UNIT OPERASI BIOPROSES (UOB)

TPE4211

UNIT OPERASI
BIOPROSES (UOB)
YUSRON SUGIARTO

KULIAH 4

MATERI KULIAH
No

Pokok Bahasan

Sub Pokok Bahasan

Waktu
(Jam)

Sistem satuan, dimensi
dan konversi

2 x 50

1.
2.

Pengantar
Satuan dimensi

3

Kesetimbangan Massa

2 x 50

4.
5.

Kesetimbangan Energi
Kesetimbangan Unsteady-State
Massa dan Energi

Dasar-dasar perpindahan
momentum

2 x 50
2 x 50

6.

7.

Lanjutan





Viskositas dan macammacam fluida, fluida
statis, aliran fluida, tipe
aliran dan faktor
gesekan

Pengukuran aliran
fluida, kebutuhan
tenaga untuk aliran,
persamaan Bernoilli dan
penerapannya

2 x 50

2 x 50

ENERGY BALANCE

• Energy is expensive….
– Effective use of energy is important task for
bioprocess engineers.

• Topics of this chapter
– Energy balance
– Energy and energy transfer
• Forms of energy : Kinetic / Potential / Internal Energy

• Energy transfer : Heat and Work

– Using tables of thermodynamic data
– Mechanical energy balances

INTRODUCTION
Unlike many chemical processes, bioprocesses are not particularly energy
intensive. Fermenters and enzyme reactors are operated at temperatures
and pressures close to ambient; energy input for downstream processing is
minimized to avoid damaging heat-labile products. Nevertheless, energy
effects are important because biological catalysts are very sensitive to heat
and changes in temperature. In large-scale processes, heat released during
reaction can cause cell death or denaturation of enzymes if it is not quickly
removed. For rational design of temperature-control facilities, energy flows
in the system must be determined using energy balances. Energy effects are
also important in other areas of bioprocessing such as steam sterilisation.

INTRODUCTION
The law of conservation of energy means that an energy
accounting system can be set up to determine the amount

of steam or cooling water required to maintain optimum
process temperatures. In this chapter, after the necessary
thermodynamic concepts are explained, an energy-conservation
equation applicable to biological processes is derived.

SYSTEM

SYSTEM

Forms of Energy
– The First Law of Thermodynamics
A balance on conserved quantity (i.e. mass, energy,
momentum) in a process system may be written as:
Input + generation - output - consumption =accumulation

Forms of Energy
– The First Law of Thermodynamics
• Forms of energy

– Kinetic energy : due to the motion of the system

mv 2
EK 
2gc

– Potential energy : due to the position of the system

Ep  m

– Internal energy : due to the motion of internal molecules
• U : from thermodynamic calculation

g
h
gc

Forms of Energy
– The First Law of Thermodynamics
Change in kinetic energy:

Change in potential energy

Change in potential energy

Forms of Energy
– The First Law of Thermodynamics
How energy can be transferred between a system and its
surroundings?
- Heat – energy that flows as a result of temperature
difference between a system and its surrounding ; heat is
defined positive when it is transferred to the system from the
surroundings.

-Work – energy that flows in response to any driving force
other than a temperature difference. ; work is defined positive
when it is done by the system on the surroundings.

Forms of Energy
– The First Law of Thermodynamics
Types of Work

• Flow work (Wfl) - energy carried across the boundaries of a
system with the mass flowing across the boundaries (i.e.
internal, kinetic & potential energy)

• Shaft work (Ws) - energy in transition across the boundaries
of a system due to a driving force other than temperature,
and not associated with mass flow (an example would be
mechanical work due to a piston, pump or compressor)

ENERGY – CONVERSION UNITS

1

Energy balance on closed systems

 There is no mass transfer into a closed system
 The only way energy can get into or out of a closed system
is by heat transfer or work

Q

Ws
a. Heat transfer (Q)
b. Work (Ws)
Note: * Work is any boundary interaction that is not heat (mechanical, electrical,
magnetic, etc.)

