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TRANSCRIPT
THERMODYNAMICS FOR CHEMICAL ENGINEERS
(EP 205)
THERMODYNAMICS (EG 207)
Outline
First law of thermodynamics
Energy balance for closed system (control mass)
Introduction of internal energy and enthalpy
Quasi-equilibrium work process
Enthalpy for ideal gas
Energy balance for open system (control volume)
First law of thermodynamics
First Law of Thermodynamics (also known as the conservation of energy principle)
Energy can neither be created nor destroyed; it can only change form.
The First Law of Thermodynamics is the relationship between heat Q, work W and the total energy E of the system and its surroundings.
∆(Energy of the system) + ∆(Energy of the surrounding) = 0
outin EE
Energy balance (system undergoing any kind of process)
outin EE
Energy In, Ein Energy Out, Eout
inout EEE
Q W
System
= (Total energy leaving the system)
- (Total energy entering the system)
(Change in the total energy leaving of the system)
Energy balance for Closed System
According to the 1st Law Ein = Eout
2
2
221
2
11
22gz
uUWgz
uUQ s
Inlet Outlet
Q
Ws
Control
Mass
U1
u1
U2
u2
z1 z2
Internal
Kinetic
Potential
Heat
Shaft work
PEKEUE
Where,
Therefore, for a steady-state closed system:-
EUWQ s
)(2/1 12 uuKE
)( 12 UUU
)( 12 hhgPE
Closed system: Stationary system – do not involve any
changes in velocity and elevation
Closed system: Stationary system – do not involve any changes in velocity and elevation
* If stationary system – do not involve any changes in velocity and elevation
∆E = ∆U = Q + W
For a closed system:
No mass flow across the boundary
Stationary system
Undergo process that only cause changes to its internal energy, U
∆E = ∆U = Q - W &
Heat transfer from a system
Work done on the system
Heat transfer to a system
Work done by the system
* At the end you have to understand the process itself instead of memorizing the sign
Internal energy, U
Arises from the random or disorganized motion
of molecules in the system
Energy of the internal molecules to the substance (sensible, latent, chemical, nuclear)
Cannot be directly measured
Work, W
Consider the compression or expansion by a cylinder piston
-ve sign ⟹ compression
=ve sign ⟹ expansion Read pg. 9 “The minus sign ….”
2
1
V
VPdVW (1.3)
)( 12 VVPW
Example:
A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During the cooling process, the fluid loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on the fluid. Determine the final internal energy of the fluid. Neglect the energy stored in the paddle wheel.
Thermodynamics for Chemical Engineers EP205, L3
Introduction to enthalpy, H
Enthalpy H, is a thermodynamic property
Appears in energy balances to calculate Q and W
For a unit mass or mole of substance, enthalpy (kJ)
where ∆H is simply H2 – H1 or Hout – Hin
H ≡ U + PV
∆H = ∆U + ∆PV = Q
Enthalpy for water/steam
It can also be expressed as:
Molar enthalpy (kJ/mole)
Specific enthalpy (kJ/kg)
* You can find data(s) for enthalpy(water) in Table F.
VΔPUΔHΔ
V~
ΔPU~
ΔH~
Δ
Example:
Calculate ∆U and ∆H for 1 kg of water when it is vaporized at the constant temperature of 100 °C and the constant pressure of 101.33 kPa. The specific volumes (v) of liquid and vapor water at these conditions are 0.00214 and 1.897 m3kg-1. For this change, heat in the amount of 2,556.7 kJ is added to the water.
Note: specific volumes v is the volume occupied by a unit of mass of a material, v = volume/mass
Enthalpy for an ideal gas
Ideal gas law Ideal gas law in above eqn. can be modified to Enthalpy in above eqn. can now be written in the form of
specific enthalpy
where P,T = absolute pressure and temperature, = molar volume (m3/mole) Z = compressibility factor (1 for ideal gas), R = ideal gas constant (8.314 Jmol-1K-1), MW = the molecular weight
MW
Rˆ TVP
MW
Rˆ ˆ
ˆˆˆ
TUH
VPUH
TZVP R~
V~
* You can find the R in Table A
Control Volume
Control volume – also known as an open system
Commonly studied control volume:-
Compressor
Turbine
Pump,
Throttle etc.
