22

CHE 333 TRANSPORT LABORATORY

Experiment No. 1

GASEOUS DIFFUSION

We have an experiment in the lab that can be used to measure gaseous diffusion coefficients. As you run the experiment you will find that it is simple to obtain and to analyze the data.

Introduction

We will be determining the gaseous diffusivities of ethyl ether and acetone (or a similar volatile material) in air. The volatile liquid compound will be contained in a capillary tube (ask Jim Rounds for the capillary tubes and for the technique of getting the liquid into the tube). The capillary tube will be contained in a test tube immersed in a constant temperature, water bath. A stream of air, at the same water bath temperature, will pass over the end of the capillary tube in an attempt to reduce the concentration of volatile material, at that point to zero. The experimental procedure is to measure the rate of drop of the liquid level in the capillary with time.

Experimental Setup

z = 0

z = z1 at t = 0

z = zt at t

pure liquid A

PA1

PA2gas B(air)

For liquids of high volatility, A. Stefan (1879) devised a convenient means of measuring the diffusivity of their vapor through a stagnant gas. If the volatile substance A (e.g. ethyl ether or ethanol) is placed in the lower part of a vertical capillary, then liquid A will evaporate and, by the mechanism of diffusion, travel to the end of the capillary. Maintaining the mouth of the capillary at a given composition automatically establishes the concentration gradient in the capillary, and the falling rate of the meniscus in the capillary provides the rate of transport. The capillary is placed in an envelope through

23

which air is passed. At the meniscus the gaseous phase composition is specified by the vapor pressure of liquid A, the diffusing constituent. At the mouth of the capillary, the gaseous phase is essentially air. The gradient in the capillary is thus obtained by circulating sufficient air to reduce the substance A concentration at the mouth to a negligible quantity. The air rate should be low, constant, and not turbulent. The falling rate of the meniscus can be observed remotely with a cathetometer. The cathetometer is like a horizontal telescope with cross hairs. It can be moved up and down on a graduated, vertical rod to measure elevation. The rod must be vertical and the cathetometer must be horizontal (use the leveling screws in the base to center the bubble in the circle). It is essential that the temperature of the envelope remain constant and that the envelope not be disturbed during observations; i.e. vibrations will give erroneous data. In the situation just described, the diffusion process is unsteady-state. Under pseudo-steady-state conditions, the diffusivity of vapor A in gas B (air) is given by

DAB = )( 21

,,

AAA

lmBLA

PPtPMRTP−

ρ

−2

21

2 zzt

where PB,lm = [(P - PA1) - (P - PA2)]/ln [(P - PA1)/(P - PA2)] PA1 = vapor pressure of liquid A at temperature T. PA2 = partial pressure of vapor A at the mouth of the capillary. R = gas law constant. t = time during which the meniscus fall from z1 to zt. z1 = distance from the mouth of the capillary to the meniscus at t = 0. zt = distance from the mouth of the capillary to the meniscus at t. P = ambient atmospheric pressure ρA,L = density of liquid A at T.

Procedure

1. Read the barometer. Measure the inside diameters of the capillary and the envelope. 2. After filling the capillary with di-ethyl ether or acetone to within 1 cm of the top,

gently lower the capillary into the envelope and submerge the envelope into the thermostat in a vertical position.

3. Initially maintain the thermostat at a convenient (room) temperature. Set an air flow rate through the envelope at about 10 cm reading. After allowing sufficient time for the envelope and its contents to come to the thermostat temperature, mark the position Yo of the mouth of the capillary and the position Y1 of the meniscus with the cathetometer as an arbitrary zero time (z = z1 = Yo - Y1, at t = 0).

4. The air temperature within the envelope must be recorded. This should be the temperature where the diffusivity is measured.

5. Then follow the fall of the meniscus as a function of time. Remember the precautions about use of the cathetometer. Take three or more readings (z = zt = Yo - Yt at t = t).

24

Plot you data as you go to evaluate its validity. From the equation for DAB , you see that a graph of (zt2 - z12) versus t will be a straight line through the origin. So your second reading will show whether your experimental data conform to a straight line or not. If not, you must take additional readings in the hopes that at least three of the later data points fall on a straight line. When the run is complete, raise the bath temperature to the next level.

6. Repeat the run (steps 2, 3, 4, 5 above) at a higher temperature of about 80-82oF. As the temperature rises the runs can be made more quickly. Make sure that bubbles of vapor do not form in the liquid phase while a run is being made. This indicates that you are exceeding the boiling point of the ether or acetone.

7. Make another run (steps 2, 3, 4, 5, above) at the same temperature of 80-82oF but use a different air flow rate. If the slope of the line representing the lab data is not the same as for (6) above, stop and investigate; especially the flow rate of the air stream.

Note: Experimental technique is very important in this somewhat crude but educational experiment. The following are critical for obtaining good data:

a. Spend some time in the determining the relative position of the test tube in the water bath and the relative position of the cathetometer, so that you can sight clearly to make good measurements. You should practice making some height readings while getting ready for your first run. Note that there is a vernier scale on the cathetometer so that you can read height to the nearest 0.1 mm.

b. The test tube should be as fully immersed in the water bath as possible (in the interest of having the contents as close to the water bath temperature as possible).

c. A thermometer should be placed in the test tube as close to the evaporating liquid as possible.

d. The water bath stirrer should run as fast as possible (for a constant bath temperature) but slow enough so that vibrations do not interfere with the cathetometer height readings.

Note: The volumetric air flow rate through the flow meter can be estimated from the equation

V = 123.6 H*P/T where V = volumetric flow rate in ft3/hr at T and P H = flow meter reading in cm T = ambient air temperature (oR) P = ambient air pressure (psia)

Analysis 1. Determine the diffusivity from the slope of the least square line of ( zt2 - z12 ) vs. t. Note: You should use the liquid density at the temperature of the system. The

specific volume Vs at any temperature can be estimated from the Rackett equation if one experimental volume VsR is available at a reference temperature TR.

Vs = VsR (ZRA)α

25

where α = (1 - Tr)2/7 - (1 - TrR)2/7, Tr = reduced temperature = T/Tcritical ZRA = 0.29056 - 0.08775ω, ω = acentric factor 2. Determine Reynolds number for the flow inside the envelope. 3. Estimate the mole fraction of the volatile species at the mouth of the capillary.

