2-Lectures Ch04a 1st Law CS

8/10/2019 2-Lectures Ch04a 1st Law CS

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The First Law of

Thermodynamics

Closed System


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So far

Weve considered various forms of energy

Q (heat) W (work)

E (Energy)

We havent tried to relate them during aprocess

They are related through the

FIRST LOW OF THERMODYNAMICS (CONSEVATION OF ENERGY PRINCIPLE)


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Recall

Experimentally observed: all adiabatic

processes between two specified states of aclosed system, the net work done is the same

regardless of the nature of the closed system

and the details of the process. This principle is called the first law

A major consequences of the first law is the

existence and the definition of the property

total energy E


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First low of thermodynamics

for Closed Systems

Reminder of a Closed System.

Closed system = Control mass

It is defined as a quantity ofmatter chosen for study.

No mass can cross its boundarybut energy can.


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Experience 1: W= 0

If we transfer 5 kJ of heat to a potato,

its total energy will increase by 5 kJ.


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Experience 3: Q=0

Heat an insulated room with electric heater.,

W= 5KJ (electric)

means E2-E1= 5KJ.

Replace the electricheater

with a paddle wheel.


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Experience 4: Q=0

If we do a 10 kJ of boundary

work on a system,

the systems internal energy

will increase by 10 kJ.

This is because (in theabsence of any heat transfer (Q

= 0), the entire boundary work

wil l be stored in the air as partof its total energy.


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Experience 5: Qin, Qout, Win

If we do 6 kJ of shaft

work on system and Transfer 15 kJ of heat

in.

Loose 3 kJ of heat

out.

Doing 4 kJ of Work

out.

Then the systeminternal energy will

increase by 14 kJ.


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Before proceeding further, let us fix

the direction of Energy Transfer

Heat transferred in to the system is positive

Qin is +ve

Heat transferred out ofthe system is negative

Qout is -ve

Work done on the system increases energy of thesystem.

Win is +ve

Work done by the system decreases energy of thesystem.

Wout is -ve


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Based on the previous experimental

observations, the conservation of energyprinciple may be expressed as:

=

energysystem

totaltheinChange

SystemtheLeaving

EnergyTotal

SystemtheEntering

EnergyTotal

sysoutin EEE =

sysoutin EEE

=Rate of net energy transfer

by heat, work and mass

Rate of change in total

energy of the system

On a rate basis


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Usually the KE and PE are small

PEKEUE ++=

umUE ==

Let us discuss the right hand side of thefirst low equation, i. e. E


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Let us turn to the Left hand side of the 1st

low equation

EEEoutin

=

E)EWQ(EWQ out,massoutoutin,massinin =++++

0

0

E)WQ(WQ outoutinin =++

Emass=0 for closed system

inE outE

EWWQQoutinoutin

=+If we rearrange, we get


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EWQ =

EWQ =

ewq =

dewq =

EWWQQoutinoutin

=+

EWQ out,netin,net =

Assuming Qin>Qout and Wout> Win

per unit time (or on a Rate basis) KJ/s= Watt

per unit mass basis (KJ/kg)

differential form

General Form (KJ)

usually we drop the subscripts, hence

The First low of thermodynamics


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Energy Change for a system

undergoing a cycle

Q-W =

U

But U = U2-U1=0

(initial state and final

states are the same)

Hence : Q=W


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Example (4-1):

Cooling of a Hot fluid in a Tank

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 800kj. Duringthe cooling process, the fluid loses 500kj ofheat, and the paddle wheel does 100 kj of work

on the fluid. Determine the final internal energyof the fluid.

Assumptions:

Tank is stationary and thus KE=PE=0. Analysis:

Q-W=U2-U1


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Example (4-2):

Electric Heating of a Gas at Constant Pressure

A piston-cylinder device contains 25 g of

saturated water vapor that is maintained at

a constant pressure of 300 kPa. Aresistance heater within the cylinder is

turned on and passes a current of 0.2 A for

5 min from a 120-V source. At the same

time, a heat loss of 3.7 kJ occurs.

Show that for a closed system the

boundary work Wb and the change in

internal energy U in the first-law

relation can be combined into one

term, H, for a constant pressure

process.

Determine the final temperature of the

steam.


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Example (4-6):

Heating of a

Gas at

Constant

Pressure

A piston-cylinder device initially contains air at 150 kPa

and 27oC. At this state, the piston is resting on a pair of

stops, as shown in the figure above, and the enclosedvolume is 400 L. The mass of the piston is such that a

350-kPa pressure is required to move it. The air now

heated until i ts volume has doubled. Determine (a) the

final temperature, (b) the work done by the air, and c)the total heat transferred to the air.

Sol:

2-Lectures Ch04a 1st Law CS

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