Thermodynamics lecture 11

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Lecture 11 FIRST LAW ANALYSIS FORLecture 11 – FIRST- LAW ANALYSIS FOR A CONTROL VOLUME

Recap: Problem 1

A piston/cylinder contains 50 kgof water at 200 kPa with avolume of 0.1 m3. Stops in thecylinder restricts the enclosedyvolume to 0.5 m3, as shown infig. The water is now heated to200oC Find the final pressure200oC. Find the final pressure,volume and the work done bythe water.

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Example 2A piston-cylinder arrangement has a linearspring and the outside atmosphere actingon the piston. It contains water at 3 MPaand 400oC with a volume of 0.1 m3. If thepiston is at the bottom the spring exerts apiston is at the bottom, the spring exerts aforce such that a pressure of 200 kPainside is required to balance the forces. The

t l til thsystem now cools until the pressurereaches 1 MPa. Find the heat transfer forthe process.

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Problem 3

An insulated piston cylinder devicecontains 5 L of saturated liquidcontains 5 L of saturated liquidwater at a constant pressure of175kPa. Water is stirred by a peddlewheel while current of 8 A flows for45 min through a resistor place inthe water. If 50% of liquid (by mass)q ( y )is evaporated during this constantpressure process and the peddlework amounts to 300kJ determinework amounts to 300kJ, determinethe voltage of the source. Also showthe on P-v diagram.


Fi 4 2 E l f k i th b d f t


Fig 4.2 Example of work crossing the boundary of a system because of electric current flow across the system boundary

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FIRST LAW ANALYSIS FOR AFIRST-LAW ANALYSIS FOR A CONTROL VOLUME

Conservation of mass and Control Volume

Control volumes:

Control Volume

Mass can cross the boundaries, and so we must keep track

of the amount of mass entering and leaving the controlof the amount of mass entering and leaving the control

volume.

Conservation of Mass Principle

Net mass transfer to or from a system during a process is

equal to the net change in total mass of the system


during that process.

Conservation of mass and Control VolumeControl Volume

∑∑VCdm&& ∑∑ −= ei

VC mmdt

..

Total mass inside the control volume

∫ ∫ +++=== .......)/1(.. CBAVC mmmdVvdVm ρ


Conservation of mass and Control VolumeControl Volume

Th l fl t iThe volume flow rate is

∫ ∨=∨= dAAV&

The mass flow rate becomes

∫ ∨=∨= dAAV local

∨=∨

=∨

=== ∫ AAdA)(VVm local ρρ&

&&


∨===== ∫ Av

dA)v

(v

Vm avg ρρ

FIRST-LAW ANALYSIS FOR A CONTROL VOLUMECONTROL VOLUMEFor a Fixed mass

12 EE − = 21Q − 21W

The instantaneous rate equation isThe instantaneous rate equation is

dtdE MC .. = Q& − W&dt Q

The amount of energy per unit mass in case of flow stream

gZue +∨+= 21


g2

FIRST-LAW ANALYSIS FOR A CONTROL VOLUMECONTROL VOLUME




FIRST-LAW ANALYSIS FOR A CONTROL VOLUMECONTROL VOLUMEThe rate of Flow Work

mPvVPAPFWflow &&& ==∨=∨=

• For the flow rate that leaves the control volume work is

being done by the control volume , eee mvP &

• For the mass that enters, surroundings do the rate of work,

eee

• the flow work per unit mass will be Pviii mvP &


p




Pve + = gZVPvu +++ 2

21

1

2

= gZVh ++ 2

21



ofRate ofRate ofRate

onAccumulatiofRate = Input

ofRate− Output

ofRate

Equation of Continuity

dtdm VC .. = ∑ im& − ∑ em&dt ∑ ∑ e



For a Fixed mass

12 EE −

For a Fixed mass

= 21Q − 21W12 21Q 21

The instantaneous rate equation isThe instantaneous rate equation is

dE MC Q& W&dtMC .. = Q − W



First Law Equation

ddE VC .. = VCQ& − W&dt ..VCQ

..VCW

+ iiem& − eeem&ii ee

+ inflowW& − outflowW&inflowW outflow

PW && PW &&


iiiinflow mvPW = eeeoutflow mvPW &=


First Law Equation

dE && )()(......

