffusion current to be 2.3µmoles/mm 2 sec. Where do you think that oxygen current is flowing? What is its direction of flow? Estimate a value of the diffusion coefficient for oxygen for this blood cell. What Data Are Given? The oxygen concentrations are given at various locations in a plane. What Data Are Implied? It is assumed that gradient and current calculations are appropriate for the scale of sizes of the various quantities given in this problem. This includes the assumption that superposition is valid for this system. What Physics Principles Are Involved? This problem involves the use of the definitions of gradient and current and their assumed linear relationship as expressed in Equation What Equations Are to be Used? concentration gradient = change in concentration/change in location =Δc/Δx (1) matter current density through an area = change in concentration/(area time)(2) matter current density = J = -D Δc/Δx (3)

6 Algebraic Solutions This is a problem without a closed algebraic solution, you need to determine numerical results from the given data. Numerical Solutions (a) Regions of maximum gradient will occur with the maximum increase in oxygen concentration while the x and y value is increasing. The maximum gradient in the x direction occurs for y = 0.0, between x = 0.4 and 0.6, then grad x c = ( )/( ) = 0.8/0.2 = 4 micromoles/ cm 3 mm = 40 µm/cm 4 = 4 x 10-5 moles/cm 4 = 4 x 10 3 moles/(meter) 4 The maximum gradient in the y - direction occurs when x = 1.0 and between y = 0.2 and 0.4. grad y c = ( )/( ) = 0.9/0.2 5 micromoles/cm3 mm = 5 x 10 3 moles/m 4 The minimum gradient will occur with the maximum decrease in oxygen concentration while the x or y value is increasing. The minimum gradient in the x - direction occurs for y = 0.4, and x is between 1.0 and 1.2 grad x c = ( )/( ) = -1.3/0.2 = -7 micromoles/cm 3 mm = 7 x 10 3 moles/m 4 The minimum gradient in the y - direction occurs for x = 0.8 and between y = 0.4 and 0.6, grad y c = ( )/( ) = -1.2/0.2 = -6 micromoles/cm 3 mm = -6 x 10 3 moles/m 4 (b) If the walls of the cell are impervious to the flow of oxygen, then the total amount of oxygen in the cell must remain a constant. There would be flow of oxygen inside the cell from regions of high concentration; e.g., (0.8, 0.4), to the regions of low concentration around the cell boundary. The final average uniform concentration would be about 0.8mm/cm 3. If the walls of the cell will allow oxygen to flow in and out, then the oxygen concentration in the cell will finally become equal to the concentration of oxygen in the environment. In both cases, the internal gradients of oxygen concentration will become zero.

7 (c) The largest gradient is absolute magnitude is grad x c for y = 0.4 and 1.0 x 1.2, so it seems most likely that the maximum flow of oxygen occurs in response to that gradient, in the positive x - direction. J x = -D Δc/Δx = -D (-7 x 10 3 moles/m 4 ) (9.16) J x = (2.3 x 10-6 moles/mm 2 sec) = (2.3 moles/m 2 sec) = 7 x 10 3 moles D/m 4 so D (3 x 10-3 m 2 )/sec.

8 PRACTICE TEST A metal rod is placed in a flame and the temperature at various locations along the rod is measured at various times The following data were obtained. 1. On graph paper plot these data as different curves so you can calculate the temperature gradient from your graph. Label your axes. 2. What is the temperature gradient at the 40 cm location at a time of 6 minutes? 3. If the Coefficient of thermal conductivity of the rod is 4W/ ø C cm, calculate the rate of flow of heat along the rod. (Cross-sectional area = 1.5 cm 2 ) ANSWERS: o C/cm 3. 3 watts