Problems and Solutions

TUTORIALS Problems & Solutions

ENERGY TRANSFER 2013 / 2014

1

PROBLEM 1 Combustion gases of 0.02kmol/s of 0.02kmol/s molar flow rate enter a compressor at 95kPa and 20OC where they are adiabatically compressed to 300kPa. Then, they are cooled to the initial temperature in a steady flow heat exchanger. Draw both processes on T-S diagram. Knowing that the compressor isentropic efficiency is equal to 80%, and neglecting changes of the gases kinetic kinetic energy, calculate (1) (1) – a power needed to drive the compressor; (2) – a rate of heat given up by the gases gases in the cooler; cooler; (3) – power losses in each of the the devices, when when the ambient ambient temperature temperature is equal to 20OC. Mixture of combustion gases can be treated as ideal one with specific heat ratio κ =1,39 and the molar specific heat at constant pressure equal to 29,5kJ/(kmolK). The universal gas constant B=8315J/(kmolK).

ENERGY TRANSFER 2013 / 2014

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PROBLE PROBLEM M 1 - SOLUT SOLUTIO ION N Data

nɺ = 0.02kmol/s; p1 = 95kPa; p2 = p3 =300kPa;  O  t1 = t3 =t 0 =20 C; ηiz = 0.8; cp = 29.5kJ/(kmolK) AD.1. AD.1. A power power needed needed to drive the the compressor. compressor.

∆ Hɺ Cr = Qɺ Cr − W ɺ C r   Qɺ C r  = 0    p  T2 = T 1  2    p1  T −T ηis = 2 1 T2 r − T 1 Wɺ Cr

κ−1 κ

→

Wɺ Cr

= − ∆Hɺ Cr = nɺ c p (T1 − T2 r  )

 

1.3−1

300  1.3  = 293.15K ⋅   =384.85K    95  384.85 85 − 293. 293.15 15) K  ( 384. T −T   → T2r = T 1 + 2 1 = 293.15K + = 407.75K 0.8 ηis

= nɺ c p (T1 − T2 r  ) =  0.02

kmol s

⋅ 29.5

kJ

 

⋅ ( 293.15-407.75 ) K = -67.614kW

kmol×K  

ɺ  = 67.614kW W  C r  ENERGY TRANSFER 2013 / 2014

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PROBLEM 1 - SOLUTION, cont. AD.2. A rate of heat given up by the gases in the cooler (HE – Heat Exchanger).

ɺ  ∆ Hɺ HE = Qɺ HE − W  HE → ɺ W HE = 0  Qɺ HE

= 0.02

kmol s

⋅ 29.5

Qɺ HE

ɺ  p (T3 − T2r  )   = ∆Hɺ HE = nc

kJ

⋅ ( 293.15-407.75 ) K   → Qɺ HE =-67.614kW

kmol ⋅ K  

AD.3. Power losses in the compressor and cooler.

ɺ p (T1 − T2 ) − nc ɺ p (T1 − T2r ) = nc ɺ p ( T2r − T2 )  ∆ Wɺ C ,loss = Wɺ C − Wɺ Cr = nc Compressor 

ɺ ∆ W  = 0.02 C ,loss

kmol s

⋅ 29.5

ɺ ∆ W  C ,loss = 13.511kW

kJ

( 407.75-384.85) K  

kmol ⋅ K  

 T3  T   p3  Qɺ HE ɺ ɺ ɺ ɺ pT0  ln 3  − Qɺ HE ln ln ∆ WHE,loss = T0 ∆S HE = nc pT0  −B = nc  − T0 ⋅ 

Cooler 

ɺ ∆ W  HE,loss = 0.02

kmol s

⋅ 29.5

T2r

p2 

kJ

⋅ 293.15K ⋅ ln

kmol ⋅ K

T0



T2r



 

293.15K  

+ 67.614kW

407.75K  

ɺ ∆ W  = 10.543kW HE,loss ENERGY TRANSFER 2013 / 2014

4

 

PROBLEM 2  An electrical heater of 1kW electric power warms up the air from 18OC to 22OC in a closed room of 3m x 5m x 2m dimensions.  Average density of air is 1.25kg/m3. It is estimated that 20% of the heat rate delivered by the heater to the air escapes to the surroundings at 0OC through the draughty windows and the  poorly insulated walls. Assuming that the heater whole power is transferred to the air determine the lost work during the warming-  up process. The room air is an ideal gas with the specific heat at constant volume equal to 715J/(kgK).

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PROBLEM 2 - SOLUTION Data

= 1kW; V=(5*5*2)m3 =50m3 ; ρair =1.25kg/m3 ; t1 = 18O C; t 2 = 22O C t 0 = 0O C ɺ = -0.2kW. cv = 715J/(kgK); Qɺ lost = −0.2 ⋅W  el.

