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[BSc TN] Summary Transport Phenomena $7.02
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[BSc TN] Summary Transport Phenomena

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  • Course
  • Fysische Transportverschijnselen (TN2786)
  • Institution
  • Technische Universiteit Delft (TU Delft)
  • Book
  • Transport Phenomena

--- Satisfied? Please don't forget to leave a rating! --- This summary covers the entirety of "Transport Phenomena - The Art of Balancing" by H. v.d. Akker and R.F. Mudde (besides the less crucial §1.5, §3.4, §4.4 and §5.5) and the lectures given of the accompanying course "TN2786 - Fysisch...

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  • July 6, 2022
  • July 13, 2022
  • 40
  • 2021/2022
  • Summary

Subjects

  • heat transport
  • mass transport
  • fluid mechanics
  • navier stokes

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  • Technische Universiteit Delft (TU Delft)
  • Technische Natuurkunde
  • Fysische Transportverschijnselen (TN2786)
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Summary Transport Phenomena
by Ruben Tol
This summary covers all useful formulas used in the book ”Transport Phenomena:
The Art Of Balancing” by H. v.d. Akker and R.F. Mudde, and sometimes refers to
the book if deemed necessary. The formulas are numbered the same as in the book,
and the most important formulas are high-lighted in light-gray . Useful equations
from the lectures are highlighted in light-blue , and are labelled with their corre-
sponding lecture (L#). The booklet ”Transport Phenomena Data Companion” by
L.P.B.M. Janssen and M.M.C.G. Warmoeskerken (further referred to as TPDC) is
needed to follow the course, and will be referred to in this summary in dark-blue.
Some quantities have their units given between square brackets.


1 Balances
1.1 The balance: recipe and form
The balance equation for a quantity G is given by
d
G = ϕG,in − ϕG,out + PG , (1.2)
dt
where ϕ is the transport rate of G, and PG the production/destruction rate of G.

A balance equation is always of the form
d
= in − out + production.
dt

1.2 The mass balance
The general form of a mass balance of a substance A is
d
MA = ϕm,A,in − ϕm,A,out + PA . (1.4)
dt
The concentration of substance A cA [kg m−3 ] in a volume V is then given by
cA = mass of A per unit of volume = MA /V. (1.7)
This then allows the mass balance to be rewritten a
d
V cA = ϕv,in · cA,in − ϕv,out · cA,out ± rA V, (1.13)
dt
where rA [kg m−3 s−1 ] is the production/reaction rate of A per unit volume.

For a n-th order reaction, the reaction rate rA is given by
rA = kr cnA , (1.24)

where kr [s−1 ] is the reaction rate constant.

1

,For a continuous stirred tank reactor (CSTR):
cA = cA (t) and therefore cA ̸= f (x, y, z); (L1)
cA,out (t) = cA (t); (L1)
ϕV,A,in = ϕV,A,out = ϕV . (L1)

For a plug flow reactor (PFR):
cA (x) = cA and therefore cA ̸= f (t); (L1)
transport only through convection (no diffusion). (L1)

1.3 The energy balance
The total energy of a system is the sum of the internal, potential, and kinetic energy.

The energy concentration e [J kg−1 ] for a control volume is defined as
1
e = u + gz + v 2 , (1.86)
2
where u [J kg−1 ] is the internal or thermal energy, with the other remaining terms
being the mechanical energy.

The general form for the total energy balance then becomes
   
d d p p
E = (ρV e) = ϕm,in e + − ϕm,out e + + ϕq + ϕw , (1.91)
dt dt ρ in ρ out

with ϕq [J s−1 ] being the heat flow and ϕw [J s−1 ] the work performed on the system
per unit of time from outside the control volume.



The total energy balance can be split up into a thermal energy balance and a me-
chanical energy balance:
dE dU dEmech
= + . (L2)
dt dt dt
The thermal energy balance (or heat balance) is then given by
d d
U = (ρV u) = ϕm,in · uin − ϕm,out · uout + ϕq + Pu . (1.119)
dt dt
For a steady-state situation, this equation can be rewritten as
 Z out  
1
0 = ϕm uin − uout − pd + ϕq + ϕm ef r , (1.122)
in ρ
where the integral is the reversible energy transformation of mechanical energy to
thermal energy, and ef r [J kg−1 ] is the irreversibly transformed mechanical energy
into thermal energy.

2

, The mechanical energy balance then simply becomes

Emech = Etotal − Etherm , (1.123)

which, for steady-state systems, yields
 Z in   
1 2 2 1
0 = ϕm (vin − vout ) + g(zin − zout ) + dp + ϕw − ϕm ef r . (1.124)
2 out ρ


In the event that no work is performed on or by the system, the dissipation is
negligible and that the density remains constant, the above equation simplifies to
the Bernoulli equation:

1 2 p1 1 p2
v1 + gz1 + = v22 + gz2 + , (1.126)
2 ρ 2 ρ
or
1 2 p
v + gz + = constant. (1.127)
2 ρ
The indices 1 and 2 imply a system going from some state 1 to some state 2, similarly
to the previous ”in” and ”out” states.

Again, Bernoulli’s equation only holds if

ϕm,1 = ϕm,2 = ϕm (steady-state); (L2)
ϕw = 0; (L2)
ef r ≈ 0; (L2)
ρ = constant. (L2)

1.4 The momentum balance
The x, y and z momentum balances are given by

d X
(ρV vx ) = ϕm,in · vx,in − ϕm,out · vx,out + Fx ; (1.139)
dt
d X
(ρV vy ) = ϕm,in · vy,in − ϕm,out · vy,out + Fy ; (1.139)
dt
d X
(ρV vz ) = ϕm,in · vz,in − ϕm,out · vz,out + Fz , (1.139)
dt
P
where F can be any other inside or outside forces acting on the system, such as
gravity Fg , the normal force Fn or friction Fw .



3

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