Download Mass Balance for Open System

ERT 206/4
Thermodynamics
CHAPTER 2
The First Law and Other Basic Concepts
Miss. Rahimah Bt. Othman
Email: [email protected]
COURSE OUTCOME 1 CO1)
1. Chapter 1: Introduction to Thermodynamics
2. Chapter 2: The First Law and Other Basic Concepts
Define, discuss, apply and analyze internal energy, first law,
energy balance-closed system, thermodynamic state and state
function, equilibrium, the Phase Rule, reversible process,
constant-V and constant-P processes, enthalpy and heat
capacity.
3. Chapter 3: Volumetric properties of pure fluids
4. Chapter 4: Heat effects
5. Chapter 5: Second law of thermodynamics
6. Chapter 6: Thermodynamics properties of fluids
2.9 CONSTANT-V and CONSTANT-P PROCESSES
For n moles of homogeneous fluid in a closed system;
d (nU )  dQ  dW
(2.6)
W meanwhile can be replaced by;
dW   Pd (nV )
Will be explained by
Carnot group next
week
Combined both equations;
d (nU )  dQ  Pd (nV )
(2.8)
Equation 2.8 is a general equation for n moles of homogeneous fluid in a
closed system experienced a mechanically reversible process
(1.2)
CONSTANT-V PROCESS
For constant-V process, the work is equal to zero
d (nU )  dQ  dW
(2.6)
W meanwhile can be replaced by;
0
d (nU )  dQ  dW
The integration of 2.9 yield
dQ  d (nU )
(2.9)
Q  nU
(2.10)
For the constant-V, closed system process, the heat transfer is equal to
internal energy change of the system.
CONSTANT-P PROCESS
For constant-P process,
d (nU )  dQ  Pd (nV )
(2.8)
Solved for dQ;
dQ  d (nU )  Pd (nV )
Since P now is a constant, therefore
dQ  d (nU )  d (nPV )
 d [n(U  PV )]
The term (U+PV) is known as enthalpy and has only mathematical definition,
which is;
H  U  PV
(2.11)
Enthalpy
Example 2.8
Constant T = 100 OC
vapor
1 kg of water
Constant P = 101.33 kPa
V water
= 0.00104 m3/kg
V vapor
= 1.673 m3/kg
For this process, heat in the amount of 2, 256.9 kJ is
added to the water
Q=2,256.9 kJ
Calculate ΔU and ΔH for 1 kg of water when it is
vaporized at the constant temperature and pressure.
Solution
Since the process is a constant-pressure process;
Q = nΔH
So,
ΔH = 2,256.9 kJ
For ΔU; from eq 2.11, H = U+PV and ΔH = ΔU+ Δ(PV)
ΔH = ΔU+ Δ(PV)
Aware of the units
ΔU = ΔH - P ΔV
= 2,256.9 kJ –[101.33 kPa (1.673 -0.001)m3]
= 2,256.9 kJ – 169.4 kPa m3
= 2,087.5 kJ
2.11 HEAT CAPACITY
Let say we have two blocks of different metals. A is a copper and
B is an aluminum. Mass for both blocks are 1 kg.
Now, we supply 5000 J of heat to the blocks. What will happen???
- We are sure that temperature will rise, since the heat will
stimulate the atomic energy in the metals.
- The question are;
a) how high the rise in temperature will occur?
b) does the rise of temperature will be the same for
both metals blocks?
•Heat capacity is the amount of heat needed to produce a
specified temperature change (∆T) in a system
C = dQ/dT
Heat Capacity – Constant Volume
 Q 
Cv  

 T V
This definition is true for both molar heat capacity and
specific heat capacity.
For constant volume process,
Q  U
So;
 U 
Cv  

 T V
This can be written as;
dU  Cv dT
(2.17)
and integratio n will yield
T2
U   Cv dT
(2.18)
T1
Comparing with Eq 2.10 Q = nΔU, thus Eq 2.18 become;
T2
Q  nU  n  Cv dT
T1
(2.19)
Heat Capacity – Constant Pressure
 Q 
Cp  

 T  P
This definition is true for both molar heat capacity and
specific heat capacity.
For constant pressure process,
Q  H
So;
 H 
CP  

