Exercises 1&2, 3&4 and 5&6 are related. Start with one of the packages and then
continue to the others. You may select the order based on your interests.
1. a) Derive an equation which describes cooling of an object which is a good thermal conductor. Assume
uniform heat transfer outside the surface of the object.
b) Spherical iron ball with diameter of 1 cm is initially at 100 oC temperature. Air at temperature 20 oC
flows outside it. Calculate temperature of the ball after one minute, when the heat transfer coefficient
between the ball and the flowing air is 50 W/m2K.
c) Calculate the same thing numerically directly from the time dependent energy balance by using ten
second time steps.
2.
Stirred tank with 2300 kg of liquid is being heated from initial temperature of 15°C to a final
temperature of 120°C. Saturated steam at 350 kPa(abs) is being fed to the jacket of the tank. Heat
capacity of the liquid is 2.0 kJ/kg K. Heat transfer area is 5m2 and overall heat transfer coefficient 300
W/m2K.
a) How long does the heating take in this batch operation
b) What is steam mass flow in the beginning and in the end?
c) If the vessel would be fully mixed and operated in a continuous mode, how big cold flow could be
heated?
d) How much of the feed could be heated with a countercurrent heat exchanger of similar area?
3. Derive an equation describing concentration changes in a batch reactor with first order irreversible
reaction to the reactant.
4.
Derive an equation for concentration changes in a steady state isothermal plug flow reactor, and
compare it to the previous result. Calculate concentration changes with Excel, and study the effect of the
step size and reaction kinetic parameter.
1
5. a) Derive compositions at the vapor-liquid interface according to the two film mass transfer theory.
Solve the compositions also numerically from the flux equations using Excel.
b) Derive an equation for the overall mass transfer coefficient in the same situation
6. Calculate instantaneous mass transfer rate (g/s) for a spherical water droplet with diameter of 2 mm,
falling in completely dry air at atmospheric pressure at terminal velocity of 6 m/s. Temperature of the
droplet is 20 oC.
Tips
Always start from the general balance equation ACCUMULATION = IN – OUT + GENERATION
Always write balances for the extensive variables describing the state of the system. Extensive variable is
such that depends on the size of the system (mass, amount of moles, total energy). After this, select proper
physical models to describe the terms in the balances. Also think which of the balance equation term (IN,
OUT or GENERATION) each rate equation describes.
If you wish, you can divide all the balance equation terms with a system size dependent variable (such as
total volume), so that you end up with an intensive variable (temperature, concentration etc.), but do this
only after writing the balance equations.
2
Some useful equations for this exercise:
Heat energy (dimension J, no phase changes)
E  c p mT  Tref 
(1.)
Heat transfer (J/s = W)
Q  hA T  Tout 
(2.)
Energy for phase change (J), (evaporation or condensation)
E  m
(3.)
Power for phase change (W), (evaporation or condensation)

Qm
(4.)
Logarithmic mean temperature difference in countercurrent heat exchangers
TLM 
T1  T2
 T 
ln  1 
 T2 
(5.)
First order reaction rate (mol/m3s)
r  kc
Mass transfer flux (diffusion + convection, mol/m2s)
ND
(6.)
dc
dc
 xN tot  D  cv
dz
dz
(7.)
Mass transfer flux according to the film model (with mass transfer coefficients)
N  k v cV  cIV   k L cIL  cL 
(8.)
Distribution coefficient in ideal vapor-liquid systems at equilibrium
y p0
K 
x p
(9.)
Correlation for mass transfer coefficient outside a spherical object
Sh  2  0,552 Re1 / 2 Sc1 / 3
(10.)
 x1  a 
dx

ln
x x  a   x 0  a 
0
x1
One perhaps useful integration formula:
3
(11.)
Solutions
Exercise 1
a) Start from the general balance equation ACCUMULATION = IN – OUT + GENERATION
Write the balance for thermal energy. Accumulation term describes the energy content changes as a
function of time (rate of change). IN-OUT describes heat transfer with the surroundings. GEN is here zero,
since heat energy is not generated nor consumed. Balance is written for the whole object thermal energy,
which is an extensive variable.
Balances can formally be written only for extensive variables (here total energy), but based on them
equations describing intensive variables can be derived (e.g. temperature).
The following equations are needed for the balances:
E  c p mT  Tref 
Thermal energy (total enthalpy, dimension J)
(12.)
where cp is the specific heat capacity per mass, m is the mass of the object and T is temperature. T ref is
arbitrary reference temperature (can be selected freely since enthalpy is a function of state)
Heat transfer (J/s = W)
Q  hA T  Tout 
(13.)
where h is the heat transfer coefficient, A is the area, and Tout is outside air temperature. Heat transfer is
here defined so, that if the object is hotter than the surroundings, sign of the heat transfer rate is positive.
Therefore it is “OUT” term in the heat balance.
Time dependent energy balance is then
dE dc p mT  Tref 