Total energy of a flowing fluid consists of:-
Enthalpy,
Kinetic,
Potential energies
Process
If the state of a system changes, then it is undergoing a process
Types of process
Isobaric – constant pressure
Isothermal – constant temperature
Isochoric – constant volume
Isentropic – constant entropy
Adiabatic – No heat transfer
Energy balance for Opened System
According to the 1st Law
Ein = Eout
2
2
221
2
11
22gz
uHWgz
uHQ s
For a steady-state opened system:-
Inlet Outlet
Q
Ws
Control
Volume
H1
u1
H2
u2
z1 z2
Enthalpy
Kinetic
Potential
Heat
Shaft work
zgu
HWQ s
2
2
Common opened system processes
As mentioned before, work done BY (such as turbine) and on (such as compressor) the system has different sign (+/-), so don’t get confuse!!
The rule of thumb is, derive your own energy balance equation using the control volume and equation Ein = Eout, and you WON’T GO WRONG
Ws
Turbine
Comp-
ressor Ws
Steam tables (SI Units) pg. 716-753
Two important steam tables
Saturation temperature table (pg.716)
Saturation pressure table (pg. 722)
What kind of information provided in the table??
Saturation Temperature, Tsat (°C/K)
Saturation Pressure, Psat (kPa)
Specific Volume, V (cm3kg-1)
Specific Internal Energy, U (kJkg-1)
Specific Enthalpy, H (kJkg-1)
Specific Entropy, S (kJkg-1K-1)
Air enters a compressor operating at steady-state at a pressure of 1 bar, a temperature of 300 K, and a velocity of 6 m/s through a feed line with a cross-sectional area of 0.1 m2. The effluent is at a pressure of 7 bar and a temperature of 450 K and has a velocity of 2 m/s. Heat is lost from the compressor at a rate of 180 kJ/min. If the air behaves as an ideal gas, what is the power requirement of the compressor in kW ?
Example:
Ideal-gas properties of air
1. Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The mass flow rate of the air is 0.02 kg/s, and a heat loss of 16 kJ/kg occurs during the process. Assuming the changes in kinetic and potential energies are negligible, determine the necessary power input (kW) to the compressor.
Assignment 2
2. Steam at 9000 kPa and 600°C passes through a throttling process so that the pressure is suddenly reduced to 400 kPa. What is the expected temperature after the throttle?
Assumptions:
1 -The throttling device is adiabatic.
2 -Changes in potential energy are negligible.
3 -Changes in kinetic energy are negligible because the cross-sectional area for flow in the feed and effluent lines have been chosen to make the fluid velocity the same at the inlet and the outlet.
*Read throttling process, pg: 264
2. Steam at 9000 kPa and 600°C passes through a throttling process so that the pressure is suddenly reduced to 400 kPa. What is the expected temperature after the throttle?
Assumptions:
1 -The throttling device is adiabatic.
2 -Changes in potential energy are negligible.
3 -Changes in kinetic energy are negligible because the cross-sectional area for flow in the feed and effluent lines have been chosen to make the fluid velocity the same at the inlet and the outlet.
*Read throttling process, pg: 264
3. Air enters a turbine and power output of the turbine is at a rate of 200 kJ/min. The following data are known for the air entering and leaving the turbine. Determine the heat required from the turbine in kW.
Inlet condition Exit condition
Pressure 1 bar 6 bar
Temperature 260 K 400 K
Velocity 360 m/min 120 m/min
Elevation above reference
plane 4 m 2 m
MW air = 28.9 kg/kmol A cross-sectional area = 0.31 m2 g = 9.8066 m/s2
Ideal-gas properties of air
END
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