ɺ W  el.

1. From the First Law of Thermodynamics for a closed system calculate the time of heating process

∆U = Q − L   → ∆U = Q  L = 0 

mcv (T2

− T1 ) = 0.8Wɺ el. ⋅ τ

 ∆U = mcv (T2 − T1 )   → Q = Qin − Qout = Wɺ el. ⋅ τ − 0.2Wɺ el. ⋅ τ = 0.8Wɺ el. ⋅ τ  m = ρair V  = 1.25 kg3 ⋅ 50m3 = 62.5kg m  → τ=

mcv (T2

− T1  )

ɺ 0.8W  el.

62.5m 3 ⋅ 0.715

=

ENERGY TRANSFER 2013 / 2014

kJ

( 4K ) 

kg ⋅ K 

0.8 ⋅1kW

= 223.44s

6

PROBLEM 2 - SOLUTION, cont.

2. From the Guoy Stodola law determine the lost work

 T V /m Qenv T ∆Wlost = T0 S gen = T0 ( ∆Sair + ∆Senv ) = T0 m  cv ln 2 + R ln +  0  T T V m /   1 0 ∆Wlost = T0 mcv ln Qout

T 2 T 1

 

− Qout

ɺ ⋅ τ = −0.2 ⋅1kW ⋅ 223.44s = −44.69kJ = −0.2 ⋅W  el.

∆W lost = 273.15K ⋅ 62.5kg ⋅ 0.715

kJ kg ⋅ K

⋅ ln

295.15 291.15

+ 44.69kJ = 166.56kJ+44.6 9kJ

∆W lost =  211.25kJ

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PROBLEM 3  A household electric heating system consists of a 300W fan and electric heating element placed in a horizontal duct with diameter of 30cm. Air flows steadily through the duct. It enters the duct at 20OC and 100kPa and leaves at the same pressure and temperature of 25OC.  A volumetric rate of air at the inlet is equal to 0.5m3 /s. The rate of heat loss from the air in the duct is estimated to be 400W. Assuming that air is a bi-atomic ideal gas with κ=7/5 and R=287J/(kgK) and neglecting kinetic energy changes, determine a power of the electric heater and  power loss in the system when the environmental temperature is equal to 20OC.

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PROBLEM 3 - SOLUTION AD.1. Power of the electric heater. The electric heater power is equal to the rate of heat delivered by the heater to

ɺ . the system:  N  heat =Q heat From the First Law of Thermodynamics: ɺ ɺW ɺ ɺ  p ∆ H=mc ( T )1 =Q+ T 2 f an ɺ ɺ Q + ɺ Q=∆ɺ H − ɺ Wfan= ɺ mc ( T )1 ɺ-Wfan  Q= T  heat . l os s p 2 ɺ  Q

 heat .

ɺ ɺQ =Q-

l os s

,

ɺ mc ( T )1 ɺ-Wfan ɺ-Q = T p 2 l os s

 p ɺ ρ =V ɺ  p ɺ =V  m  1 1 1  R T   1 1  N  5 5 N   1 0  3 3  2  k  g kg  m m  m m2 ɺ =0. 5 =0. 5946  m  J J  s s   287 ×293K   kgK  J  c  7 004. 5  p p = R=1   2 kgK

 kg  J J ɺ ɺ mc ( T )1 ɺ-Wfan =0. )0 K-300W =268  Q= T 5946 ×1004. 5 ( 25-2  4 6. 4 W W  p 2  s kgK ɺ ɺ ɺ Q =2686.  Q 4W -( 400) W  heat =Ql os s

ɺ  Q 086. 4W  heat =3

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PROBLEM 3 - SOLUTION, cont. AD.2 Power loss in the system. ɺ ɺ ɺW ∆ W  l os t =W t , r ev. t ɺ ɺW  W  t =f an

 1 ɺ  T T  1 ɺ ɺ ɺW = ɺW = ɺm ɺm  c ∆ W ( T) -T ( 1s-2s )  + ( 1T-2T ) -0Tcl +W  pc  l os t =W t , r ev. t p 1 T 2 0 f un p n f un  T T    2 2 OR Use the Guoy Stodola Law ɺQ    T T m) ɺ Q ɺ S= ɺ  (  2 2  Q 2 ɺ ɺ ɺ ɺ mcTl ∆ W ∆ ∆ ∆ T S + S T mcl n -Q =T =T = =T ( 0 ) 0 p  T T =  l os t 0 n 0 0 p 0 n T    1 0 1

 kg  J J 298 ɺ ∆ W ⋅K l . 5946 ⋅ 1004. 5 ⋅ 293 n -2686. 4W  l os t =0  s

kgK

293

ɺ ∆ W 74. 8W  l os t =2

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PROBLEM 4 Water of 38OC is flowing out from a kitchen tap at the volume flow rate of 10 liter/min. The water arises through mixing two streams of water: the cold one at temperature of 10OC and the hot one at 80OC. Determine mass flow rates of the cold and hot water streams knowing that during the mixing process 500W of heat is lost to the ambient air. Calculate the total entropy generation and the power loss. Water  density is 1000 kg/m3 and its specific heat is equal to 4200J/(kg K). The ambient air is at 20OC.