 hA T  Tout 
dt
dt
(14.)
Let’s assume density and volume constant, so that we get terms outside the time derivative:
c p V
dT  Tref 
 hA T  Tout 
dt
(15.)
4
Reference temperature is naturally also a constant, so that we get
cp
dT
 ha T  Tout 
dt
(16.)
where a is the ratio of surface area and volume. This equation can be separated (move dT and dt to the
different sides of =)
dT
ha

dt
T  Tout  c p 
(17.)
Let’s assume the coefficient on the right hand side as constant (heat transfer coefficient and other terms
are here assumed to be time and temperature independent), and we can integrate
T1
t
dT
ha 1
 T  Tout    c p  0 dt
T0
(18.)
For which the solution is
 T  Tout  
ha
  
ln  1
t1
cp
 T0  Tout  
(19.)
From this, also temperature T1 at any time moment t1 can be solved:
 ha 
T1  Tout  T0  Tout  exp  
t 
 c  1
 p

(20.)
This solution is also called as Newton’s law of cooling.
b) Fill in the known values in the equation. Heat capacity and density was not given, but it was stated that
the material is iron. Search for iron material properties: cp = 450 J/kgK,  = 7874 kg/m3.
5
Specific area can be calculated from the diameter, a = A/V = 6/d, for spherical objects. Remember to check
that all the variables are in SI units!
The final temperature is obtained T1 = 68.13 oC.
c) Chemical engineering models are often nonlinear, and analytical solutions are not available. This would
for example result from the situation where heat transfer coefficient depends on the temperature
difference (as is the case in natural convection). In the previous example, the rate of change of temperature
was
dT
ha
T  Tout 

dt
cp
(21.)
This can be solved numerically by evaluating the right hand side at t=0 and estimate what would be the
temperature after a short time interval. Here time step was specified to be 10 s. The r.h.s. of the equation is
initially -0.67733 K/s. Assume this as constant during the first time step, so that after 10 s the temperature
would be 100 – 10*0.67733 = 93.2267 oC. Calculate the r.h.s. again with this value and from this,
temperature after 20 s. Continue in a similar way until t=60s. With this numerical result, the final
temperature is 67.05 oC, and the error compared to the analytical result is about 1 oC. This error could be
reduced by using shorter time steps. Another option is to use more advanced time integration methods,
where it is not assumed that the rate of change remains constant during each time step.
In the following table, the calculation process is shown in more details.
time (s)
T (C)
T-Tout (C)
dT/dt (C/s)
0
100
80
-0.677334688
10
93.22665312 73.22665312 -0.619986903
20
87.02678409 67.02678409 -0.567494574
30
81.35183835 61.35183835 -0.519446604
40
76.15737232 56.15737232 -0.475466703
50
71.40270528 51.40270528 -0.435210442
60
67.05060087 47.05060087 -0.398362551
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Exercise 2
a) Use the equation derived earlier (eq. 8), but in such a form where mass and area are shown instead of
specific surface area
 T  Tout  
hA
  
ln  1
t1
cpm
 T0  Tout  
(22.)
Outside (tank jacket) temperature is not given, but it is known that it is heated with 350 kPa saturated
steam. Steam temperature can be found from the vapor pressure correlations or from the steam tables. In
either way, Tout = 139 oC. Let’s solve the previous equation for time t1:
 T  Tout  
 120o C  139o C 


ln  1
ln 
T0  Tout  

15o C  139o C 


t1  

 5752 s
hA
300W / m 2 K  5m 2
cpm
2000J / kgK  2300kg
(23.)
Which is approximately 96 minutes.
b) Calculate first the heating power given by the steam. This can be calculated from the heat transfer
equation
Heat transfer (J/s = W)
Q  hA T  Tout 
(24.)
Insert initial and final temperatures separately, and we get Q(initial) = 186000 W and Q(final) = 28500 W.
Look at the heat of evaporation (latent heat) for water at 350 kPa. This is 2148.5 kJ/kg. Solve steam
consumption at the beginning and at the end. It is assumed here that only latent heat is available from the
steam.