26

Report

1. Present your experimental results in the form of a graph of diffusivity vs. temperature. 2. Compare the experimental values of DAB with those appearing in the literature. Note: The diffusivity obtained from literature should be corrected for the

temperature and pressure of your system by the following equation.

DAB(T1, P1) = DAB(T2, P2)

1

2

75.1

2

1

PP

TT

where: T and P are absolute temperature and pressure respectively. 3. Estimate the value of the diffusion coefficient, for your system, using the empirical

equation of Fuller, Schettler, and Giddins (Ref. 1, p. 99 or Ref. 3, p. 435). Note: Your report should present three values of diffusity at your system

conditions: experimental value, literature value, and empirical value. 4. Estimate the accuracy of your results. Answer the following questions: (Support your answers with numerical values if

possible) - What should be the optimum air flow rate? - During the course of the experiment, when do you think the concentration of the

volatile species will be highest at the capillary mouth. - What should be the optimum height for the capillary?

References 1. Seader J. D., and Henley E. J., Separation Process Principles, Wiley, (1998) 2. Sherwood, T. F., R. L. Pigford and C. R. Wilke, Mass Transfer, McGraw-Hill, (1975) 3. Welty, J. R., Wicks, C. E., Wilson, N. E., and Rorrer, G. Fundamentals of Momentum,

Heat and Mass Transfer, John Wiley and Son, (2001), p. 479.

27

CHE 333 TRANSPORT LABORATORY

Experiment No. 2

HEAT TRANSFER ALONG A CYLINDRICAL FIN

In this heat transport lab you will study and perform calculations for extended surface heat transfer. As part of the experiment you will be using automated data collection instruments and thermocouples. We will be interested in the performance of an aluminum pin fin available in our laboratory. You should determine the temperature distribution for both free and forced convection flows and compare the experimental measurements with the predicted values.

Introduction

Consider the area A on the surface shown in Figure 1 where heat is being transfer from the surface at a fixed temperature Ts to the surrounding fluid at a temperature T∞ with a heat transfer coefficient h. The heat transfer rate may be increased by increasing the convection coefficient h, reducing the fluid temperature T∞, or adding materials to the area A.

Surface

A

AsPlate

L

A

As

Figure 1. Use of extended surface or fin to enhance heat transfer.

Look on the plane side-view of the surface and the surface with fin. The heat transfer rate without the fin from area A to the surrounding fluid is qc = hA(Ts − T∞) With the fin attached to the area A, the heat transfer to the surrounding fluid must first be transferred by conduction from area A to the fin

qf = − kA0=∂

∂

xxT = ∫ ∞−

sA

sdATxTh0

))((

where dAs = Pdx and P = perimeter of the fin.

28

For the extended surface to enhance the heat transfer rate, the ratio of heat transfer with and without the fin must be greater than one

εf ≡ c

f

qq

= )(0

∞

=

−∂∂

−

TThxTk

s

x = )(

))((0

∞

∞

−

−∫TTh

dATxTh

s

A

ss

(1)

εf is called the fin effectiveness. For the fin to be cost effective, the fin effectiveness should be greater than 2. The temperature profile along the fin must be determined before the fin effectiveness can be calculated. Consider the cylindrical extended surface with diameter D shown in Figure 2. To simplify the analysis, we will assume one-dimensional heat transfer in the x direction, steady state, no heat generation, no radiation, constant heat transfer coefficient, and constant physical properties.

x

Tb

A

dx

T(x)

L

T

Figure 2. A cylindrical fin with convective end.

An energy balance will be applied to a differential control volume, ∆xA, shown in Figure 2. Since temperature is dependent on x, a differential distance along x must be chosen. The surface area of the control volume is ∆As = ∆xP = ∆xπD From the energy balance applied to the control volume ∆xA qx – (qx+∆x + ∆qc) = 0 Divide the equation by ∆x and take the limit as ∆x→ 0

0

limit→∆x

∆∆

−∆− ∆+

xq

xqq cxxx = 0

0

limit→∆x

∆− ∆+

xqq xxx –

0limit→∆x

∆∆

xqc = 0

29

– dxdqx –

dxdqc = 0

dqc = hdAs(T(x) − T∞) The energy equation becomes

– dxdqx – h

dxdAs (T(x) − T∞) = 0

From the Fourier's law

qx = – kAdxdT

where A is the cross-sectional area normal to the x-direction. The energy equation becomes

dxd

dxdTkA – h

dxdAs (T(x) − T∞) = 0

since As = P⋅x, dxdAs = P

For constant k and A, the energy equation becomes a second order ordinary differential equation (ODE) with constant coefficients.

2

2

dxTd –

kAhP (T(x) − T∞) = 0 (2)

The above equation is a non-homogeneous ODE which can be made homogeneous by introducing a new variable θ = T(x) − T∞

2

2

dxd θ –

kAhP θ = 0 (3)

Let m2 = kAhP , the solution to the homogenous ODE can be written as

θ = B1sinh(mx) + B2cosh(mx) (4) The constants B1 and B2 can be evaluated using the following boundary conditions at x = 0, T = Tb ⇒ θ = θb (5a)

30

at x = L, dxdTk− = h(T - T∞) ⇒ −

dxdθ = hθ (5b)

to obtain the temperature distribution along the pin fin

bθθ =

mLmkhmL

xLmmkhxLm

sinhcosh

)(sinh)(cosh

+

−+− (6)

and the fin heat transfer rate

qf = mL

mkhmL

mLmkhmL

Msinhcosh

coshsinh

+

+ (7)

where θ = T - T∞ θb = Tb - T∞

m = kD

h4 M = θb(hPkA)0.5

P = perimeter = πD A = 4

2Dπ

The heat transfer coefficient for a long, horizontal cylinder can be estimated from appropriate empirical correlations for free and forced convection flow1. Fin performance is assessed by two factors: Fin Effectiveness, εf, and the Fin Efficiency, ηf. Fin effectiveness is defined as the ratio of the fin heat transfer rate to the heat transfer rate that would exist without the fin as given by equation (1) earlier.

εf = bc

f

hAqθ

(1)

Fin efficiency is defined as the ratio of the actual amount of heat transferred to the amount of heat that would be transferred if the entire fin was at the base temperature.