ENERGY TRANSFER 2013 / 2014

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PROBLEM 4 - SOLUTION AD.1. Mass flow rates of the cold and hot water streams.  3 3 m  1 mi n  4 m ⋅  =1, 6⋅610  3 3  mi  n mi n 10 l i t r  60s s  kg  ɺ  3 ρɺ V  m  = = 00⋅0 1, 6⋅61 04 =0. 1666 3  1  s s ɺ  3 ɺ 1 m ɺ m2 mass balance:  m  m + m = = ɺ ɺ QɺW + ɺH ɺ H ɺ∆ U=  +  1 2 H 3

 l i t r l i t r ɺ ⋅  V = 1 0 = 1 0  3 3

First Law of Thermodynamics for an open steady-flow system (energy balance)

ɺ ɺ mh+ ɺ m2 h ɺ m3 3  0=  Q+ h  1 1 2 3 ɺ  + ɺ 3 ɺɺ m3 3 )1 h  0=  Q mh m h Qɺ  1 1 (+ m 2 3 ɺ Q+ ɺ  1 ɺ m3 2 ( h )1 =  m h h ɺ-m3 3 h 2 3

ɺ  + ɺ Q+ ɺ  3 ɺ m3 c ( ) ( T )3  Q  m h h T Q 2  3 w 2  3 3 ɺ  1  m  = = 1 = ( T )1  h h c T  2 1 w 2  kg ɺ  2 ɺ ɺ  m  = m m = ( 0 . 1 6 6 6 6 0 . 0 9  2 8  6 6 ) 2  3  1 2 3 1  s s ENERGY TRANSFER 2013 / 2014

 kg ɺ 1  =  m . 09826 ==0 9  s s  kg ɺ  2  m  =0. 06834 m2  s s 12

PROBLEM 4 - SOLUTION, cont. AD.2. Total entropy generation and the power loss. Q) ɺ ɺ ∆ S ∆ ɺ m2 ∆+ɺ  (  S S  gen =  m1 + S 0

 T T ɺ  + ɺ S =  3 ɺ 3 ɺ ∆ S ∆ cl n + m cl n m  m1 m2 1w 2 w  T  1 1

T  311. 15    1 311. 15 ɺ  + ɺ S =0. ∆ S ∆ 0 9 8 2 6 ×4 2 0 0 ×l n + 0 . 0 6   8  4 3 ×4  2 2 0 0×l  n + n  m1 m2  283 5 . 1 353. 15  W W ɺ  + ∆ S ∆ ɺ m2 =2 573 = 5 2.  m1 + S  K K ɺ ɺ  Q Q  Q 500 W Q ) ɺ (  0 0 ∆ S  = =- = = = 7 1  0 .  5 7 06  0 0  T T  9 2  3 . 15  K 9 1 K  0 0  0 0  W W Q) ɺ ɺ ɺ S ∆+ɺ  ( ∆ ∆  S =  S + S = ( 2 . 5 7 3 + 1 . 7 0 5  ) 6 =  2 4 . 2 7 9 )  gen  m1 m2 0  K K The loss power from Gouy-Stodola Law:

ɺ W lost

 2 2

= T Sɺ

0 gen.

= 293.15K ⋅ 4, 366

ENERGY TRANSFER 2013 / 2014

W K 

= 1254.3 W 13

PROBLEM 5  A turbocharger of an internal combustion engine consists of a turbine, a compressor and a cooler. All these devices can be treated as adiabatic ones. Hot exhaust gases enter the turbine at a mass flow rate of 0.02kg/s and at 400OC and leave at 350OC. 95% of thus produced power drives the compressor ( 5% of turbine work is lost during its transmission to the compressor). Air enters the compressor at a mass flow rate of 0.018kg/s, at 70OC and 95kPa and leaves at 135kPa. For simplicity assume that the exhaust gases and the air are ideal gases of the same cp=1kJ/(kgK) and κ=1.4. What is an isentropic efficiency of the compressor?  To avoid the possibility of an engine knock (due to ɺa a side effect of the air temperature increase in the m ɺg m  p3 , t3 compressor), a cooler is placed between the  p 2 , t 2r  ηm N tT compressor and the engine suction manifold to comp. turbine T S decrease the air temperature to 80OC. Cold ambient air is used as a cooling fluid. Its ɺa m ɺg m  p 4 , t 4 temperature rises from 20OC to 40OC between the  p1, t1 ɺ c , t c2 inlet and the outlet of the cooler. What is a total m ɺ c , t c1 m cooler  lost power in the turbocharger assuming that the ɺa m  pressure in the turbine decreases 1.7 times, and  p 4 , t 5 the ambient temperature is equal to 20OC?  ENERGY TRANSFER 2013 / 2014

