chapter 1 and 2 (contd)

7/28/2019 Chapter 1 and 2 (Contd)

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CHEMICAL ENGINEERINGTHERMODYNAMICS

Course no: CHE C311 / F213

Dr. Srinivas Krishnaswamy1st Semester 2012

2013

DEPT. OF CHEMICAL ENGG.

BITS PILANI K. K. BIRLA GOA CAMPUS


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Contents of lectures 2

Joules experiment

First law of thermodynamics for a systemundergoing a thermodynamic cycle

Energy balance: Law of conservation ofenergy

First law of thermodynamics for a process in aclosed system

Perpetual motion machine of the first kindThe concept of internal energy

The concept of enthalpy


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Process and Cycle

A cycle as discussed previously consistsof processes

A cycle may comprise of any number ofprocesses

The minimum number of processesrequired to complete a cycle is 2

One can have a cycle with only onecyclic process with no discontinuities


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Joules experiment

This experiment led to the formulationof the first law of thermodynamics

A two process cycle carried out on asystem with a fluid

W

Insulated system

Insulated system

Q


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Joules experimentJoule repeated his experiment for varioussystems and for different amount of work

interactions and measuring the correspondingamount of heat transfer in each case forbringing system back to original state.

He found that net work input W was always

equal to the net amount heat transfer Qfrom the system (equal in magnitude whensame units are used)


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First Law for a system undergoing a

thermodynamic cycleSince system at the end is restored to its

original state he concluded that the

algebraic sum of heat and work interactionsduring a thermodynamic cycle is zero, thus

During a cycle, a system undergoes, the cyclic

integral of heat added is equal to the cyclicintegral of work done

WQ


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First Law for a system undergoing athermodynamic cycle

The symbol is used to indicate that

Wand Qare inexact differentialsNote the constraints of this law:applies to a closed system and only to

a thermodynamic cycleBoth heat and work have same units


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Energy balance: Law of conservation

of energyEverything discussed so far was for closedsystems undergoing a thermodynamic cycle.

What happens in open systems?Remember the section on control volume taughtearlier. A boundary is drawn around the wholeequipment when analyzing such systems

In all heat engines and refrigerants (opensystems), the working system undergoes achange of state and is finally restored to its initialstate


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Energy balance: Law of conservationof energy

Q= WQb Qc = WtWp

Qb = Qf- Qflue

Boiler

Turbine

Condenser

Pump

Qf

Qflue

Wt

Qc

Wp

Whole system treated as

closed whereas individual

equipment separately

require analysis as an open

system


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Energy balance: Law of conservationof energy

The basis of first law is thus the law ofconservation of energy

Net energy added to a system = Energyin Energy out = increase in stored

energy

But for a cycle no change in state sostored energy increase is zero


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First Law for a process in a closedsystem

2

B

E

P

1

A

C

E2E1

X2

X1 B1-2A2-1B1-2A2-1

WWQQ Path 1-A-2-B-1

Path 1-A-2-C-1

C12B12

WQWQSubtracting and rearranging

C1-2A2-1C1-2A2-1

WWQQ


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First Law for a process in a closed

systemIt is seen that for any change in the stateof a system, the quantity (Q W) is

always the sameIt is independent of the path followed forthis change of state

It is a point or state function and hence aproperty

This property is called Stored energy


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First Law for a process in a closed

system

dEWQ

dEWQ

dE used because it is an exact dif ferential. Integrating

211221 WEEQFIRST LAW FOR A GENERAL PROCESS

WHERE E2 E1 IS STORED ENERGY

INCREASE


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First Law of thermodynamics for a

process in a closed systemNote heat is transferred to a system when work isdone by the system during which stored energy

increasesHeat added to a system goes to increase storedenergy and can result in work

Representation of law of conservation of energy

For isolated system Wand Q= 0. Hence E2 = E1The stored energy of an isolated system remainsconstant


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Perpetual motion machine of the

first kindSimply put it is a device that produceswork from nothing

It creates its own energyIt violates the first law ofthermodynamics

All attempts to make such a machinehave failed, thus providing experimentalproof of the validity of the First Law


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ProblemDuring a process a system receives 10 Jof energy by the work mode and 15 J of

energy by the heat mode? What is theincrease in stored energy?