Qm
(25.)
Instantaneous steam flow in the beginning is about 312 kg/h and in the end 48 kg/h.
7
c) In a continuous well mixed vessel temperature is the same as the temperature in the outflow, here 120
o
C. Heat transfer thus corresponds to part b) end value. Although the feed is colder, it is assumed to be
mixed very fast with the content inside the tank.
Mass flow can be calculated from the energy balance
Q  cpm
 Tout  Tin 
(26.)
From here, the mass flow is 0.136 kg/s, or 489 kg/h.
d) In order to calculate countercurrent heat exchanger, logarithmic mean temperature is needed
Q  hA
T1  T2
 T 
ln  1 
 T2 
(27.)
Where T1 and T2 are the cold and hot end temperature differences; does not matter which is which
(why?). These are 124 oC and 19 oC. Insert numerical values:
Q  300 W / m 2 K  5m 2
124 o C  19 o C
 83962 W
 124 o C 
ln  o 
 19 C 
(28.)
In the same way as in part c), the mass flow is 0.3998 kg/s, or 1440 kg/h. Countercurrent heat exchanger is
considerably more effective than the mixed tank since the driving force (temperature difference) stays
higher over the whole heat transfer area.
8
Exercise 3
Balance ACCUMULATION = IN – OUT + GENERATION is written for the amount of moles, which is extensive
variable. Batch reactor is time dependent (contains ACCUMULATION term), but in this example nothing is
fed or removed from it before the end. Generation term is the reaction rate, which basically describes
change in the intensive variable (here amount of moles per unit volume, i.e. concentration). Therefore it
needs to be multiplied by the total reacting volume.
Note! Raction rate is often given in other forms, e.g. mol/s per unit mass of catalyst. This needs to be
always checked and multiplied with the correct term (e.g. total mass of catalyst) so that the correct
generation of the extensive variable can be included in the material balance.
Note 2! Here symbols n and r are for a certain chemical component (amount of moles and reaction rate);
subscript i is not separately shown.
dn
  rV
dt
(29.)
Amount of moles can be expressed by using concentration. Reaction rate is here first order, and also that
can be expressed with concentration rather simply.
dn dcV 

 rV  k r cV
dt
dt
(30.)
where kr is the reaction rate constant (for an isothermal reaction it can be assumed as constant). Let’s
assume that the total reaction volume does not depend on time, so that we end up with
dc
 k r c
dt
(31.)
this can again be separated and then solved from the initial time to the batch time t1.
c1
t
1
dc
 c  k r 0 dt
c0
(32.)
concentration in the end is
9
c1  c 0 exp  k r t 1 
(33.)
Exercise 4
Now accumulation term is zero (time independent situation). Let’s draw a picture:
Control volume is a short ”slice” of the reactor. It starts after length z from the feed (this length is an
independent variable in the model), and the length of the control volume is infinitesimally small, dz. There
are two mass transfer mechanisms to the control volume: diffusion and convection (there are no other
mechanisms but these two for any situation). Here again symbols n, N, D, c, and x without subscripts refer
to a certain chemical component, subscript tot refers to the sum (e.g. Ntot is the total flux, or sum of
individual component fluxes). Similar balance equation should be written for each chemical component.
 dc

 dc

n  NA   D  xN tot A   D  cv A
 dz

 dz

(34.)
The first term in the parenthesis describes diffusion (Fick’s law) and the second convection, or in other
words flow of material with the flow. Convection can be given in different ways, for example mole fraction
multiplied by the total flux, or component concentration multiplied with the flow rate (m/s).
Now a plug flow is assumed, so diffusion contribution can be neglected. This is often a good assumption in
cases where velocity across the control volume boundary is reasonably high.
Material balance for the control volume is
0  n in  n out  rV
(35.)
10
Left hand side describes accumulation which was assumed zero (time independent situation). Mole flow
(mol/s) to the control volume and out from it is the same as the concentration at the control volume
boundary multiplied with volumetric flow rate. Volumetric flow rate is linear velocity (m/s) multiplied by
the reactor cross-sectional area (m2). Control volume size (m3) is the same as reactor cross-sectional area
multiplied by the control volume length dz. From these we get:
0  A p cvz  A p cvzdz  rA p dz
(36.)
Cross-sectional area Ap is assumed constant (tubular reactor with constant tube diameter), so it can be
eliminated from the equation. Also the flow velocity v (m/s) is assumed constant. Finally we get:
v
c zdz  c z   r
(37.)
dz
From the definition of a derivative
v
dc
r
dz
(38.)
For simplicity, let’s assume again the same first order reaction kinetics as earlier, r=krc, so that we can
compare results.
k
dc
 r c
dz
v
(39.)
This can be solved again by separating the variables (dc and dz).
c1
z
kr 1
dc