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PROBLEM 5 - SOLUTION ɺa m  p3 , t 3

ɺg m  p 2 , t 2r 

ɺ  g = ɺ ma =0.  m 0. 02kg/s; 018kg/s; =0

ηm N tT

turbine T

ɺg m  p1 , t1

Data

comp. S

ɺa m  p 4 , t 4 ɺ c , t c1 m

cooler 

ɺa m  p 4 , t 5

ɺ c , t c2 m

 O O O O  t= 40 0 C;t 50 C;t 0 C; t= 80 C t=  1 2r =3 3 =7 5  p p . 7;p 5kPa; p 30kPa  1 / 2 =1 3 =9 4 =1 η mm =0. 95  O O  t 0 C; t 0 C  c1 =2 c2 =4

c p

= 1kJ/(kgK); κ = 1.4; t 0 = 20O C

AD.1 – an isentropic efficiency of the compressor 

 T 70+273. 15) K=343. 15K  3 3 =(  κ 1  T T  4 3  0. 4/1. 4 η C  = κ  ,s    i .=    p  1 1 3 5 p  4 4 -T  4  T  4 T =T =3 15 = 379. 4K =343. =3 3  K   r r 3  T    p 9 5   33  ENERGY TRANSFER 2013 / 2014

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PROBLEM 5 - SOLUTION, cont. ɺg m  p 2 , t 2r  turbine T

ɺg m  p1 , t1

ɺa m  p3 , t 3

 f r om t heFi r s tLaw ofTher modynami cs  T  4r

ηm N tT

ɺ η ɺ T,r  W  C, r =m W

comp. S

ɺ ɺ ma c ∆ɺ H = ( T )3  W T  C, r = Cr  p 4r -

ɺa m  p 4 , t 4 ɺ c , t c1 m

cooler 

ɺa m  p 4 , t 5

ɺ c , t c2 m

ɺ ɺ gc ( T)r  W  T, r =m p 1 T 2

 k g  k kJ ɺ ⋅ ( 400-35 )0K=1kW  W = 0 . 0 2 1  T, r ⋅gK  s k

 m mg  g ɺ ɺ η m ⇒ η ( ) (p T) ( T)r =395.  mg c T = c T T  T T = T + 93K m 1 2r a p 4r 3 4r 3 m 1 T 2 ɺ ɺ

 m ma  a

( 379. )5K 4-343. 1  T T  4 4 3 η C  = = ,s  i . ( )5K  T T 395. 93-343. 1  4  r r 3

η C  = . 6868( 68. 68%) ,s  i . 0

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PROBLEM 5 - SOLUTION, cont. AD.2. Total lost power in the turbocharger - calculations based on an isentropic process and the G.S. Law

ɺ ∆ W ∆ ɺ los)t  l os t (= W ɺg m  p 2 , t 2r  turbine T

ɺg m  p1 , t1

 T

∆ ɺ W ) (+ ∆ ɺW) (+ l os t

l os t

C

cool .

Lost power in the turbine

ɺ ) ( ∆ W

 l os t  T T

κ 1 κ

  p  p  2 2  T 1   2 =T   p   1 1 ɺ ) ( ∆ W

 l os t  T T

 W ɺ ɺ ( T)r p 1 T 2   T,r =mg c ɺ W ɺ-W ⇒  ɺ W = ɺ mg c (p T)2 = T Tr T 1 T  ɺ ( T )2 T ( ∆ W )t TT= ɺ mg c  l os p 2r  0 0. 4  1. 4

  1  1   1. 7

( 673. )K = 15 

O =578. 45K . 3 ( 305 O )C

 kg  k kJ =0. 02 ⋅ 1 ( 350-305). 3K= 0. 894kW 3K=0  s kgK

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PROBLEM 5 - SOLUTION, cont. Lost power in the compressor 

ɺa m  p3 , t 3 ɺ ) ( ∆ W

 l os t  C C

ɺ W ɺ-W = C, r C

comp. S

ɺa m  p 4 , t 4 ɺ ) ( ∆ W

= 395.93K;  T  4r

 l os t  C C

 W ɺ ɺ ( T )3 T p 4r   C,r =ma c ɺ ɺ ma c (p T )3 WC = T  W 4  ɺ ɺ ma c ( T )4 T ( ∆ W )t CC =  l os p 4r = 379, 4K   T 4

 kg  k kJ =0. 018 ⋅ 1 ( 395. 93-379). 4K=0 298kW 4K=0.  s kgK

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PROBLEM 5 - SOLUTION, cont. ɺa m t 4 ,p 4

ɺ c , t c1 m

Power loss in the cooler 

ɺ c , t c2 m

cooler 

lost cool.