ηf = bf

f

hAqθ

(8)

31

For this experiment you will determine the temperature distribution, the amount of energy transferred to the air, the fin effectiveness, and the fin efficiency for both forced and free convection.

Procedure: For Free Convection : 1. Turn on the Varriac. Check to make sure that the heater connected to the fin is

connected with the varriac. 2. Wait until the system reach steady state and record the temperatures along the fin. 3. Read and record the ambient air temperature and pressure. Record the humidity

using the wet bulb thermometer. For Forced Convection : 4. Turn the air blower on. Adjust the varriac so that the based temperature of the fin has

approximately the same value as in free convection. 5. Measure the air velocity by placing the wind velocity meter near the fin. It is

suggested that at least 5 readings over different x position along the fin be taken to obtain an average value.

6. Wait until the system reach steady state and record the temperatures along the fin. Repeat steps 1-6 for at least two more settings. Turn everything off and clean up. Note: a. Check the identification of the thermocouples. Some were mislabeled in the

past. b. What is the criterion for a steady state temperature? Report : Your report should: 1] Derive equations 6 & 7. 2] Determine the temperature distribution for both free and forced convection. 3] Compare the predicted values with the experimental values. Note: the predicted and

experimental values have the same base temperature. 4] Determine the fin effectiveness and fin efficiency for both free and forced

convection. References: 1) Incropera and De Witt, Fundamentals of Heat and Mass Transfer, Wiley 2002.

32

2) Chapman, A. J., Heat Transfer, McMillan Publishing Co., 1985. 3) Any CRC Handbook of Physics and Chemistry.

33

CHE 333 TRANSPORT LABORATORY

Experiment No. 3

GAS SEPARATION MEMBRANE EXPERIMENT

http://www.medal.com/en/membranes/nitrogen/index.asp

Figure 1. Hollow-fiber module used for air separation.

Gas separation with polymer membrane is becoming an important component of separation technology1. Examples of common used membrane separations are enrichment of nitrogen from air, hydrogen separation in ammonia plants an refineries, removal of carbon dioxide from natural gas, and removal of volatile organic compounds from mixtures with light gases. Gas separation membranes are often packaged in hollow fiber modules depicted in Figure 1. As air flows under pressure into the module through the bores of the hollow fiber, some of the air gases permeate through the wall of the fibers into the shell of the hollow fiber. The gas in the shell side of the fibers leaves the module as the permeate stream. Since oxygen, water, and carbon dioxide are more permeable than nitrogen and argon, the gas in the fiber bore is enriched as it moves from the feed to the residue end of the module.

tNO2 NN2

x, P

y, p 1-y

1-x

P>pP PO2 N2>

Figure 2. Schematic of a membrane with thickness t used to separate O2 from N2.

The flux NO2 of oxygen across the membrane shown in Figure 2 is given as

NO2 = tO2P (xP − yp) (1)

where PO2 is the permeability of the membrane to oxygen, x is the mole fraction of oxygen on the upstream, or high pressure P, side of the membrane, and y is the mole fraction of oxygen on

34

the downstream, or low pressure p, side of the membrane. The ratio of permeability to membrane thickness is called the permeance Q’O2 of the membrane to oxygen. The permeance can be viewed as a mass transfer coefficient that connects the flux with the driving force for transport, which is the partial pressure difference between the upstream and downstream sides of the membrane. We now need to consider the fact that as the feed gas travels through the hollow fibers, its composition changes as selective permeation depletes the more permeable components from the feed gas mixture. Figure 3 illustrates the ideal countercurrent flow pattern for the binary mixture of oxygen and nitrogen moving through the fiber module.

Permeatey , n = nP P Fθ

Feedx , nF F

xn xn+d(xn)dA P

pd(yn) yi

Retentatex , nR R

Figure 3. Ideal countercurrent flow pattern through the separator.

The total mole and O2 species balances around the separator are2

nF = nR + nP (2) xFnF = xRnR + yPnP (3) where nF, nR, and nP are the molar flow rates of the feed, retentate, and permeate streams, respectively, and xF, xR, and xP are the feed, retentate, and permeate O2 mole fraction, respectively. The molar flux of oxygen through a differential area dA in the membrane is given by equation (1) or by

NO2 = dAxnd )( =

tO2P (xP − yp) = Q’O2(xP − yp)

Therefore d(xn) = Q’O2(xP − yp)dA = d(yn) (4) The above equation is just the O2 species balance around the differential volume element in the membrane. The reduction in the O2 molar flow rate d(xn) of the retentate stream provides the same O2 molar flow rate d(yn) through the membrane. P and p are the average retentate and permeate side pressures, respectively. Similar species balance for nitrogen around the differential volume element in the membrane yields d[(1−x)n] = Q’N2 [(1−x)P − (1−y)p)]dA (5)

35

Dividing equation (4) by equation (5), we obtain

])1[(

)(nxd

xnd−

= 2

2

''

N

O

QQ

pyPxypxP

)1()1( −−−− (6)

The ratio ])1[(

)(nxd

xnd−

is just the molar flow rate of oxygen over that of nitrogen in the

permeate stream, therefore it is equal to the ratio of the mole fraction of oxygen over that

of nitrogen y

y−1

as shown schematically in Figure 4. Let α* = 2

2

''

N

O

QQ , equation (6)

becomes

y

y−1

= α*

pyPxypxP

)1()1( −−−− (7)

Permeate

y , n = nP P Fθ

Feedx , nF F

xn xn+d(xn)

d(yn)

yi

Retentatex , nR R

P

p

Figure 4. Molar flow rate ratio is equal to mole fraction ratio.