The law of conservation of energy
can be written as:

energy in through
system boundaries

energy out through
energy accumulated
=
system boundaries
within the system

1

Energy balance on closed systems

• Balance equation
(Final System Energy) – (Initial System Energy)
= (Net Energy Transfer)
• Initial System Energy U  E  E
• Final System Energy U  E  E
The first law of
thermodynamics
Q

W
• Net Energy Transfer
for closed systems
i

f

pi

ki

pf

kf

U  E p  Ek  Q  W

1

Energy balance on closed systems
U  E p  Ek  Q  W

For a closed system what is ∆E equal to?
 Is it adiabatic? (if yes, Q = 0)
 Are there moving parts, e.g. do the walls move? (if no, Ws = 0)
 Is the system moving? (if no, ∆KE = 0)
 Is there a change in elevation of the system? (if no, ∆PE = 0 )
 Does temperature, phase, chemical composition change, or
pressure change less than a few atmospheres ? (if no to all, ∆U = 0)

Energy Balance Procedures

• Solve material balance  Get all the
flow rate of streams
• Determine the specific enthalpies of each
stream components
– Using tabulated data
– Calculation

• Construct energy balance equation and
solve it. H  E p  Ek  Q  Ws

Exercise
A closed system of mass 5 kg undergoes a process in which
there is work of magnitude 9 kJ to the system from the
surroundings. The elevation of the system increases by 700m
during the process. The specific internal energy of the system
decreases by 6 kJ/kg and there is no change in kinetic energy
of the system. The acceleration of gravity is constant at
g=9.6m/s2.
Determine the heat transfer, in kJ!

2 Energy balance on open systems

 How are open systems control volumes different from closed
systems
 What effect does this have on the energy balance?

Some common open system steady flow devices

Some common open system steady flow devices

2 Energy balance on open systems
*Steady State

• Flow work and shaft work
– Flow work : work done on system by the fluid itself at the
inlet and the outlet
– Shaft work : work done on the system by a moving part
within the system W  W  W
s

Vin (m3 / s )
Pin ( N / m 2 )

f

Process
Unit

W f  PinVin  PoutVout

Vout (m3 / s )
Pout ( N / m 2 )

Application of First Law - Conservation of
energy for a control volume

The Steady-State Open System Energy Balance
U  E p  Ek  Q  W

W  Ws  W f

W f  min PinVˆin  mout PoutVˆout

mv2
EK  
2 gc
EK  m

Hˆ  Uˆ  PVˆ

g
h
gc

U  mUˆ
H  E p  Ek  Q  Ws

Exercise
Air at 300oC and 130 kPa flows through a horizontal 7 cm
ID pipe at a velocity of 42 m/s

a) Write and simplify the energy balance
b) Calculate the rate of kinetic energy, if the air is heated to 400oC at
constant pressure, assuming ideal gas behaviour
c) Why would be correct to say that the rate of heat transfer to the gas equals
the rate of change of kinetic energy? Why?

Enthalpy
• It is convenient to define the following property
for the calculation of energy balance for
flowing systems.
– Enthalpy
H  U  PV
– Specific Enthalpy Hˆ  Uˆ  PVˆ

Exercise
What is the enthalpy of 150 g formic acid at
70oC and 1 atm relative to 25oC and 1 atm?

Tables of Thermodynamic Data
• U, H, S, V,…  Thermodynamic function
• Tables of Thermodynamic Data
– Tabulation of values of thermodynamic functions (U, H,
V,..) at various condition (T and P)
– It is impossible to know the absolute values of U , H for
process materials  Only changes are important ( U,
H,…)
– Reference state
• Choose a T and P as a reference state and measure changes of U
and H from this reference state  tabulation

Steam Tables

• Compilation of physical properties of water

(i) Steady-stateflow process. At steady state, all properties
of the system are invariant. Therefore, there can be no
accumulation or change in the energy of the system: AE = 0

(ii) Adiabatic process. A process in which no heat is
transferred to or from the system is termed adiabatic; if
the system has an adiabatic wall it cannot receive or
release heat to the surroundings. Under these conditions Q0 and becomes:

Exercise
Water at 25oC enters an open heating tank
at a rate of 10 kg h-1. Liquid water leaves
the tank at 88 oC at a rate of 9 kg h-1;
1 kg h-1 water vapor is lost from the system
through evaporation. At steady state, what is
the rate of heat input to the system?

Exercise
1. Assemble
(i) Select units for the problem. kg, h, kJ, oC
(ii) Draw the flow sheet showing all data and units.
(iii) Define the system boundary by drawing on the flowsheet.

THANK YOU...