2-Lectures Ch04b 1st Law OS

8/3/2019 2-Lectures Ch04b 1st Law OS

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The First Law of

Thermodynamics

Open System(Control Volume)


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First low of thermodynamics foropen Systems

Reminder of an open System.

Open system = Controlvolume

It is a properly selected regionin space.

Mass and energy can crossits boundary.


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Control volume involves two

main processes

Steady flow processes. Fluid flows through the control volume

steadily.

Its properties are experiencing nochange with time at a fixed position.

Unsteady flow processes. Fluid properties are changing with time.


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Mass balance for steady flow

processes

We already showed that for steady flow

Many engineering devices involve a singlestream (one inlet and one exit only).

= mmei

21

mm =

22211VAVAor

1 =


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Energy Balance for Steady-

Flow Systems

2 21 1

( ) ( )2 2e e i iQ W m h V gz m h V gz = + + + +

sysoutinEEE =

outinEE =

in in i i out out e eQ W m Q W m + + = + +

2 21 1( ) ( )2 2

in out in out e e i iQ Q W W m Pv u V gz m Pv u V gz + = + + + + + +

2 21 1( ) ( )2 2

e e i iQ W m Pv u V gz m Pv u V gz = + + + + + +

0

, ,in in mass in out out mass out Q W E Q W E + + = + +


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Let us look at some common steady

flow devicesOnly one in and one out

More than oneinlet and exit


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For single stream steady flow

devices, the 1st low becomes

( ) ( )

++=

+++

eeeeiiii zgV

hmzgV

hmWQ 22

22

Often the change in kinetic energy and potential energy is small.

ei i eQ W m h m h

+ =

iehhwq = Per unit mass


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Nozzles

A nozzle is a device that

increases the velocity of a

fluid at the expense ofpressure

222111 AVAV =

2

1

2

112

A

AVV

=A1 A2

gasspeedlowfor

liquidsfor

21

21

=

12 VV >

21 AA >


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A1 A2

Diffusers

A diffuser is a device that

slows down the velocity of a

fluid causing an increase in its

pressure

12VV


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Diffusers


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Nozzles and Diffusers

(1st low analysis)

( )

+

+=

ie

ie

ie zzg

VV

hhmWQ 2

22

Is there work in this system?NO

Is there heat transfer?

In fact, it depends on the problem!

Does the fluid change elevation?NO

( )2

022

+= ieie

VVhh

?mthetohappenedWhat:Q

Ans: It is divided out

let us say: NO


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22

22

e

e

i

i

Vh

Vh +=+

How can you find the mass flow rate in a nozzle?

In a nozzle, enthalpy is converted into kinetic energy

torearrangedbecanwhich

222111AVAVm ==

2

22

1

11

vAV

vAVm ==


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Example (4-9): Deceleration of

Air in a DiffuserAir at 10oC and 80 kpa enters the diffuser of a jetengine steadily with a velocity of 200 m/s. The inletarea of the diffuser is 0.4 m2. The air leaves thediffuser with a velocity that is very small compare tothe inlet velocity.

Determine(1) The mass flow rate of the air and(2) The temperature of the air leaving the

diffuser.


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Example (4-10): Acceleration

of Steam in a NozzleSteam at 1.72 Mpa (250 psia) and 371C (700F) steadily enters a nozzle whose inlet areais 0.019 m2 (0.2 ft2.) The mass flow rate ofthe steam through the nozzle is 4.54 kg/s(10 lbm/s) . Steam leaves the nozzle at 1.38Mpa (200 psia) with a velocity of 274.3 m/s(900 ft/s). The heat losses from the nozzleper unit mass of the steam are estimated tobe 2.8 KJ/kg (1.2 Btu/lbm).Determine:(a) the inlet velocity and

(b) the exit temperature of the steam.

Answers: a)41.0 m/sor 134.4 ft/s,b)350Cor 661.9 F


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Solution of Example: Acceleration of Steam in a Nozzle

Note that there is heat transfer (Q).So, you have to go back to the general form

of the 1st low for single stream devices andget the following:

22

22e

ei

i

Vh

Vhq +=++


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TurbinesA turbine is a device that produces work at theexpense of temperature and pressure.

As the fluid passes through the turbine, work is doneagainst the blades, which are attached to a shaft. As a

result, the shaft rotates, and the turbine produces work.


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CompressorsA compressor is a device that increases thepressure of a fluid by adding work to the system.

Work is supplied from an external source through arotating shaft.

Compressor

in


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( )

++=

ieie

ie zzgVV

hhmWQ2

22

Turbines and Compressors

Is there work in this system? Yes!

Is there heat transfer? Negligible because of insulation. Exception: Internalcooling in some compressors.