The separation factor α* is assumed to be constant. The permeate composition at the capped end of the hollow fibers is obtained from equation (7) by replacing y with yi and x with xR.

i

i

yy−1

= α*

pyPxpyPx

iR

iR

)1()1( −−−− (7)

When the change in feed mole fraction of oxygen is less than 50%, the driving force for diffusion across the membrane, ∆ = xP − yp, is assumed to be a linear function of the change in the molar flow on the feed side of the membrane

d(xn) = FR

FR xnxn∆−∆

− )()( d∆ (8)

From the species balance around the separator xFnF = xRnR + yPnP (xn)R − (xn)F = − (yn)P (9)

36

Combine equations. (8) and (9) with equation (4) d(xn) = Q’O2(xP − yp)dA, we obtain

− yPnP FR

d∆−∆∆ = Q’O2∆dA

Separate the variables and integrate

− yPnP ∫∆

∆ ∆∆R

F

d = Q’O2 (∆R − ∆F) ∫mA

dA0

yPnP

∆∆

F

Rln = Q’O2 (∆R − ∆F)Am (What happens to the minus sign?)

yPnP = Q’O2∆lm Am (10) where the log mean average ∆lm is defined as

∆lm =

F

R

FR

∆∆∆−∆

ln = (xP − yp)lm =

F

R

FR

ypxPypxP

ypxPypxP

)()(ln

)()(

−−

−−− (11)

Equation (10) expresses the molar flow rate yPnP of oxygen as a function of the permeance Q’O2 or mass transfer coefficient, area of membrane Am for mass transfer, and an average driving force ∆lm across the membrane. Similarly, the molar flow rate of nitrogen in the permeate stream can be found (1−yP)nP = Q’N2 [(1−x)P − (1−y)p)]lmAm (12) The oxygen species balance, xFnF = xR( nF − nP) + yPnP, can be written in dimensionless form using the definition of the cut θ = nP/nF, xF = xR( 1 − θ) + yPθ (13) Similarly, equations (7), (10), and (12) in dimensionless forms are

i

i

yy−1

= α*

)1()1( iR

iR

yrxyrx−−−

− (14)

yPnPF

R

nn

2'1

NQ =

F

R

nn

2'1

NQQ’O2∆lm Am

37

yP pAQn

mN

R

2' F

P

nn =

−

F

P

nn1

2

2

''

N

O

QQ (xr − y)lm

yPKRθ = (1 − θ)α*(xr − y)lm (15) (1−yP) KRθ = (1 − θ)[(1−x)r − (1−y))]lm (16)

where r = P/p and KR = pAQ

n

mN

R

2'

The algebraic model equations (13-16) represent a system with four equations in eight variables: xF, xR, yP, r, yi, θ, α*, and KR. The system can be solved with measured values of xF, xR, yP, and r, leaving yi, θ, α*, and KR as unknowns in the solution. For better convergence of these nonlinear algebraic equations, the log-mean average can be replaced by the Chen’ approximation

F

R

FR

∆∆∆−∆

ln =

3/1

2)(

∆+∆

∆∆ FRFR

The algebraic equations can be solved by Newton’s method presented next. Newton’s Method for Systems of Nonlinear Algebraic Equations Consider two equations f1(x1, x2) and f2(x1, x2) for which the roots are desired. Let 0

1p , 02p be the guessed values for the roots. f1(x1, x2) and f2(x1, x2) can be expanded about point

( 01p , 0

2p ) to obtain

f1(x1, x2) = f1( 01p , 0

2p ) + 1

1

xf∂∂ (x1 − 0

1p ) + 2

1

xf

∂∂ (x2 − 0

2p ) = 0

f2(x1, x2) = f2( 01p , 0

2p ) + 1

2

xf∂∂ (x1 − 0

1p ) + 2

2

xf∂∂ (x2 − 0

2p ) = 0

Let 0

1y = (x1 − 01p ) and 0

1y = (x2 − 02p ), the above set can be written in the matrix form

∂∂

∂∂

∂∂

∂∂

2

2

1

2

2

1

1

1

xf

xf

xf

xf

02

01

yy

= −

),(),(

02

011

02

011

ppfppf

38

or J(p(0))y(0) = − F(p(0)) In general, the superscript (0) can be replaced by (k−1) J(p(k-1))y(k-1) = − F(p(k-1)) J(p(k-1)) is the Jacobian matrix of the system. The new guessed values x at iteration k are given by x = p(k) = p(k-1) + y(k-1) Example Use Newton’s method with the initial guess x = [0.1 0.1 −0.1] to obtain the solutions to the following equations2

f1(x1, x2, x3) = 3x1 − cos(x2 x3) − 21 = 0

f2(x1, x2, x3) = 21x − 81(x2 + 0.1)2 + sin x3 + 1.06 = 0

f2(x1, x2, x3) = 21xxe− + 20x3 + 3

310 −π = 0

Solution The following two formulas can be applied to obtain the roots J(p(k-1))y(k-1) = − F(p(k-1)) J(p(k-1)) is the Jacobian matrix of the system.

J(p(k-1)) =

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

3

3

2

3

1

3

3

2

2

2

1

2

3

1

2

1

1

1

xf

xf

xf

xf

xf

xf

xf

xf

xf

F(p(k-1)) is the column vector of the given functions

2 Numerical Analysis by Burden and Faires

39

F(p(k-1)) =

),,(),,(),,(

3213

3212

3211

xxxfxxxfxxxf

The new guessed values x at iteration k are given by x = p(k) = p(k-1) + y(k-1) Table 1 lists the Matlab program to evaluate the roots from the given initial guesses. Table 1 Matlab program for the example ------------- % Newton Method % f1='3*x(1)-cos(x(2)*x(3))-.5'; f2='x(1)*x(1)-81*(x(2)+.1)^2+sin(x(3))+1.06'; f3= 'exp(-x(1)*x(2))+20*x(3)+10*pi/3-1' ; % Initial guess % x=[0.1 0.1 -0.1]; for i=1:5 f=[eval(f1) eval(f2) eval(f3)]; Jt=[3 2*x(1) -x(2)*exp(-x(1)*x(2)) x(3)*sin(x(2)*x(3)) -162*(x(2)+.1) -x(1)*exp(-x(1)*x(2)) x(2)*sin(x(2)*x(3)) cos(x(3)) 20]'; % dx=Jt\f'; x=x-dx'; fprintf('x = ');disp(x) end

Experimental Procedure

Compressed air at about 110 psig is supplied to the membrane module through an air regulator. The supplied air pressure can be controlled by turning the knob on top of the regulator. The oxygen concentration is measured by a portable oxygen analyzer model GPR-30. You can calibrate the oxygen analyzer by turn it on while in the ambient air and set the oxygen concentration to 21.0 %. Adjust the inlet pressure of the membrane module to 30 psig. Read the flow rate on the permeate side of the membrane and set the same flow rate for the retentate. Record the oxygen concentrations on both sides of the membrane when the system reaches steady state. The permeate pressure p is assumed to be the ambient pressure and the retentate pressure P is the average of the feed and retentate pressures as measured by the pressure gages.