 T  P
This can be written as;
dH  C P dT
(2.21)
and integratio n will yield;
T2
H   C P dT
(2.22)
T1
Comparing with Eq 2.13 Q = nΔH, thus Eq 2.22 become;
T2
H  n  CP dT
(2.23)
T1
Try examples 2.9 and 2.10
2.12 Mass and Energy Balances for Open System
Since most of the bioprocess eng problems are in open system
situation, it is compulsory for those students to understand the
mass and energy balances for open system.
Measures of Flow



m, q, n, u

m  mass flowrate

q  volumetric flowrate

n  molar flowrate
u  velocity
The measure of flow are interrelated;



m, q, n, u



m  M n and q  uA


Importantly, m and n relate to velocity, u

m  uA
(2.24a)

n  uA
(2.24b)
A  Cross sec tional area
  specific or molar density
Try examples 2.11
Mass Balance for Open System
Control volume – a space identified for analysis in open system
Control surface – surface where it is separated from surrounding
Fluid within the control volume is where thermodynamics analysis
will be done.

dmCV
 (m) fs  0 where
dt



(2.25)

(m) fs  m3  m1  m 2

replacing m with 2.24a eq 2.25 becomes
dmCV
 ( uA) fs  0
(2.26) (continuity equation)
dt
Mass Balance for Open System- cont’
The flow process known as steady state – condition within the
control volume do not change with time.
Therefore
0
dmCV
 ( uA) fs  0
dt
( uA) fs  0
2.26
 2u2 A2  1u1 A1  0 or

m  cons   2u2 A2  1u1 A1
1
Because specific volume is reciprocal of density   V

u2 A2 u1 A1 u A
m


V1
V1
V
(2.27)
General Energy Balances for Open System.
The rate of change of energy within the control volume equals
the net rate of energy transfer into the control volume.
Stream flowing in and out of the control volume are associated
with energy.
Each stream will have total energy of;
1
U  u 2  zg where u  average velocity,
2
z  elavation above datum level
g  local gravity accelaration
Each stream transport energy at rate;

1 2
(U  u  zg ) m
2
Net energy transport into the system by flowing streams is

1 2
 [(U  u  zg ) m] fs
2
The total rate of energy in the control volume including this quantity

and heat transfer rate Q and work rate.


d (mU ) cv
1 2
 [(U  u  zg ) m] fs  Q  work rate
dt
2
work rate – may include work of several forms and has a set of average
properties of P, V, U, H, etc.
Piston, supply the constant pressure done the work, PV, and work rate is;

( PV ) m

The net work done on the system is  [( PV ) m] fs

Another form of work is the shaft at the rate = W
The equation now can be written as;




(mU ) CV
1 2
 


d
   U  u  zg  m  Q   PV  m  W
dt
2

 fs
  fs

replacing H  U  PV to the equation



(mU ) CV
1 2
 
d
    H  u  zg  m  Q  W
dt
2
  fs

usually written as



(mU ) CV
1 2
 
d
   H  u  zg  m  Q  W
dt
2
  fs

(2.28)
velocity, u , is the bulk  mean velocity

u
m
A
for la min ar flow cases, kinetic energy can be assumed as u 2
for turbulent flow cases, kinetic energy can be assumed as u 2 2
Usually, for many applications of bioprocess engineers, kinetic and
potential energy are negligible, and simplify equation 2.28 to



(mU ) CV
d
 ( H m) fs  Q  W
dt
(2.29)
Try examples 2.12, 2.13 and 2.14
Energy Balances for Steady-State Flow Processes
At steady state;
d(
mU CV
)0
dt
Thus, equation 2.28 becomes



1 2
 
  H  u  zg  m  Q  Ws
2
  fs

(2.30)
Assume system has one entrance and one system


1 2


 H  u  zg  m  Q  W s
2



devide by m gives


1

 Q Ws
 H  u 2  zg     
2

 m m
u 2
H 
 gz  Q  W
2
(2.32a)
So far, all the energy unit are presume as Joule. For the English unit,
the kinetic and potential energy unit have to be divided by the constant,
gc. Therefore, Eq 2.32a becomes
u 2 g
H 
 z  Q  W
2gc gc
(2.32a)
Usual unit for ΔH and Q is Btu.
Kinetic, potential energy, work are in (ft lbf).
Try examples 2.15, 2.16 and 2.17
Thank you