1. Chemical Engineering Thermodynamics (CHPE208) Summer 2014-15
Chapter 2
First law of thermodynamics
Chapter 2 – First Law of Thermodynamics

2. Chapter -2 First law of thermodynamics
Joule’s experiment
Internal Energy
The first law of thermodynamics
Energy balance for closed systems
Thermodynamic state and state functions
Equilibrium
The Phase Rule
The reversible processes
Constant-v and Constant-p processes
Enthalpy
Heat Capacity
Mass and Energy Balances for Open Systems
Chapter 2 – First law of thermodynamics

3. Chapter 2 – First Law of Thermodynamics
Joule’s experiment
Known amounts of water, oil, and mercury taken in insulated containers
Agitated the fluids with rotating stirrer
Measured the temperature changes
Amount of work done were measured accurately
Temperature changes were noted
Findings:
(i) for each fluid a fixed amount of work was required per unit mass for every degree of temperature rise caused by stirring.
(ii) Original temperature could be restored by transfer of heat through simple contact with a cooler object
(iii) A quantitative relationship exists between work, and heat and therefore, heat is a form of energy
Chapter 2 – First Law of Thermodynamics

4. Chapter 2 – First Law of Thermodynamics
Internal Energy (U)
In Joule’s experiment, energy added to a fluid as work is later transferred from the fluid as heat. Where is this energy between its addition to and transfer from the fluid?
It is contained in the fluid in another form, called internal energy
Internal energy of a substance does not include energy that it may possess as a result of its macroscopic position or movement. Rather it refers to energy of the molecules internal to its substance.
Internal energy of a substance includes kinetic energy of molecules resulting from their ceaseless motions such as rotation, vibration, and translation, and the potential energy resulting from inter molecular forces.
Internal energy cannot be directly measured
Chapter 2 – First Law of Thermodynamics

5. Change in Internal Energy (U )
E = Total energy of the System = internal + external (unit: joules)
The internal energy, U = potential energy from intermolecular forces
External energy: kinetic/potential energy -macroscopic
When external energy = 0, E = U
Although the absolute value of the internal energy of a system cannot be measured, the changes in the internal energy can be measured.
The internal energy U of a system will change if :
Heat, Q passes into or out of the system
Work, W is done on or by the system
Mass enters or leaves the system
Chapter 2 – First Law of Thermodynamics

6. Changes in Internal Energy
Closed system: no transfer of mass is possible: internal energy may only change due to heat and work.
Isolated system: no change in the internal energy is possible: heat, work and mass transfer are all impossible.
Open system: the internal energy may change due to transfer of heat, mass and work between system and surroundings.
Chapter 2 – First Law of Thermodynamics

7. The First Law of Thermodynamics
The recognition of heat and internal energy as forms of energy makes possible a generalization of the law of conservation of mechanical energy to include heat and internal energy in addition to work and external potential and kinetic energy. Generalization can be extended to still other forms, such as surface energy, electrical energy, and magnetic energy.
First Law of Thermodynamics: Although energy assumes many forms, the total quantity of energy is constant, and when energy disappears in one form it appears simultaneously in other forms.
First Law of Thermodynamics (Conservation of Energy): Energy can neither be created nor destroyed, only changed from one form into another.
Note: heat and work represent energy in transit across the boundary dividing the system from its surroundings, and are never stored or contained in the system.
Potential, kinetic, and internal energy reside with and are stored with the matter.
Chapter 2 – First Law of Thermodynamics