ɺa m  p 4 , t 5

( ∆Sɺ a )

= T0∆Sɺ cool.  = T0 ( ( ∆Sɺ a )cool. + ∆Sɺ c )

( ∆Wɺ )

  p 4   T5 T ɺ ɺ ∆ = − ( Sa )cool. ma  c p ln T R ln  p   = mɺ a cp ln T 5  4  4r 4r 

= 0.018 cool.

∆Sɺ c = mɺ cc p ln

kg s

⋅1

kg ⋅ K

⋅ ln

(80 + 273.15) 395.93

= −2.06 ⋅ 10−3

kW K  

Tc2 Tc1

Mass flow rate of the cooling water from the First Law of Thermodynamics

kJ

 

ɺ ∆Hɺ = Qɺ − W  cool.  ɺ =W ɺ Q

cool.

=0

 ∆Hɺ = ( ∆Hɺ a )cool.+ ∆Hɺ c = 0 

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PROBLEM 5 - SOLUTION, cont.

( ∆Hɺ a ) = mɺ a c p ( T5 − T4r  ) 

T −T ⇒ mɺ c = mɺ a 4r 5  Tc2 − Tc1 ∆Hɺ c = mɺ cc p ( Tc2 − Tc1 )  395.93 − ( 80 + 273.15) ) ( kg kg ɺ c = 0.018 = 0.0385 m s ( 40 + 273.15 − ( 20 + 273.15) ) s cool.

∆Sɺ c = mɺ cc p ln

Tc2 Tc1

= 0.0385

kg s

⋅1

kJ kg ⋅ K

ln

( 40 + 273.15) kW = 2.54 ⋅ 10−3 K   ( 20 + 273.15)

( ∆Wɺ )

= + ∆ T0 ( ( ∆Sɺ a ) Sɺ c ) = 293.15K ⋅ ( −2.06 ⋅ 10−3 + 2.54 ⋅ 10−3 ) cool.   cool.

( ∆Wɺ )

cool.

lost

lost

kW K 

= 0.1407kW

ɺ = ( ∆W ɺ ) + ( ∆W ɺ ) + ( ∆W ɺ ) = 0.894kW + 0.298kW + 0.1407kW ∆W lost lost T lost C lost cool.

∆Wlost = 1.3327kW ENERGY TRANSFER 2013 / 2014

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PROBLEM 6 Liquid water of 20OC and the mass flow rate of 2.5kg/s is heated to 60OC by mixing it with superheated steam of 150OC in a chamber working at constant pressure of 200kPa. It is estimated that during the process the chamber loses 20kW of heat to the surrounding at temperature 25OC. Determine a lost power in the mixing chamber. Specific heat of liquid water is equal to 4.22kJ/(kgK). Specific enthalpy and specific entropy of the superheated steam, read from the data table for 150OC and 200kPa, are 2769kJ/kg and 7.28kJ/(kgK), respectively. Specific enthalpy and specific entropy of a liquid water can be approximated as hw=cwtw and sw=cwln(Tw /273K), respectively.

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21

PROBLEM 6 - SOLUTION

− t w ) + mɺ v ( c wt m − h v ) = Qɺ

ɺ wcw ( t m m ɺ  ∆Hɺ = ∆Hɺ w + ∆Hɺ v = Qɺ − W  ɺ W=0   ∆Hɺ w = mɺ w c w ( t m − t w )   ∆Hɺ v = mɺ v ( cw t m − h v ) 

   ɺS = Sɺ − ( Sɺ + Sɺ ) + ∆Sɺ  gen m w v 0 ɺ = T Sɺ ∆W 0 gen lost

ɺv m

=

ɺ −m ɺ wcw ( t m Q

− tw )

( cw t m − h v ) −20kW − 2.5

kg s

ɺv m

=

Sɺ m

ɺ v ) c w ln = ( mɺ w + m

Sɺ w

ɺ w c w ln =m

Sɺ v

= mɺ v s v

∆Sɺ 0 = −

⋅ 4.22

kJ

( 60 − 20 ) K   kgK  

 kJ kJ   4.22 kgK 60K − 2769 kgK    

= 0.16kg/s

Tm

273K 

Tw

273K 

ɺ Q T0

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PROBLEM 6 - SOLUTION, cont. Sɺ m