40

Measure the oxygen concentrations and the retentate pressures again at the retentate flow rates of twice and four times the permeate flow rate. Repeat the procedure at 40, 50, 60, 70, and 80 psig.

Analysis The Matlab program to solve equations (13-16) can be download from T.K. Nguyen website (http://www.csupomona.edu/~tknguyen/che333/home.htm). Make sure you understand the program. The system can be solved with measured values of xF, xR, yP, and r, leaving yi, θ, α*, and KR as unknowns in the solution. (Appendix A gives instructions to solve equations 13-16 using Excel Solver) xF = xR( 1 − θ) + yPθ (13)

i

i

yy−1

= α*

)1()1( iR

iR

yrxyrx−−−

− (14)

yPKRθ = (1 − θ)α*(xr − y)lm (15) (1−yP) KRθ = (1 − θ)[(1−x)r − (1−y))]lm (16) 1. Plot the experimental and calculated separation factor α* as a function of r (= P/p) and

discuss the results. 2. Plot the experimental and calculated cut θ (= nP/nF) as a function of r and discuss the

results. 3. Present one iteration of the Newton’s method at 30 psig and θ = 0.5 using the guessed

values yi = 0.2, θ = 0.5, α* = 6, and KR = 2. Clearly indicate how you evaluate the Jacobian matrix.

4. Explain the difference in the diffusion rates of gases through the membrane.

References 1. Coker, D. T., Prabhakar, R., Freeman, ”Tools for Teaching Gas Separation Using

Polymers,” Chemical Engineering Education, 36, Winter 2002, 60 2. Davis, R. A., Sandall, O. C., “A Simple Analysis for Gas Separation Membrane

Experiments,” Chemical Engineering Education, 36, Winter 2002, 74

41

3. Welty, J. R., Wicks, C. E., Wilson, N. E., and Rorrer, G. L., Fundamentals of Momentum, Heat and Mass Transfer, John Wiley and Son, (2001)

42

% Gas membrane separation calculation % Ph=[15 25 35 45 55]; xf=.229; ypv=[.342 .394 .415 .431 .427;.367 .411 .456 .472 .476;.374 .436 .478 .510

.518]; xrv=[.115 .101 .081 .075 .060;.178 .153 .135 .125 .111;.200 .189 .180 .174 .161]; Kr=2;a=6;yi=0.2;theta=0.5; dy=.01; f1='yi*((1-xr)*r-(1-yi))-a*(1-yi)*(xr*r-yi)'; f2='xr*(1-theta)+yp*theta-xf'; dt1='((xf*r-yp)*(xr*r-yi)*(xf*r-yp+xr*r-yi)/2)^(1/3)'; f3='yp*Kr*theta-(1-theta)*a*eval(dt1)'; dt2='(((1-xf)*r-(1-yp))*((1-xr)*r-(1-yi))*((1-xf)*r-(1-yp)+(1-xr)*r-(1-yi))/2)^(1/3)'; f4='(1-yp)*Kr*theta-(1-theta)*eval(dt2)'; % df1dk=0;df1dt=0; df2dk=0;df2da=0;df2dy=0; df4da=0; for ni=1:3; for nj=1:5; xr=xrv(ni,nj);yp=ypv(ni,nj);r=(Ph(nj)+14.7)/14.7; % x=[Kr a yi theta]; disp('Guessed values') fprintf('KR = %8.3f, alfa = %8.3f, yi = %8.3f, theta = %8.3f\n',Kr,a,yi,theta) for i=1:20 % f=[eval(f1) eval(f2) eval(f3) eval(f4)]; df1da=(1-yi)*(xr*r-yi); df1dy=(1-xr)*r-(1-yi)+yi+(xr*r-yi)+a*(1-yi); df2dt=-xr+yp; dlm1=eval(dt1);dlm2=eval(dt2); df3dk=yp*theta; df3da=-(1-theta)*dlm1; df3dt=yp*Kr+a*dlm1;

43

df4dk=(1-yp)*theta; df4dt=(1-yp)*Kr+dlm2; yi=yi+dy; df3dy=(eval(f3)-f(3))/dy; df4dy=(eval(f4)-f(4))/dy; yi=yi-dy; Jt=[df1dk df1da df1dy df1dt; df2dk df2da df2dy df2dt; df3dk df3da df3dy df3dt; df4dk df4da df4dy df4dt;]; dx=Jt\f'; x=x-dx'; Kr=x(1);a=x(2);yi=x(3);theta=x(4); a_exp=yp*(1-xr)/xr/(1-yp); if max(abs(dx))<.001, break, end end fprintf('For P/p = %8.3f,Calculated values from %g iterations\n',r,i) fprintf('KR = %8.3f, alfa = %8.3f, alfa exp = %8.3f, yi = %8.3f, theta =

%8.3f\n',Kr,a,a_exp,yi,theta) disp(' ') end end

44

INTERCOMPANY MEMORANDUM CAL CHEM CORPORATION To: CHE Juniors Date: Spring Quarter File: CHE 333 From: CHE faculty Laboratory Managers Subject: Double-pipe Heat Exchanger We have recently purchased a double-pipe heat exchanger and wish to check on its performance. We plan to use the exchanger to study the variation of the overall heat transfer coefficient with flow rate. The exchanger has been piped to permit counter and parallel flow operation. The tube outer diameter is 15 mm with a wall thickness of 0.7 mm. The shell outer diameter is 22 mm with a wall thickness of 0.9 mm. The length for heat transfer is 1.5 m. Determine the overall heat transfer coefficient for the exchanger using both experimental data and generalized correlations. Would you expect the overall heat transfer coefficient to be different for counter and parallel flow operation? The overall heat transfer coefficient can be based on the inside or the outside surface area of the tubes according to the following equation

1

UiAi = 1

UoA0 = 1

hiAi + 1

hdiAi + ro - ri

kAlm + 1

hdoAo + 1

hoAo You can use the correlations given in Incropera for flow inside a pipe (Ref. 1, pg. 508). Be sure to use the equivalent diameter for the flow in the annular space.