eeeeiiiiVCVCVC vPemvPemWQ

dtdE

+−++−= &&&&

1 )21( 2

.... iiiiVCVC gZhmWQ +∨++−= &&&2

)1( 2eeee gZhm +∨+− &


)2

( eeee g

FIRST-LAW ANALYSIS FOR A CONTROL VOLUMECONTROL VOLUMEIn general

1dE )gZ21h(mWQ

dtdE

i2iii.V.C.V.C

.V.C +∨++−= ∑ &&&

)21( 2

eeee gZhm +∨+−∑ &2

∑∑ −+−= e,totei,toti.V.C.V.C.V.C hmhmWQ

dtdE

&&&& ∑∑ e,totei,toti.V.C.V.CQdt

gZ1hh 2 +∨+Wh


gZ2

hhtot +∨+=Where

The Steady State Process

Assumptions:• The control volume does not move relative to the

coordinate frame• The state of the mass at each point in the control volumeThe state of the mass at each point in the control volume

does not vary with time i.e. no properties within the controlvolume changes with time.

• As for the mass that flows across the control surface the• As for the mass that flows across the control surface, themass flux and the state of this mass at each discrete areaof flow on the control surface do not very with time

• The rates at which the heat and work cross the controlsurface remains constant



Continuity Equation :

0dt

dm .V.C =

∑ & ∑ &

dt

∑ im = ∑ em



)gZ1h(mWQdEi

2iiiVCVC

.V.C +∨++−= ∑ &&& )gZ2

h(mWQdt iiii.V.C.V.C +∨++∑

)21( 2

eeee gZhm +∨+−∑ &

=>= 0dtdE .V.C

2

..VCQ& + ∑ ⎟⎠⎞

⎜⎝⎛ ++ iiii gZVhm 2

21

&⎠⎝

= ∑ ⎞⎜⎛ ++ gZVhm 21

& + VCW&


∑⎠

⎜⎝

++ eeee gZVhm2

+ ..VC


Continuity Equation :

im& = em& = m&eeeiii VAVA ρρ =

First Law:

Q& ⎞⎜⎛ 21

..VCQ +⎠⎞

⎜⎝⎛ ++ iii gZVhm 2

21

&

⎞⎛= ⎟⎠⎞

⎜⎝⎛ ++ eee gZVhm 2

21

& + ..VCW&



First in modified form

wgZV21hgZV

21hq e

2eei

2ii +++=+++ g

2g

2q eeeiii

Where,&

,

Qq VC

&

&..= m

Ww VC

&

&..=

mq

& m

The units are in kJ/kg


/ g


Examples of Steady State ProcessExamples of Steady State Process• Heat Exchanger• Nozzle• Diffuser• Throttle• Turbine• Pump

C• Compressor


Heat Exchanger

Heat exchangers are devices where two moving fluid streams

exchange heat with or without mixing. Heat exchangers are

widely used in various industries, and they come in various

designs.


Heat Exchanger


Heat Exchanger

The heat transfer associated with a heat exchanger may be zero


g y

or nonzero depending on how the control volume is selected.

Heat Exchanger


Heat Exchanger

h3 =452.34kJ/kg

h4 =249.10kJ/kg


Heat ExchangerTwo steady flows of air enters a control volume, shown in Fig One is 0 025 kg/s flow at 350 kPa 150°C state 1 andFig. One is 0.025 kg/s flow at 350 kPa, 150 C, state 1, and the other enters at 450 kPa, 15°C, both flows with low velocity. A single flow of air exits at 100 kPa, −40°C, state 3. y g , ,The control volume rejects 1 kW heat to the surroundings and produces 4 kW of power. Neglect kinetic and potential

i d d i h fl 2energies and determine the mass flow rate at state 2. Assuming constant Specific heat


Heat Exchanger

2321 025.0 mmmm &&&& +==+

CV332211loss WhmhmhmQ &&&& +=++−Substitute the work and heat transfer into the energy equation and

15005.1m150005.1025.0 2 ××+×× &

gy quse constant heat capacity

( ) 0.10.4)40(005.1m025.015005.1m150005.1025.0

2

2

++−××+=××+××

&

( )( ))40(15005.1

15040005.1025.00.10.4m2 −−−−××++

=&


s/kg0041.0m2 =&

Nozzle and Diffuser

Nozzles and diffusers are commonly utilized in jet engines, y j g ,

rockets, spacecraft, and even garden hoses.

f fA nozzle is a device that increases the velocity of a fluid at the

expense of pressure.