8. Energy Balance for closed systems
First law applies to both system and surroundings
(Energy of the system) + (Energy of Surroundings) = 0
Closed system: no material (mass) is transferred across the system boundaries. Only heat and work interactions occur between the system and its surroundings.
As the gas is being compressed, the work is done by the surrounding, as a result, the system will cool down. W > 0 (work is positive)
As a real gas expands, the work is done by the system and as a result the system will heat up W < 0 (work is negative)
We balance the changes in the internal energy of the gas with the amount of heat transferred to/from the gas and work done by/on the gas. This energy balance is called the first law for a closed system. It is written in
differential form (infinitesimal change): dU = dQ + dW
integrated form: U = Q + W
The sign conventions for heat transfer and work are
Heat transferred IN the system is POSITIVE, heat transferred OUT of the system is NEGATIVE Heat
Work done by the surrounding is POSITIVE Work, work done by the system is a NEGATIVE Work
As the surrounding done work on the system( real gas), the temperature and pressure of the system(real gas) increases, some heat will be trans
Chapter 2 – First Law of Thermodynamics

9. State and Path Functions
State function: is one whose change on going from initial to final is independent of the route taken. (thermodynamic function)
Path function: if the change in a function is dependent on the route taken. (Not thermodynamic function)
Some familiar state functions are volume, V, pressure, P, and temperature, T. For example, if we take a state from 0oC to 100oC, the change in the temperature is +100oC whether we go straight up the temperature scale or we first cool the system for a few degrees then take the system to the final temperature. The change in temperature is independent of the route taken so temperature is a state function.
Question: Are these state or path functions: U and W?
Answer: W is a path function; U is a state function
Chapter 2 – First Law of Thermodynamics

10. Thermodynamic State and State Functions
Note: For a closed system undergoing the same change in state by several processes, experiment shows that the amount of heat and work required differ for different processes, but that the sum Q + W is the same for all processes. This is the basis for identification of internal energy as a state function. The U is same for a change in system between same initial and final states regardless of the process.
Chapter 2 – First Law of Thermodynamics

11. Chapter 2 – First Law of Thermodynamics
The Phase Rule 
Chapter 2 – First Law of Thermodynamics

12. Reversible and Irreversible Processes
A reversible process is one that can be halted at any stage and reversed. In a reversible process the system is at equilibrium at every stage of the process.
An irreversible process is one where these conditions are not fulfilled.
If pint > pext in an expansion process then the process is irreversible because the system does not remain at equilibrium at every stage of the process. (There will be turbulence and temperature gradients, for example.) 
Chapter 2 – First Law of Thermodynamics

13. Chapter 2 – First Law of Thermodynamics
Reversible Process
Reversible process: direction of process can be reversed at any point by an infinitesimal change in external conditions
It is frictionless and when reversed, retraces its forward path and restores initial state of system and surroundings
“Loss of work” occurs when dissipative mechanisms exist through friction, viscous fluids, etc. Real processes involve lost work.
If mechanically reversible:
If irreversible, pint > pext :
The reversible work as the limiting value can be combined with appropriate efficiency to yield a reasonable approximation of the work of an actual process
Chapter 2 – First Law of Thermodynamics

14. Example: Reversible process
A horizontal piston-cylinder arrangement is placed in a constant Temperature bath. The piston slides in the cylinder with negligible friction, and an external force holds it in place against an initial gas pressure of 1000 kN m-2. The initial gas volume is 0.1 m3. The external force on the piston is to be reduced gradually, allowing the gas to expand until its volume doubles. Under the conditions it has been determined that the volume of the gas related to its pressure in such a way that PV is constant. Calculate the work done by the gas in moving the external force:
Solution: Process is reversible
Chapter 2 – First Law of Thermodynamics

15. Example: Irreversible process
How much work will be done if the external force were suddenly reduced to half its initial value instead of being gradually reduced?
Solution:
Here external force is constant and is 500 kN/m2. There will be a sudden expansion and final volume of gas is 0.2 m3. Thus DV is the same as before. This second process is irreversible.
Chapter 2 – First Law of Thermodynamics