= ( 2.5 + 0.16 )

Sɺ w

= 2.5

Sɺ v

= 0.16

∆Sɺ 0 = Sɺ gen

kg s

⋅ 4.22

kg s

kg s

⋅ 4.22

kJ kgK

⋅ 7.28

ln

kJ kgK

kJ kgK

ln

(60 + 273)K

273K

( 20 + 273)K

273K

= 1.165

= 0.746

= 2.23

kW K  

kW K  

kW K  

− ( −20kW ) kW = 0.067 K   ( 25 + 273) K

= 2.23

kW kW  kW kW  −  0.746 + 1.165 + = . . 0 067 0 386  K  K K  K K

kW

ɺ = T Sɺ = ( 25 + 273) K ⋅ 0.386 ∆W lost 0 gen

 

kW K 

ɺ = 115kW ∆W lost ENERGY TRANSFER 2013 / 2014

23

PROBLEM 7 Superheated steam enters the turbine of 4MW power at 2.1MPa and temperature of 475OC. The water vapor leaving the turbine is at the saturated state and at pressure of 10kPa. It is then directed to the heat exchanger where it condenses and is cooled to 30OC by the stream of cooling water, which enters the condenser at 15OC and leaves at 25OC. Determine the lost power in the turbine and in the condenser, assuming that there is not heat loss in the turbine and condenser, and the ambient temperature is equal to 15OC. Specific heat of liquid water is 4.19kJ/(kgK) and its specific enthalpy and specific entropy can be approximated as hw=cwtw and sw=cwln(Tw /273K ), respectively. From the steam tables the following data are given: for 2.1MPa & 475OC specific enthalpy and specific entropy are 3411.3kJ/kg and 7.34kJ/(kgK ), respectively; for 10kPa specific enthalpy and specific entropy of the saturated water vapor are: 2584kJ/kg and 8.15kJ/(kgK), respectively. ENERGY TRANSFER 2013 / 2014

24

PROBLEM 7 - SOLUTION Mass flow rate of the steam from the First Law of Thnermodynamics for the turbine

ɺ  ∆Hɺ = Qɺ − W t ɺ  W 4000kW t ɺ Q=0 = = 4.835kg/s  → {mɺ v = h h . − 3411 3 − 2584 kJ/kg ( ) ( ) 1 2 ɺ v ( h 2 − h1 )  ∆Hɺ = m  Lost power in the turbine from the Guy Stodola law

ɺ ɺ v ( s2" − s1 ) = 288.15K ⋅ 4.835 ∆W = T0 ∆Sɺ v1,2 = T0 mɺ v ( s2 − s1 ) = T0 m t1, 2 Power loss in the turbine

kg s

⋅ (8.15 − 7.34 )

kJ kgK

 

ɺ ∆W = 1128.5kW t1,2

Mass flow rate of cooling water from First Law of Thermodynamics for the condenser

  ɺ ɺ Q = Wtc = 0   ∆Hɺ c = ∆Hɺ vc + ∆Hɺ cw  ɺ ∆Hɺ c = Qɺ c − W tc

where

∆Hɺ vc = mɺ v ( h3 − h 2 )  ɺ cw ⋅ c w ( t w 2 − t w1 ) ∆Hɺ cw = m

ENERGY TRANSFER 2013 / 2014

25

PROBLEM 7 - SOLUTION, cont. ɺ v ( h3 − h 2 ) + m ɺ cw ⋅ cw ( t w 2 m h2

= 2584

kJ kg

;

h3

− t w1 ) = 0 → mɺ cw =

= cw t 3 = 4.19

kJ kgK

kg kJ ( 2584 − 125.7 ) s kg kJ 4.19 ( 25 − 15 ) K   kgK 

=

cw ( t w 2

⋅ 30K = 125.7

4.835

ɺ cw m

ɺ v ( h 2 − h3 ) m

= 283.67

− t w1 )

kJ kg

kg s

Lost power in the condenser from the Guy Stodola law TW 2  ɺ ɺ ɺ ) = T m ɺ ɺ ɺ ∆W = = ∆ + ∆ − + T S T S S s s m c ln ( ) (  0 cond. 0 0  3 2 tcond v 2 ,3 cw v cw w Tw1   T3 ( 273 + 30)K kJ kJ = 4.19 = 0.437 s 3 = c w ln ln

273K

kgK

273K

kgK  

 kg kJ kg kJ 298K    ɺ ∆W = 288K  4.835 ( 0.437 − 8.15) + 283.67 ⋅ 4.19 ln tcond  s kgK s kgK 288K     ɺ ∆W = 943.88kW tcond ENERGY TRANSFER 2013 / 2014

26

PROBLEM 8 Pure nitrogen at 0.1MPa and 25OC is transferred along a 10m distance through a pipe of 3cm diameter, made of 2mm thickness rubber. How many kmols of the nitrogen is lost per second to the ambient air, whose pressure and temperature are equal to those in the pipe and a molar fraction of nitrogen in the air is equal to 79%. Compare this nitrogen loss with the one that occurs in the case when the  pipe is placed in a vacuum. Diffusivity and solubility of nitrogen in the rubber at temperature

25OC

are, respectively, 1.5·10-10m2 /s

and

0.00156 kmol/(m3bar).