Does the fluid change elevation? NO

Does the kinetic energy change? Usually it can be ignored

( ) )W(hhmieW =

( ) )kg/kJ(hhw ie =


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Example (4-12): Power

Generation by a Steam TurbineThe power output of an adiabatic steamturbine is 5 MW, and the inlet and the

exit conditions of the steam are asindicated in the figure on the right.

a) Compare the magnitude of h, ke,and pe.b) Determine the work done per unitmass of the steam flowing through theturbine.

c) Calculate the mass flow rate of the

steam.

Answers: a)h = -885.9 kJ/kg,ke = 14.95 kJ/kg, pe = -0.04 kJ/kg,b) 871.0 kJ/kg, and c) 5.74 kg/s


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Throttling ValveA throttling valve reducesthe fluid pressure.

P1>P2

It is small device and thus

the flow through it may beassumed adiabatic (q=0)since there is neithersufficient time nor large

enough area for any effectiveheat transfer to occur.


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( )

++=

ieie

ie zzgVV

hhmWQ2

22

Throttling Valve

Is there work in this system? NO

Is there heat transfer?

Usually it can be ignoredDoes the fluid change elevation? NO

Does the fluid change velocity? Usually it can be ignored

( )0 e i e ih h h h

isenthalpicdevice

= =


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What happens to the fluid temperature

a cross throttling Valves ?

21 hh=

222111vpuvpu +=+

Tuuvpvpif121122

>< Tuuvpvpif121122


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Throttling Valves

(incompressible substance )For incompressible substance (like water), is constant

21 vv = 21 PP>

risewillT

uuvpvp

>


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Throttling Valves (compressible

substance: Vapor)Example (4-13): Expansion of Refrigerant-134a in aRefrigerator

Refrigerant-134a enters the capillary tube of arefrigerator as saturated liquid at 0.8 MPa and isthrottled to a pressure of 0.12 MPa.

Determine the quality of the refrigerant at the final stateand the temperature drop during this process.

S l ti f E l E i f R f i t 134 i R f i t


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Solution of Example: Expansion of Refrigerant-134a in a Refrigerator

Notice that T2


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Throttling Valves (Compressible

substance: an Ideal Gas) What happens if the gas is ideal?

For ideal gases h = Cp T But h = 0

So T = 0

The inlet and outlet temperatures are the same!!!


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Mixing ChamberMixing two or more fluids is a

common engineering process

The mixing chamber does not have tobe a distinct chamber. An ordinary T-

elbow, or a Y-elbow in a shower, forexample, serves as the mixingchamber for the cold- and hot-water

streams as shown in the figure (Left).

Mixing

Chamber


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Mixing Chamber

++

++=

ii

iiee

eegzVhmgzVhm

netnet WQ 22

22

We no longer have only one inlet and one exit stream

Is there any work done? No

Is there any heat transferred? No

Is there a velocity change? No

Is there an elevation change? No

( ) ( )

= iiee hmhm0


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Mixing Chamber

Material Balance

Energy balance

eim m

= 321 mmm

=+

Mixing

Chamber

Mixing

Chamber

i e eim h m h

= 332211 hmhmhm

=+


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Example(4-14): Mixing of Hot and ColdWaters in a Shower

Consider an ordinary shower wherehot water at 140oF is mixed with cold

water at 50oF. If it is desired that asteady stream of warm water at 110oFbe supplied, determine the ratio ofthe mass flow rates of the hot to cold

water. Assume the heat losses fromthe mixing chamber to be negligibleand the mixing to take place at apressure of 20 psia.


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Heat Exchanger

A heat exchanger is a

device where two moving

fluids exchange heat

without mixing.


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Heat Exchangers

Your analysis approach will depend on how you define your system


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Heat Exchangers

System (a): entire HX

Mass Balance

Divide into two separatestreams with equal inletand outlet flow rates

Energy balance

Two inlets

Two outlets

1 1 3 3 2 2 4 4m h m h m h m h+ = +

1 2 3 4,m m m m= =


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Heat Exchangers

System (b): Single stream

Mass Balance

Considering one singlestream with one inlet andone outlet flow rates

Energy balance

One inlet

One outlet Plus heat transfer

2 2 1 1

Q m h m h=

1 2m m=


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Example: Cooling of Refrigerant-134a by Water

Refrigerant-134a is to be cooled bywater in a condenser. The refrigerantenters the condenser with a massflow rate of 6 kg/min at 1 MPa and70oC and leaves at 35oC. The coolingwater enters at 300 kPa and 15oC andleaves at 25oC. Neglecting any

pressure drop, determine(a) the mass flow rate of the coolingwater required and (b) the heattransfer rate from the refrigerant towater.

2-Lectures Ch04b 1st Law OS

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