16. Chapter 2 – First Law of Thermodynamics
Constant-V Process
For n moles of a homogeneous fluid contained in a closed system, the energy balance is:
d(nU) = dQ + dW
The work for a mechanically reversible, closed system is given by:
dW = -P d(nV)
Combining the two equations, the general balance for n moles of homogeneous fluid in a clsoed system undergoing a mechanically reversible process: d (nU) = dQ – Pd(nV)
Constant volume process: work is zero
dQ = d (nU) (const V)
Integration yields Q = nU (const V)
Chapter 2 – First Law of Thermodynamics

17. Constant-pressure process and Enthalpy (H)
General energy balance: d (nU) = dQ – Pd(nV)
For a constant-pressure process
dQ = d(nU) + P d(nV)
dQ = d(nU) + d (nPV) = d[n(U + PV)]
The term (U+PV) is called as enthalpy: H = U + PV
dQ = d(nH) (const P)
Q = nH (const P)
Thus the heat transferred in a mechanically reversible, constant-pressure, closed-system process equals the enthalpy change of the system.
Change in enthalpy:
dH = dU + d(PV), Upon integration H =  U + (PV)
Chapter 2 – First Law of Thermodynamics

18. Example Problem: calculations of U and H
Solution:
Chapter 2 – First Law of Thermodynamics

19. Chapter 2 – First Law of Thermodynamics

20. Heat capacity at constant volume (CV)
The constant-volume heat capacity of a substance is defined as
dU = CVdT (const V)
For a mechanically reversible, constant-volume process
Q = n U
Chapter 2 – First Law of Thermodynamics

21. Heat capacity at constant pressure (CP)
The constant-pressure heat capacity of a substance is defined as
dH = CPdT (const P)
For a mechanically reversible, constant-volume process
Q = n H
Chapter 2 – First Law of Thermodynamics

22. Chapter 2 – First Law of Thermodynamics
Example: Cv and Cp
Chapter 2 – First Law of Thermodynamics

23. Chapter 2 – First Law of Thermodynamics
Solution:
Chapter 2 – First Law of Thermodynamics

24. Chapter 2 – First Law of Thermodynamics

25. Chapter 2 – First Law of Thermodynamics

26. Chapter 2 – First Law of Thermodynamics

27. Chapter 2 – First Law of Thermodynamics
Open system
Open system: material (mass) can flow across the boundaries of the system, as well as heat and work.
eg: gas enters a system and is heated where its T, P and V increase. The gas passes through an expansion turbine, where it cools down and the P decreases. During the expansion process work is extracted from the gas.
The first law written for an open system:
Hout – Hin = Q + W
where Hin = enthalpy flowing in through the dotted boundary line
Hout = enthalpy flowing out through the dotted boundary line.
The sign conventions are:
Heat transferred to the system is POSITIVE
Work done on the system is POSITIVE
The first law for a flow process is written in terms of the enthalpy H, because the enthalpy includes the PV work done on the system by the gas flowing in and the PV work done by the system as the gas flows out.
Chapter 2 – First Law of Thermodynamics

28. Measures of flow in open systems
Measures of flow rate
Name Notation Units
Mass flow rate kg/s
Molar flow rate mol/s
Volumetric flow rate q m3/s
Velocity u m/s
Relationship between various measures
Chapter 2 – First Law of Thermodynamics

29. Open Systems: Mass and Energy Balance
For detailed derivation, refer to textbook – open systems: pg 45 ( 7th Edition)
Chapter 2 – First Law of Thermodynamics

30. Example: open system calculations
Water at 93.5ºC ( K) is pumped from a storage tank at the rate of 3.15×10-3 m3s-1. The motor for the pump supplies work at the rate of 1.5 kW. The water goes through a heat exchanger, giving up heat at the rate of 700 kW, and is delivered to a second storage tank at an elevation 15 m above the first tank. What is the temperature of the water delivered to the second tank?
Answer:
Given Data: Water inlet temperature = K
Volume flow rate = 3.15×10-3 m3s-1
Input Pumping work = 1.5 kW
Heat loss, Q = 700 kW
Elevation = 15 m
Chapter 2 – First Law of Thermodynamics