A diffuser is a device that increases the pressure of a fluid by

slowing it down.slowing it down.


Nozzle and Diffuser

The cross-sectional area of a nozzle decreases in the flow direction for subsonic flows and increases for supersonic flowsdirection for subsonic flows and increases for supersonic flows. The reverse is true for diffusers.

Nozzles and diffusers areNozzles and diffusers areshaped so that they causelarge changes in fluid

l iti d th ki tivelocities and thus kineticenergies.


Nozzle and Diffuser

Superheated vapor ammonia enters an insulated nozzle at 20°C 800 kPa shown in Fig with a low velocity and at the20 C, 800 kPa, shown in Fig., with a low velocity and at the steady rate of 0.01 kg/s. The ammonia exits at 300 kPa with a velocity of 450 m/s. Determine the temperature (or quality, ifvelocity of 450 m/s. Determine the temperature (or quality, if saturated) and the exit area of the nozzle


Superheated vapor ammonia enters an insulated nozzle at 20°C, 800 kPa, shown in Fig., with a low velocity and at the steady rate of 0.01 kg/s. The ammonia exits at 300 kPawith a velocity of 450 m/s. Determine the ytemperature (or quality, if saturated) and the exit area of the nozzle

s/kg01.0mm 21 == && Given inlet velocity =0, insulated nozzle,

∑∑ +∨++=+∨++ )gZ21h(mW)gZ

21h(mQ e

2eee.V.Ci

2iii.V.C &&&&

1 2eei 2

1hh ∨+=⇒

k/kJ651363hk/kJ91464h kg/kJ65.1363hkg/kJ9.1464h ei =⇒=Q

At 300 kPa, it is Sat. Liquid–sat. vapour mixture, so Te=-9.24 oC

and9471.0xe =∴From Continuity equation


26eeeee m10x56.8AvVAm −=⇒=&

Nozzle and Diffuser

Adiffuser, shown in Fig. has air entering at 100 kPa, 300 K,with a velocity of 200 m/s. The inlet cross-sectional area ofthe diffuser is 100 mm2. At the exit, the area is 860 mm2,and the exit velocity is 20 m/s Determine the exit pressureand the exit velocity is 20 m/s. Determine the exit pressureand temperature of the air.


A diffuser, shown in Fig. has air entering at 100 kPa, 300 K, with a velocity of 200 m/s. The inlet cross-sectional area of the diffuser is 100 mm2. At the exit, the area is 860 mm2, and the exit velocity is 20 m/s. Determine the exit ypressure and temperature of the air.

VAmVAm eee

iii === &&

2020011hh

V21hV

21h

vv

2222

2ee

2ii

ee

ii

⇒+=+

( ) K72.319005.18.19300

ChhTT

1000220

10002200V

21V

21hh

p

ieie

2e

2iie

=+=−

+=×

−×

=−=−

PRT

VAVA

PRT

VAVAvv

lawgasidealandequationcontinuityuseNow

e

e

ii

ee

i

i

ii

eeie

p

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

kPa92.12320860200100

30072.319100

VAVA

TTPP

ee

ii

i

eie

eiiiii

=⎟⎠⎞

⎜⎝⎛

××

⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎠⎝⎠⎝⎠⎝⎠⎝


Throttling valve

Throttling valves are any kind of flow-restricting devicesg y g

that cause a significant pressure drop in the fluid.

f fThe pressure drop in the fluid is often accompanied by a large

drop in temperature, and for that reason throttling devices

are commonly used in refrigeration and air-conditioning

applications.


Throttling valve


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Thermodynamics lecture 11