45

Double-pipe Heat Exchanger

The heat transfer between a hot and a cold streams in a concentric tube heat exchanger is Q = UA∆Tlm = UoAo∆Tlm = UiAi∆Tlm (1) where U = overall heat transfer coefficient A = surface area normal to direction of heat transfer ∆Tlm = average driving force for heat transfer = average temperature difference between two streams.

Thi Thi

ThoThoTco

Tco

TciTci

(a) Parallel flow (b) Countercurrent flow

Fig.1 Flow arrangements in heat exchanger

For parallel flow, ∆Tlm is defined by the following equation

∆Tlm =

coho

cihicohocihi

TTTT

ln

)TT()TT(

−−

−−− (2a)

For countercurrent flow, ∆Tlm is defined by the following equation

∆Tlm =

ciho

cohicihocohi

TTTT

ln

)TT()TT(

−−

−−− (2b)

If there is no heat loss to the surrounding, all the energy leaving from the hot stream will be transferred to the cold stream, then Q can also be evaluated from Q = MhCph(Thi - Tho) = McCpc(Tco - Tci) (3) where Mh = mass flow rate of the hot stream Mc = mass flow rate of the cold stream

46

Experimental value of U can be calculated from Eq.(1). The overall heat transfer coefficient can also be estimated from the following relation

1 U i A i

= 1 U o A 0

= 1hiAi

+ 1hdiAi

+ ro - rikAlm

+ 1hdoAo

+ 1hoAo (4)

where hi = heat transfer coefficient for the inner tube hdi = fouling coefficient for the inner tube hdo = fouling coefficient for the outside surface of the inner tube ho = heat transfer coefficient for the annular space ri = inside radius of the inner tube ro = outside radius of the inner tube Ai = inside surface area of the inner tube Ao = outside surface area of the inner tube Alm = (Ao - Ai)/ln(Ao/Ai)

hi

hdi

hdo

ho

rori

Annular space

Fig.2 Concentric tube heat exchanger

For laminar flow in a tube, the average heat transfer coefficient might be estimated from the following correlation

lDNu , = 3.66 + [ ] 3/2PrRe)/(4.01

PrRe)/(0668.0

D

D

LDLD

+ (5)

For turbulent flow, the heat transfer coefficient might be estimated from the following correlation tDNu , = 0.023ReD

4/5Prn (6)

47

where n = 0.4 for heating (surface temperature > fluid temperature) and 0.3 for cooling (surface temperature < fluid temperature). For transitional flow 2,300 < ReD < 10,000

( DNu )10 = ( lDNu , )10 + ( ( )2,

]365/)Re200,2exp[(

lD

D

Nu− + ( )2,

1

tDNu)-5 (7)

where lDNu , and tDNu , are the laminar and turbulent Nusselt numbers given in equations (5) and (6). For noncircular cross section, the above correlations may be applied by using an effective or hydraulic diameter Dh = 4Ac/P (8) where Ac and P are the flow cross-sectional area and the wetted perimeter, respectively. This diameter should be used in calculating ReD and NuD. For flow in an annular space, the effective diameter is

Dh = )()( 22

oi

io

DDDD

+−

ππ = Do − Di (9)

Experimental Procedure

Turn on the heater for the water. Set the temperature of the water to about 75oC; do not exceed 80oC. While the water is heating calibrate the cold-water flow meter. The flow meter is read from the top of the float. Set the valves for counter flow using the diagram on the exchanger or T.K. Nguyen website (http://www.csupomona.edu/~tknguyen/che435/heat.htm). Set the hot water flow rate at half the maximum allowable value and vary the cold water flow rate. At each setting, record the inlet, outlet, and middle temperatures of the hot and cold streams when the system reaches steady state. At least five cold water readings should be measured. Repeat the procedure at half the maximum allowable value for the cold water flow rate while varying the hot water flow rate. Repeat the entire procedure for parallel flow.

Minimum Data Analysis 1. At a fixed hot water flow rate, plot a graph of experimental U versus ReD using Eq. (1)

for both parallel and counter flow. 2. At a fixed cold water flow rate, plot a graph of experimental U versus ReD using Eq.

(1) for both parallel and counter flow.

48

3. Look up the values of hdi and hdo from reference and repeat (1) and (2) for calculated U using Eq. (4).

5. Plot the effectiveness of the heat exchanger for both parallel and counter flow as a

function of ReD as in (1) and (2). 6. Discuss the limitations of Eqs. (5) and (6).

49

References 1. Incropera, F. P. and DeWitt D. P, Fundementals of Heat and Mass Transfer, Wiley,

2002. 2. Hanesian, D. and Perna A. J., “A Laboratory Manual for Fundamentals of Engineering Design”, NJIT. 3. Walas S. M., “Chemical Process Equipment, Selection and Design”, Butterworths, 1988.

50

CHE 333 TRANSPORT LABORATORY EXPERIMENT NO. 5

Boiling Heat Transfer

Introduction

This experiment may appear very crude. And it does not provide very accurate information. But it does illustrate the three modes of heat transfer. It also provides some insight into the complex area of nucleate boiling. This is an experiment where the data are more qualitative than quantitative. However you can investigate the following areas of the heat flux versus ∆T (difference between the surface temperature and the boiling temperature of the liquid) relationship3: a. the free convection zone b. the nucleate boiling zone c. the critical heat flux point, i.e. the maximum in a typical boiling curve d. the transition zone between nucleate and film boiling e. the Leidenfrost point, i.e. the minimum in a typical boiling curve f. the film boiling zone Carefully determining the transition between free convection and nucleate boiling zone, the critical heat flux, and Leidenfrost points is very difficult and will require perseverance in your experimental technique. The experimental setup consists of: a hotplate connected to a variable transformer a convex metal surface, sitting on the hotplate and having a thermocouple, to contain and boil drops of liquid a syringe to supply drops of liquid a stopwatch.