211221 WEEQ

JUST UNDERSTAND THE

CONVENTIONS (Ans: 25 J)


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Summary

The first law is nothing but the law ofenergy conservation

In the absence of any work interactions,

energy change of a system is equal to thenet heat transfer

The work done (electrical or shaft) on anadiabatic system is equal to the increase inthe stored energy of the system

It says the total quantity of energy in theUniverse is constant


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The concept of internal energy

Stored energy represents energy present inmacroscopic and microscopic forms

Macroscopic: KE and PE

Microscopic: Energy associated with motion,position and chemical bonding andconfiguration of individual molecules

In absence of electrical, magnetic, surfacetension effects

E =U+ KE + PE


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The concept of internal energy

For stationary systems KE and PE = 0 andE= U

Uis known as internal energy

Q W= Uor Qnet in Wnet out = UsystemNo process in nature has been known to violatethe first law of thermodynamics

Remember: Energy is not an absolute quantity, butis rather defined only relative to a reference statewhich must be identified first


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Problem

Consider a stone having a mass of 10 kg and abucket containing 100 kg of liquid water. Initiallythe stone is 10.2 m above the water and the stoneand the water are at the same temperature, State1. The stone then falls into the water. Determine

U, KE, PE, Qand W for the following changesof state assuming g =9.80665 m/s2

a. The stone is about to enter the water, state 2

b. The stone has just come to rest in the bucket state

3c. Heat has been transferred to the surroundings in

such an amount that the stone and water are at thesame temperature


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Solution Procedure

The first law for any stepQ =U+ KE + PE +W

For case a stone has fallen from Z1 to Z2 and U = Q = W =0 (no change of state)

KE + PE = 0 (hence find KE and PE)For process 2 3 with zero kinetic energy

PE = Q = W =0

KE + U= 0 (hence find internal energy)

In the final state, there is no kinetic nor potential energy andthe internal energy is the same as in state 1, Q3-4 = U


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Internal energy of a pure substance

The internal energy units are the sameas the units of heat and work (Joule)

Internal energy per unit mass is specificinternal energy (Tables give this value)

First Law on a per unit mass basis canbe written as q w= e

Calculations are same as that forspecific volume


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Enthalpy is defined as the total energycontent in a system

It is a property like internal energyEnthalpy per unit mass is specificenthalpy

Enthalpy has same units as that ofinternal energy or work and is denotedby Hor h


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The concept of enthalpyMost processes in nature take place atconstant pressure as compared to

constant volumeSo if the volume is not constant duringa process, some provision must bemade to accommodate this change involume, i.e. some energy must be spenton this (increasing or decreasingvolume)


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So for example, if heat is supplied to aprocess (in which volume is changing), allthe heat will not go to increase the internal

energy. Some of it will be used to accountfor change in volume (as work)

The same applies if in a process, volumedecreases

Sounds complicated. So to simplify thingsand account for this work, the termenthalpy is used


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Consider decomposition reactions for whichheat is supplied. Some of this heat will go toincrease volume and some will cause thereaction to take place (internal energyincrease). The increase in internal energy willbe lower by an amount equal to that used tochange volume

For an exothermic reaction in which there isdecrease in volume, work has to be done onthe system to cause a contraction. This workgets converted into heat and hence thedecrease in internal energy will be less becauseof heat obtained from the work done


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So in case 1

Total heat (energy) supplied = Reaction(stored energy increase) + expansion work

In case 2Total heat (energy) given out = Heat from

reaction (expected to decrease storedenergy) + heat from work of compression

This total heat is defined as enthalpy andwritten as

H= U+ PVor h= u+ P

Fi l i f


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First law equation for areversible process

Work done by a system of fixed mass during areversible process is Pd

Substituting in the first law

q= du+ Pd

------- (1)Enthalpy = h= u+ P

dh =du+ d(P) = du+ Pd+ dP

Substituting in 1

q= dh dP(Important result)General result and must be used when bothpressure and volume are changing. Note how

enthalpy is distributed


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hbeing defined in terms of state functions P,, and u is a state function

The change in enthalpy between states 1 and

2 is given as

H=

U+ P

VEnthalpy is always greater than the internalenergy of a system and increases withtemperature

Tables of thermodynamic properties givevalues ofh(specific enthalpy)

Calculations are same as that for specificvolume and specific internal energy


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Enthalpy at constant pressure

For a constant pressure processH= U+ PV= Q - PV+ PV = Q

For a constant pressure process the change in

enthalpy is equal to heat absorbed by a system.In other words the heat supplied to a system atconstant pressure goes to increase the enthalpy

of a system

Thus if 10 kJ is supplied to a system that is free tochange volume at constant pressure, theenthalpy changes by 10 kJ regardless of how

much energy goes to do work


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Objective assessmentDifference between conservation andBalance

First law for a cycle

First law for a process

Sign conventions

Internal energy and enthalpy

Problem solving


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Beliefs are what

divide people. Doubtunites them.

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chapter 1 and 2 (contd)

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