From Y.A. Cengel, Heat and Mass Transfer , The Third Edition. ENERGY TRANSFER 2013 / 2014

27

PROBLEM 8 - SOLUTION Molecular diffusion of nitrogen through a cylindrical wall C A

A + B

 L

A

C A,1 r 1

B C A,2 r 

nɺ A

≡ J A = 2πLD AB

where L = 10m; r1 CA1

C A1 − C A 2  kmol  ln ( r2 / r1 ) 

= 0.015m;

= ℜp A1 = 0.00156

r 2

r2

s

 = const.

= 0.017m

kmol m3bar

×1bar = 0.00156

kmol m3

AD.1 – molar flow rate of the lost nitrogen for ambient air 

CA 2

= ℜp A 2 = 0.00156

kmol m3 bar

⋅ 0.79 ⋅1bar = 0.001232

kmol

m3  kmol  . . − 0 00156 0 001232 ( ) 2  m3  −10  m  ⋅ J A = 2π ⋅ 10[m] ⋅1.5 ⋅10   ln ( 0 .017 / 0.015 )  s 

kmol  = 2.47 ⋅10−12   s 

AD.2 – molar flow rate of the lost nitrogen for vacuum

CA 2

= ℜp A 2 = 0

 kmol 

( 0.00156 − 0 )  3  2   m  m  = 1.17 ⋅10 −11  kmol  J A = 2π ⋅10[m] ⋅1.5 ⋅10 −10   ⋅  s   s  ln ( 0.017 / 0.015 ) ENERGY TRANSFER 2013 / 2014

28

PROBLEM 9

 A 230 mm diameter pan of water at 22OC has a mass loss rate 1.5·10-5 kg/s when the ambient air is dry and at 22OC. Determine the convection mass transfer coefficient. Estimate the evaporation mass loss rate when ambient air has a relative humidity of 50% and remains at 22OC. Water vapour saturation pressure at 22OC is equal to 2.617kPa.

From Y.A. Cengel, Heat and Mass Transfer , The Third Edition. ENERGY TRANSFER 2013 / 2014

29

PROBLEM 9 - SOLUTION

= ∆mɺ (A1) = 1.5 ⋅ 10−5 kg/h; O ps ( 22 C) = 2.617kPa; ϕ1 = 0.0; ϕ2 = 0.5; M A = M H O = 18kg/kmol; B = 8315J/(kmol ⋅ K) O

d = 0.23m; t = 22 C = 297K; J A

Data A – water vapour  B - dry air 

2

Solution

= ∆mɺ (A1) = A ⋅ k m  ( ρAS − ρ(A1∞) ) ⇒ (1) (1) (1) ɺ A / A = k m ( ρAS − ρA∞ )   jA = ∆m  (1)

JA

k m

=

∆mɺ A / A ( ρAS − ρ(A1∞) )

where A=π

d

2

4

=π

(0.23m )

4

2

= 0.0415m 2 o

ρ

(1) A∞

= 0.0; ρAS =

ps ( 22 C) R AT

o

=

ps ( 22 C) BT

MA

= 19.2 ⋅10−3

ENERGY TRANSFER 2013 / 2014

kg m3

30

PROBLEM 9 - SOLUTION,cont.

k m

=

1. 5 ⋅10−5

kg

∆mɺ A / A s = (1) (ρAS − ρA∞ ) 0.0415m2 ⋅ (19.2 ⋅10−3 − 0.0) kg

m3

km

ρ

( 2) A∞

( 2)

JA

= 1.883 ⋅10−2

 m s

ϕ2 ⋅ ps ( 22oC) ϕ2 ⋅ ps ( 22oC) (1) −3 = = M A = ϕ2 ⋅ρ A∞ = 0.5 ⋅ 19, 2 ⋅ 10 R AT

BT

= ∆mɺ (A2 ) = A ⋅ k m ( ρ AS − ρ(A2∞) ) = 0.0415m 2 ⋅ 1.883 ⋅ 10 −3 ( 2)