Experimental Procedure

Note: It is important to carefully and thoroughly clean the convex surface of the stainless steel heating surface. 1) Insert three thermocouples into the holes in the side of the stainless steel plate and place the plate on top of the resistance heater. Place the heater on the tripod stand. A Labview program may be used to read and display the temperature of the plate. Appendix B gives instructions to run this program. 2) Measure the volume of the droplet generated from the microsyringe or burette. It is important to maintain the same size and shape of the water drops so be careful to avoid heating the microsyringe or burette

51

3) Plug the heater into the powerstat variable transformer and plug the transformer into a 120 V A.C. source. 4) Initially adjust the powerstat so that the temperature of the plate is about 2oC higher than the boiling point of water. Wait for the temperature to stabilize. 5) Deliver a drop of distilled water to the hot plate. If necessary, adjust the tripod stand to level the apparatus so that the drops of water do not bounce off the heating plate. As the temperature differential becomes greater this is more critical. 6) Measure the time required for a drop to completely vaporized. If the time is more than one minute with the hot plate at 102oC raise the initial temperature to 103oC and, if necessary, to 104oC. Make at least five experimental runs at each temperature. 7) Repeat steps 4 to 6 at another temperature. See figure 10.4 of reference (3) for some idea of the temperature settings. You should plan to make many runs at ∆T's about 5-10oC until the plate temperature reaches about 300-400oC. ∆T can be larger, 10-20oC, if the temperature is not near the three important points aforementioned. 8) Repeat the experiment with another liquid. You will need to know the boiling and critical points of the liquid. Take the data with the plate temperature between the boiling point and the critical point. Investigate placing the drop on different areas of the disc and the height and manner of delivering the drop. Investigate the effect of the initial drop temperature on the breakage of the drop. Aim to keep the water temperature and the size of the liquid drop constant. Try to obtain data that show the maximum heat flux and the minimum heat flux versus ∆T. You can chart the course of your experimental work by preparing a crude graph in the laboratory. If everything else is constant, heat flux is proportional to the reciprocal time (of boiling).

Theory

The authors of reference 1 have modeled the film boiling of drops of liquid on a hot surface and reduced the differential momentum and diffusions equations to the following two equations:

( ) ( ) ( )

( )[ ]bppl

bpbpb

p

TTC

TTTTrRTDMpCk

dtdr

−+

−+

+−

+

−=5.04

14

23 44

*

λρε

σεδ

(1)

Where Tp = temperature of the plate (K) Tb = boiling point temperature of the drop (K) Cp = specific heat of the vapor (cal/g-C)* k = thermal conductivity of the vapor (cal/cm-s-C)*

( ) ( ) ( ) ( ) ( )8 9

0 541

33 4 4ρ ρ ρ δ µλ δ

σεε

λv l v

p p bp b p b

b

g rC T T

k T T T T DMprRT

− =+ −

× − ++

− −

.

*

(2)

52

L = latent heat of vaporization of the drop (cal/g)** D = molecular diffusivity of the vapor species in air (cm2/s)** M = molecular weight of the vapor p* = vapor pressure of the drop (dynes/cm2)** ρl = density of the liquid (g/cm3)** ρv = density of the vapor (g/cm3)* µ = viscosity of the vapor (g/cm-s)* ε = thermal emissivity of the liquid r = radius of the drop (cm) δ = distance the drop is levitated above the plate (cm) t = time (s) R = ideal gas constant (dyne-cm/gmole-K) σ = Stefan-Boltzman constant (cal/cm2-s-K4) g = gravitational constant (cm/s2) Note: "*" denotes evaluation at (Tp + Tb)/2 and "**" denotes evaluation at Tb. Equation (1) accounts for molecular diffusion and radiation heat transfer. If molecular radiation and thermal radiation are neglected, the drop vaporization time is given by

τ =

π5

12 12/7

34

π

4/1

33

33

8*)(9

∆ gTk v

l

ρλµρ Vo

5/12 (3)

where Vo is the initial droplet volume. Other parameters are the same as in equation (1). In order to estimate a droplet’s boiling time, Eqs. (1) and (2) must be solved. A Matlab program is provided to solve for the boiling time. Most of the physical properties required for the calculation can be obtained from the PROP program. Eqs. (1) and (2) can only predict the evaporation time of the liquid drop in the film boiling zone. (See Figure 3, Reference 1)

Report 1) For the experimental data, plot the boiling curve (expressed as reciprocal time) vs. (Tp

- Tb). 2) In the boiling regime, plot the evaporation time vs. (Tp - Tb) using experimental

values and values obtain from equations (1) and (3). Discuss the agreement between the experimental values and the values obtained from the two equations.

3) For the experimental data, plot the heat flux (W/m2) vs. (Tp - Tb). The heat flux can

be estimated to be the heat required evaporating the liquid drop divided by the initial projected area of the drop. How does this compare with figure 10.4 of Incropera?

4) Discuss the sources of error in this experiment.

53

5) Look up the values and the equations for the physical properties from other sources

(not from the PROP program) and compare them with those used in the Matlab program. You need to compare them side by side in a table that lists the values and the equations used.

References

1. Gottfried, Byron and Bell, Kenneth J., Film Boiling of Spheroidal Droplets, I&EC Fundamentals, Vol. 5, No. 4, November, 1966, pg. 561-568. 2. Reid, Robert C., et al, The Properties of Gases & Liquids, Fourth Ed., McGraw Hill, 1987. 3. Incropera and De Witt, Fundamentals of Heat and Mass Transfer, Wiley 2002, pg. 598.

