Open Flow Systems

Open Flow Systems

Chapter 2. The First Law and

Other Basic Concepts

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

m1

m2

m3

n M m  

uA q 





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

(m _fs  m m m



0 )

( 



 _fs

cv uA

dt

dm



0 )

( 





0 )

( )

(









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





Q m

zg u

dt U mU d

fs

cv       



 



 



 



  





 ( )

2 1 )

( ₂

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 ( )    

) (

W

– 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…

Energy Balances for Steady-State Flow Processes

 Steady state : no accumulation term

s fs

W Q

m zg u

H     



 



 



 



  ²  2



Ws

Q zg

u

H   



 



  ²  2



Ws

Q H  



kinetic and potential energy changes are negligible

Only shaft work

No stirring work for steady-state

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withdrawn from the tank at a steady rate of m = 0.2 kg/s, how long will it take for the temperature of the water in the tank to drop from 60 to 35 ^oC?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 C_v = C_p = C, independent of T and P.

Example 2.14 - solution

Energy balance become

W Q m

dt H mU d

fs

cv ( )    

)

( m(H H₁ ) 0

dt

m dU 

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?

Example 2.16 - solution

Energy balance

Ws

Q zg

u

H   



 



  ²  2

 1 Q  u₂² W_s

2 1

Chapter 5. The Second Law of

Thermodynamics

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

C C H

H

toal T

Q T

S  Q 





C

H Q

Q W  



 



 









H C H

toal

C T

Q T S

T

W 1

Example 5.4

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

quenched in 150 kg of oil (C_p = 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   



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} S_G dt

dS dt

mS m d

S  







j j

j t

surr

T Q dt

dS 

) 0 ) (

(    





j j

cv j

fs S

T Q dt

mS m d

S  



Page 176 for description

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 C_p= 7/2R, that the surroundings are at 300 K and that kinetic- and potential-energy changes are negligible.

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 }

^{ Q }







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









 



 

 

 

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.

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



H Q Q'Q  2,000Q 0 . 676 , 2 0 . 0 H

H

H  ₂  ₁  



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

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

H Q Q'Q

2. The entropy balance

) process reversible

( 0 S

assume T

) Q m S (

S _G

j fs

G  ^ 













T Q '

T ' S  Q 

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

5.8 Calculation of Ideal Work

5.9 LostWork