c c
0 dz
v
0
(40.)
Concentration at the end is
11
 k

c  c 0 exp   r z1 
 v 
(41.)
When this is compared to the result in the previous exercise, it can be seen that the solutions look similar. If
it is further considered, that the residence time tres = z1 / v, the two equations are identical and can be
solved with the same mathematical tools.
Exercise 5
a) Let’s again first draw a schematic picture:
y is mole fraction in the bulk vapor, yI is mole fraction in the interface on the vapor side, xI in the liquid side,
and x in the liquid bulk.
Interface is assumed to be infinitesimally thin, so there is no material accumulation in it. The material
balance is simply IN=OUT. Mass transfer fluxes at both sides of the interface needs to be equal. Let’s
assume here further that convection across the interface is negligible.
N vapor  N liquid
(42.)
Mass transfer fluxes are expressed with mass transfer coefficients and the driving forces. Here the
dimension of the mass transfer coefficient is m/s and driving force is concentration difference.
k v cV  cIV   k L cIL  cL 
(43.)
12
where V refers to vapor and L to liquid. Subscript I refers to the interface. Let’s write the previous by using
mole fractions instead of concentrations
k vctV y  yI   k LctL x I  x 
(44.)
Next insert phase equilibrium equation (definition of the distribution coefficient K), now valid at the
interface:
yI  KxI
(45.)
and solve the equation for interface composition on the liquid side:
xI 
k v c tV y  k L c tL x
k v c tV K  k L c tL
(46.)
On the vapor side the interface composition can be calculated from the phase equilibrium equation y=Kx
(interface is in equilibrium).
b) Insert the previous result in the liquid side mass transfer equation
 k c y  k L c tL x

N liquid  k L c tL x I  x   k L c tL  v tV
 x 
 k v c tV K  k L c tL

(47.)
After a couple of intermediate steps, we end up with
N liquid 
1
K
1

k L c tL k v c tV
y  Kx 
(48.)
13
The factor in front of the parenthesis on the right hand side is the requested vapor side overall mass
transfer coefficient. This coefficient must be used with such driving force (the term in parenthesis), where
compositions are expressed as if they were vapor phase mole fractions: (y-y*) = (y-Kx)
Exercise 6
Droplet is pure water, so there is no mass transfer resistance in the liquid (there cannot be a composition
profile in the film on the liquid side). Therefore also the liquid droplet surface is pure water. Water mole
fraction at the interface on the gas side can be obtained from the following, where gas (air) is assumed
ideal
yI p0
K

xI
p
(49.)
Vapor pressure p0 of water in the droplet temperature 20 oC is 2341 Pa. K-value is thus 2341 Pa / 101325 Pa
= 0.023104, which is the same as water mole fraction at the interface, since corresponding liquid phase
mole fraction xI is 1.
In addition to the previous, we need the mass transfer coefficient. A correlation for it was given above:
Sh  2  0,552 Re1 / 2 Sc1 / 3
(50.)
Next we need to calculate Reynolds and Schmidt numbers:
Re 
vd

(51.)
Sc 

D AB
(52.)
At 20 oC air viscosity is 1.83710-5 kg/ms. Water diffusion coefficient in air is assumed be 0.282 cm2/s, or
2.8210-5 m2/s. Air is assumed to be ideal gas, so its density can be calculated from the ideal gas law:
14

pM
RT
(53.)
Average molar mass of air is 28.97 g/mol, so that density at the given temperature is 1.204 kg/m3. Insert
known values and we get Re = 196, Sc = 0.54 and from the mass transfer coefficient correlation Sh = 8.3.
From this, the mass transfer coefficient in the film outside the droplet is k=0.469 m/s. Insert these values
into the mass transfer equation
N  k v c tV y I  y  k v c tV y I  0
(54.)
The mass transfer flux N is 3.5310-7 mol/s. After multiplying this with the molar mass of water (the
transferring component), we get the mass transfer in the requested dimension (g/s). Evaporation rate is
6.36 micrograms per second.
Let’s think if this result makes sense. Let’s calculate time needed for the droplet to evaporate completely
provided the mass transfer rate would be constant. The droplet mass can be calculated from the given
diameter and water density. From these, the evaporation tame would be 10 seconds, which is probably a
reasonable value. In reality, the droplet size would change during evaporation, and the droplet diameter
should be calculated as a function of time, if precise evaporation time would be asked. In practice air is
neither completely dry. Air humidity is close to saturation if there are lots of droplets, so that evaporation
rate is reduced due to smaller driving force. For completely saturated air there is no evaporation.
15