Open Flow Systems

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Open Flow Systems

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Chapter 2. The First Law and

Other Basic Concepts

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2.12 Mass and Energy Balance for Open Systems

Measures of flow

u

q n

m

velocity

rate flow

volumetric

rate flow

molar

rate flow

mass

Control Volume

control surface

m1

m2

m3

n M m  

uA q

uA m 

uA n 

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Mass Balance for Open Systems

Mass balance for open system

Steady State : no accumulation term

Only single entrance and single exit stream

0 )

( 

fs

cv m

dt

dm

2 1

) 3

(mfsm m m

0 )

( 

fs

cv uA

dt

dm

Often called “Continuity equation”

0 )

( 

uA fs

0 )

( )

(

uA 2

uA 1m (

uA)2 (

uA)1 const

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The General Energy Balance

Net rate of change of energy within control volume equals the net rate of energy transfer into control volume

rate work 2

1 )

( 2

 

 

 

 

  

U u zg m Q

dt mU

d cv  

PV work Stirring work Shaft work

PV m

W

Q m

zg u

dt U mU d

fs

cv       

 

 

 

  

 ( )

2 1 )

( 2

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The General Energy Balance

W Q

m zg u

dt H mU

d cv     

 

 

 

  

2

2 ) 1

(

W Q

m zg u

dt H mU d

fs

cv     

 

 

 

  

2

2 1 )

(

kinetic and potential energy changes are negligible

W Q

m dt H

mU d

fs

cv ( )    

) (

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W

s

– Shaft work?

Shaft work : work done on the system by a moving part within the system

Components such as turbines, pumps, and compressors – all operate by energy

transfer to or from the working-fluid

Energy transfer usually through blades rotating on a shaft

Also fluid dynamics problem…

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Energy Balances for Steady-State Flow Processes

Steady state : no accumulation term

s fs

W Q

m zg u

H     

 

 

 

  2  2

1

Ws

Q zg

u

H   

 

  2  2

1

Ws

Q H  

kinetic and potential energy changes are negligible

Only shaft work

No stirring work for steady-state

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Example 2.14

An insulate, electrically heated tank for hot water contains 190 kg of liquid water at 60 oC when a power outage occurs. If water is

withdrawn from the tank at a steady rate of m = 0.2 kg/s, how long will it take for the temperature of the water in the tank to drop from 60 to 35 oC? Assume cold water enters the tank at 10 oC, and negligible heat losses from the tank. For liquid water let Cv = Cp = C, independent of T and P.

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Example 2.14 - solution

Energy balance become

W Q m

dt H mU d

fs

cv ( )

)

( m(H H1 ) 0

dt

m dU

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Example 2.16

Air at 1 bar and 25 oC enters a compressor at low velocity, discharges

at 3 bar, and enters a nozzle in which it expands to a final velocity of 600 m/s at the initial conditions of P and T.

If the work of compression is 240 kJ per kg of air, how much heat must be removed during compression?

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Example 2.16 - solution

Energy balance

Ws

Q zg

u

H

2 2

1 Q u22 Ws

2 1

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Chapter 5. The Second Law of

Thermodynamics

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5.6 Mathematical Statement of the Second Law

Mathematical Statement of the Second Law

Every process proceed in such a direction that the total

entropy change associated with it is positive, the limiting value of zero being attained only by a reversible process.

No process is possible for which the total entropy decreases.

Work produced by engine

 0

S

total

C C H

H

toal T

Q T

SQ

C

H Q

Q W  



 

 

H C H

toal

C T

Q T S

T

W 1

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Example 5.4

A 40-kg steel casting (Cp = 0.5 kJ/kgK) at a temperature of 450 oC is

quenched in 150 kg of oil (Cp = 2.5 kJ/kgK) at 25 oC. If there are no heat losses, what is the change in entropy of (a) the casting, (b) the oil, and (c) both considered together?

From the energy balance, Q = 40×0.5×(T – 450) + 150×2.5×(T – 25) = 0

 T = 46.5 oC (a) S of the casting

K / kJ 33 . 450 16

15 . 273

52 . 46 15

. ln 273 5

. 0 T 40

C dT m

S T

T p

0

(b) S of the oil

K / kJ 13 . 26 2 5

15 . 273

52 . 46 15

. ln 273 25

T 150 C dT

m

S T

T p

0

(c) Total S change

K / kJ 80 . 9 13 . 26 33

. 16

S

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5.7 Entropy Balance for Open Systems

Entropy Balance Equation

Important characteristics

- Entropy is not conserved

- Entropy generation term is required

) 0 ) (

(    

fs cv surrt SG dt

dS dt

mS m d

S  

j j

j t

surr

T Q dt

dS

) 0 ) (

(    

G

j j

cv j

fs S

T Q dt

mS m d

S  

Page 176 for description

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Example 5.5

In a steady-state flow process, 1 mol/s of air at 600 K and 1 atm is continuously mixed with 2 mol/s of air at 450 K and 1 atm. The product stream is at 400 K and 1 atm. A schematic representation of the process is shown in Fig. 5.7. Determine the rate of heat transfer and the rate of entropy generation for the process. Assume that air is an ideal gas with Cp= 7/2R, that the surroundings are at 300 K and that kinetic- and potential-energy changes are negligible.

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Example 5.5 - solution

From the energy balance on this process,

s fs

2 zg m Q W

2u

H 1

From the entropy balance,

  H m

fs

Q

1 (400 600) 2 (400 450)

8,729.7J / s

2 R

Q 7

s K / J . .

ln ln

) .

)(

/ (

T Q T

ln T C T n

ln T C n

T ) Q S S ( n ) S S ( T n

) Q m S ( S

B P

B A

P A

B B

A A

j fs G





446 300 10

7 8729 450

2 400 600

1 400 314

8 2 7

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Example 5.6

An inventor claims to have devised a process which takes in only saturated steam at 100 oC and which by a complicated series of steps makes heat

continuously available at a temperature level of 200 oC, where 2,000 kJ of energy as heat is liberated for every kg of steam taken into the process. Show whether or not this process is possible. To give this process the most favorable conditions, assume cooling water available in unlimited quantity at a temperature of 0 oC.

Use the basis as 1 kg of steam.

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Example 5.6 - solution

- The values of H and S for saturated

steam at 100 oC and liquid water at 0 oC can be found in the steam table.

1. The energy balance

HQQ'Q  2,000Q 0 . 676 , 2 0 . 0 H

H

H21  

kJ 0 . 676 Q  

2. The entropy balance

For steam, S = 0.0 – 7.3554 = -7.3554 kJ/K For the heat reservoir at 200 oC,

For the heat reservoir at 0 oC,

K / kJ 2270 .

15 4 . 273 200

000 ,

S 2

K / kJ 4748 .

15 2 . 273 0

S 676

Therefore, Stotal = -7.3554 + 4.2270 + 2.4748 = -0.6536 kJ/K < 0 (impossible!)

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Example 5.6 - solution

In this process, what is the maximum amount of heat that can be transferred to the heat reservoir at 200 oC?

1. The energy balance

HQQ'Q

2. The entropy balance

) process reversible

( 0 S

assume T

) Q m S (

S G

j fs

G

T Q '

T ' SQ

Since T’ = 473.15, T = 273.15, H = -2676.0 kJ, and S = 7.3554 kJ Solving for Q’ gives Q’ = -1577.7 kJ/kg

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5.8 Calculation of Ideal Work

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5.9 LostWork

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