JA

ɺ (A2 ) = 7. 5 ⋅10 −6 = ∆m

m s

kg m

3

= 9.6 ⋅ 10−3

⋅ (19.2 − 9.6 ) ⋅10 −3

kg m

kg m3

 kg s

ENERGY TRANSFER 2013 / 2014

31

3

PROBLEM 10  A cylindrical jug of 8cm internal diameter and 30cm height is half filled with water and left in a dry air at 15OC and pressure of 87kPa. Its top is open. Saturated pressure at 15OC is 1.705kPa and mass diffusivity of water vapor in dry air is 2,6·10-5m2 /s. Determine the amount of water in kg, which will evaporate from the jug after 24 hours.  Assume that changes of the water level in the jug are negligible during this period of time. Water density is 1000kg/m3, water molar weight is 18kg/kmol, and the universal gas constant has a value of 8315J/(kmolK). From Y.A. Cengel, Heat and Mass Transfer , The Third Edition. ENERGY TRANSFER 2013 / 2014

32

PROBLEM 10 - SOLUTION Assumptions It is Stefan flow. Assumption of constant water level in the pitcher means a steady state with constant mass flux

Solution A – water vapour 

ɺ Aτ = J Aτ ∆m A = m

where J A

= A ⋅ jA = A ⋅ M A ⋅ jA

B - dry air 

x A,L Stefan law

 jA

=

C ⋅ DAB L

ln

1 − x A,L 1 − x A ,0

=0 O

x A,0

=

ps (15 C ) p

where

C=

p BT

=

=

1.705kPa 87kPa

87 ⋅10

3

= 19.6 ⋅10−3

 N m2

= 36.33 ⋅10−3

J ⋅ 288K   8315 kmol×K 

ENERGY TRANSFER 2013 / 2014

kmol m

3

33

PROBLEM 10 - SOLUTION, cont.

 jA

=

A=

36.33 ⋅10

πD 2

2

=

−3

2 kmol −5 m ⋅ 2.6 ⋅10 3 1− 0 m s ln 0.15m 1 − 19.6 ⋅10 −3

π (8 ⋅10 4

−2

m

)

= 12. 47 ⋅10 −6

kmol m2 ⋅ s

2

= 5.027 ⋅10−3 m 2

ɺ A τ = J A τ = A ⋅ M A ⋅ jA ⋅ τ ∆m A = m

∆m A = 5.027 ⋅10 −3 m 2 ⋅18

kg kmol

⋅12.47 ⋅10 −6

kmol m 2 ×s

⋅ (24 ⋅ 3600 ) s

∆m A = 9.75 ⋅ 10−2 kg

ENERGY TRANSFER 2013 / 2014

34

PROBLEM 11 To keep a can of beverage at 6OC in hot dry ambient air at temperature of 27OC, the can is continuously moistened with a highly volatile liquid of molecular mass equal to 200kg/kmol. Saturated vapour pressure of this wetting agent at 6OC is equal to 5kPa. Thermophysical properties of dry air at 27OC are as follows: thermal conductivity k=0.026W/mK; density ρ=1.16kg/m3; specific heat cp=1kJ/kgK. The diffusivity of the agent vapour in air is 2·10-5m2 /s. Knowing that heat and mass transfer between the wetting liquid and the ambient air occurs only by forced convection (neglected thermal radiation), take advantage of the Chilton Colburn analogy and calculate a latent heat of the liquid agent vaporization. Hint: In such case the energy conservation principle states that the convective heat flux delivered to the can surface from the air equalizes the heat flux released from the surface due to the liquid evaporation. The liquid

agent

vapour

is

an

ideal

gas.

Universal

gas

constant

B=8315J/(kmolK). From F. Incropera et al. Principles of Heat and Mass Transfer , Seven Edition. ENERGY TRANSFER 2013 / 2014

35

PROBLEM 11 - SOLUTION A - vapour of the liquid agent; B - dry ambient air 

Energy balance

−h(TS − T∞ ) = jArA   ⇒  jA = h m ( ρA,S − ρA ,∞ )   rA

=

h hm

h hm Chilton Colburn analogy

hm

− TS ) = hm ( ρA,S − ρA,∞ ) ⋅ rA

− TS ) ( ρA ,S − ρA,∞ ) (T∞

= ρcp Le 2/ 3

Le =

h

⋅

h(T∞

α D AB

= 1.16

=

k 

ρcp DAB

= 1.16

kg

J

m

kgK

⋅103 3

0.026

W mK 

kg

J

m

kgK

3 ⋅ 10 3 2/3

⋅ 2 ⋅ 10−5

⋅ (1.121) = 1251.78

ENERGY TRANSFER 2013 / 2014

m

= 1.121

2

s

J m3K

  36