54

% This program determines the evaporation time of a liquid droplet with a % given initial radius. % The physical properties must be evaluated at the conditions given in the lab manual. % disp('Water droplet') R = .13888; % Drop radius (cm) L = 539.08; % Latent Heat (cal/g) DAB = .371; % Diffusivity of vapor species in air (cm2/s) M = 18; % Vapor molecular weight P = 992450; % Vapor Pressure(dyne/cm2) DL = .95785; % Liquid Density(g/cm3) EM = .96; % Liquid emissivity IGC = 8.314E+07; % Ideal gas constant (dyne-cm/mol-K) SBC = 1.3543E-12; % Stefan-Boltzman constant (cal/cm2-s-K4) GA = 980.54; % Gravitational constant (cm/s2) DV = M*P/(IGC*Tf); % Vapor Density(g/cm3) TB = 372.57;Tstart=TB+120; Tp=Tstart:10:700;np=length(Tp); dT=Tp-TB; for i=1:np TP=Tp(i); Tf=.5*(TP+TB); CP=32.24+1.924e-3*Tf+1.055e-5*Tf^2-3.596e-9*Tf^3; CP=CP/(4.18*M); % Vapor Cp(cal/g-C) K=7.341e-3-1.013e-5*Tf+1.801e-7*Tf^2-9.1e-11*Tf^3; K=K/(418); % Vapor k(cal/cm-s-C) V=1.315e-4+(Tf-436)*(1.617-1.315)*1e-6; % Vapor viscosity(g/cm-s) %fprintf('Vapor viscosity (g/cm-s) = %g\n',V) %fprintf('Vapor k (cal/cm-s-C) = %g\n',K) %fprintf('Vapor density (g/cm3) = %g\n',DV) %fprintf('Vapor Cp (Cal/g-C) = %g\n',CP) TE = TP - TB; A = K * TE; B = 3 * CP * DAB * M * P * TE / 2 / IGC / TB; C = 4 * SBC * EM * (TP ^ 4 - TB ^ 4) / (1 + EM); D = 4 * DL * (L + CP * TE / 2); E = 8 * DV * (DL - DV) * GA; F = 9 * V / (L + CP * TE / 2); G = 3 * DAB * M * P * L / IGC / TB; E4 = 4 * E; BF = B * F; RF = R * F; RFC = RF * C; FG = F * G; DEL = .01; for I = 1:10 D3 = DEL ^ 3; FD = RF * (A + C * DEL) - DEL * (E * D3 + FG); DFD = RFC - E4 * D3 - FG; ERRO = FD / DFD; DEL = DEL - ERRO; if abs(ERRO / DEL) < .0001, break, end

55

end if I>9, fprintf ('exceed 10 iterations \n'), end Z = [.1488743 .4333954 .6794096 .8650634 .9739065]; W = [.2955242 .2692667 .2190864 .1494514 6.667135E-02]; CI = DEL / 2; S = 0; for I = 1:5 X = Z(I) * CI + CI; XC = X ^ 3; CA = E * XC + FG; CB = A + C * X; NUM = X * CA * (C * X * CA - CB * (E4 * XC + FG)); FF = NUM / (CA + BF) / CB ^ 3; FS = FF; X = -Z(I) * CI + CI; XC = X ^ 3; CA = E * XC + FG; CB = A + C * X; NUM = X * CA * (C * X * CA - CB * (E4 * XC + FG)); FF = NUM / (CA + BF) / CB ^ 3; S = S + W(I) * (FF + FS); end T = -S * CI * D / F; fprintf('del(cm) = %g, evaporation time (sec) = %8.2f, DT(C) = %8.2f\n',DEL,T,dT(i)) end disp(' ');disp('Isopropanol droplet') % Isopropanol droplet R = .122; % Drop radius (cm) L = 158.454; % Latent Heat (cal/g) DAB = .139; % Diffusivity of vapor species in air (cm2/s) M = 60.094; % Vapor molecular weight P = 992450; % Vapor Pressure(dyne/cm2) DL = .778; % Liquid Density(g/cm3) EM = .96; % Liquid emissivity IGC = 8.314E+07; % Ideal gas constant (dyne-cm/mol-K) SBC = 1.3543E-12; % Stefan-Boltzman constant (cal/cm2-s-K4) GA = 980.54; % Gravitational constant (cm/s2) DV = M*P/(IGC*Tf); % Vapor Density(g/cm3) TB = 355.4;Tstart=TB+120; Tp=Tstart:10:700;np=length(Tp); dT=Tp-TB; for i=1:np TP=Tp(i); Tf=.5*(TP+TB); CP=32.43+.1185*Tf+6.404e-5*Tf^2-9.261e-8*Tf^3; CP=CP/(4.18*M); % Vapor Cp(cal/g-C) K=-7.931e-3+3.987e-5*Tf+11.93e-8*Tf^2-5.021e-11*Tf^3; K=K/(418); % Vapor k(cal/cm-s-C) V=1.22e-4+(Tf-403)*(1.51-1.22)*1e-6; % Vapor viscosity(g/cm-s) %fprintf('Vapor viscosity (g/cm-s) = %g\n',V) %fprintf('Vapor k (cal/cm-s-C) = %g\n',K) %fprintf('Vapor density (g/cm3) = %g\n',DV) %fprintf('Vapor Cp (Cal/g-C) = %g\n',CP)

56

TE = TP - TB; A = K * TE; B = 3 * CP * DAB * M * P * TE / 2 / IGC / TB; C = 4 * SBC * EM * (TP ^ 4 - TB ^ 4) / (1 + EM); D = 4 * DL * (L + CP * TE / 2); E = 8 * DV * (DL - DV) * GA; F = 9 * V / (L + CP * TE / 2); G = 3 * DAB * M * P * L / IGC / TB; E4 = 4 * E; BF = B * F; RF = R * F; RFC = RF * C; FG = F * G; DEL = .01; for I = 1:10 D3 = DEL ^ 3; FD = RF * (A + C * DEL) - DEL * (E * D3 + FG); DFD = RFC - E4 * D3 - FG; ERRO = FD / DFD; DEL = DEL - ERRO; if abs(ERRO / DEL) < .0001, break, end end if I>9, fprintf ('exceed 10 iterations \n'), end Z = [.1488743 .4333954 .6794096 .8650634 .9739065]; W = [.2955242 .2692667 .2190864 .1494514 6.667135E-02]; CI = DEL / 2; S = 0; for I = 1:5 X = Z(I) * CI + CI; XC = X ^ 3; CA = E * XC + FG; CB = A + C * X; NUM = X * CA * (C * X * CA - CB * (E4 * XC + FG)); FF = NUM / (CA + BF) / CB ^ 3; FS = FF; X = -Z(I) * CI + CI; XC = X ^ 3; CA = E * XC + FG; CB = A + C * X; NUM = X * CA * (C * X * CA - CB * (E4 * XC + FG)); FF = NUM / (CA + BF) / CB ^ 3; S = S + W(I) * (FF + FS); end T = -S * CI * D / F; fprintf('del(cm) = %g, evaporation time (sec) = %8.2f, DT(C) = %8.2f\n',DEL,T,dT(i)) end