Problems and Solutions

(1)

1 1 1 1 1 1 1 1 ENERGY TRANSFER 2013 / 2014 ENERGY TRANSFER 2013 / 2014

TUTORIALS

Problems

&

Solutions

(2)

(3)

2 2 2 2 2 2 2 2 ENERGY TRANSFER 2013 / 2014 ENERGY TRANSFER 2013 / 2014

PROBLEM 1

Combustion gases of

Combustion gases of 0.02kmol/s0.02kmol/s molar flow rate enter a molar flow rate enter a compressorcompressor at

at 95kPa95kPa andand 2020OO_C_{C where they are adiabatically compressed to}_{where they are adiabatically compressed to 300kPa.}_300kPa.

Then, they are cooled to the initial temperature in a

Then, they are cooled to the initial temperature in a steady flow heatsteady flow heat exchanger.

exchanger.

Draw both processes on

Draw both processes on T-ST-S diagram. Knowing that the diagram. Knowing that the compressorcompressor isentropic efficiency is equal to

isentropic efficiency is equal to 80%,80%, and neglecting changes of theand neglecting changes of the gases kinet

gases kinetic energy, ic energy, calculate (1calculate (1) –) – a power a power needed to needed to drive thedrive the compressor;

compressor; (2) –(2) – a rate of a rate of heat given up heat given up by the gasby the gases in the coes in the cooler;oler; (3) –

(3) – power losses power losses in each of tin each of the devices, whehe devices, when the amn the ambient temperabient temperatureture is equal to

is equal to 2020OO_C_C..

Mixture of combustion gases can be treated as ideal one with specific Mixture of combustion gases can be treated as ideal one with specific heat ratio

heat ratio κ κ ==1,391,39 and the molar specific heat at constant pressure equaland the molar specific heat at constant pressure equal

to

(4)

3 3 3 3 3 3 3 3 ENERGY TRANSFER 2013 / 2014 ENERGY TRANSFER 2013 / 2014 Data Data 1 1 2 2 33 O O 1 1 33 00 iizz pp 0

0..0022kkmmooll//ss; ; 9955kkPPaa; ; ==330000kkPPaa;; t t ==t t ==220 0 CC; ; 00..88; ; 2299..55kkJJ//((kkmmoollKK)) n n pp pp pp t t cc

=

==





₌

_η

₌



ɺɺ AD.1

AD.1.. A poA power needewer needed to drive thd to drive the comprese compressor.sor.

PRO

PROBLE

BLEM 1

M 1 -- SO

SOLUT

LUTIO

ION

N

( (

))

r r r r r r r r rr 11 22rr r r 00 C C C C C C C C C C pp C C H H Q Q W W W W H H nnc c T T T T Q Q



∆

=

−−

_

→

=

−− ∆

∆ =

=

−−



==

_

ɺɺ

ɺ

ɺɺ

ɺ

ɺɺ

_ɺɺ

ɺɺ

( (

))

1 1 _{1.3 1}_{1.3 1} 1.3 1.3 2 2 2 2 11 1 1 2 2 11 22 11 2r 2r 11 2 2r r 11 300 300 2 29933..1155K K ==338844..8855K K 95 95 38 384.4.85 85 29293.3.15 15 K K 2 29933..1155K K 440077..7755K K 0.8 0.8 is is is is p p T T T T p p T T TT TT TT T T T T T T T T κ− κ− ₋₋ κκ





_

_

=

_

_

=

⋅⋅

_

_









−−

−

−−

η

=

→

=

+

=

+

==

−

η

(

)

(

))

r r 11 22rr k kmmool l kkJJ 0 0..002 2 2299..5 5 229933..1155--440077..775 5 K K --6677..661144kkWW s s kkmmooll××K K C C pp W W

ɺɺ

=

nc n

ɺɺ

c T

−

T T

−

T

=

⋅

==

r r 67.614kW 67.614kW C C W W

ɺɺ

==

(5)

4 4 4 4

ENERGY TRANSFER 2013 / 2014

PROBLEM 1 - SOLUTION, cont.

AD.2. A rate of heat given up by the gases in the cooler (HE – Heat Exchanger).

(

)

(

)

HE HE HE HE HE 3 2r HE HE 0 kmol kJ 0.02 29.5 293.15-407.75 K s kmol K p H Q W Q H nc T T W Q



∆

=

−

_{ → =}

∆ =

−



=

_

=

⋅

→

⋅

ɺ

_ɺ

ɺ

Q

ɺ

_HE=-67.614kW

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

Compressor

(

)

(

)

(

)

(

)

,loss r 1 2 1 2r 2r 2 ,loss kmol kJ 0.02 29.5 407.75-384.85 K s kmol K C C C p p p C W W W nc T T nc T T nc T T W

∆

= − =

− −

− =

−

∆

=

⋅

ɺ ɺ ɺ _ɺ _ɺ _ɺ ɺ ,loss 13.511kW C W

∆

ɺ

=

Cooler 3 3 HE 3 HE,loss 0 HE 0 0 0 HE 2r 2 0 2r HE,loss ln ln ln kmol kJ 293.15K 0.02 29.5 293.15K ln 67.614kW s kmol K 407.75K p p T p Q T W T S nc T B T nc T Q T p T T W









∆

=

∆

=

_

−

_

−

⋅

=

_

_

−









∆

=

⋅

+

⋅

ɺ

_ɺ

ɺ

HE,loss 10.543kW W

∆

ɺ

=

(6)

5 5 5 5

An electrical heater of 1kW electric power warms up the air from

18

O

_{C to 22}

O

_{C in a closed room of 3m x 5m x 2m dimensions.}

Average density of air is 1.25kg/m

3

_{. It is estimated that 20% of the}

heat rate delivered by the heater to the air escapes to the

surroundings at 0

O

_{C 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).

(7)

6 6 6 6 ENERGY TRANSFER 2013 / 2014 Data

PROBLEM 2 - SOLUTION

3 3 3 O O O el. air 1 2 0 v lost el. 1kW; V=(5*5*2)m =50m ; =1.25kg/m ; t 18 C; t 22 C t 0 C 715J/(kgK); 0.2 -0.2kW. W c Q W

=

ρ

=

−

⋅

=

ɺ

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

(

)

(

)

(

)

( )

v 2 1

in out el. el. el.

3 air 3 3 v 2 1 v 2 1 el. el. 0.2 0.8 0 kg 1.25 50m 62.5kg m kJ 62.5m 0.715 4K kg K 0.8 223.44s 0.8 0.8 1kW U mc T T U Q L U Q Q Q Q W W W L m V mc T T mc T T W W



∆ =

−



∆ = −





→ ∆ = →

= − = ⋅ τ −

⋅τ =

⋅ τ





=

_

_

 =

ρ =

⋅

=



⋅

−

_⋅

−

=

⋅ τ

→

τ

=

⋅

ɺ

(8)

7 7 7 7

PROBLEM 2 - SOLUTION, cont.

2. From the Guoy Stodola law determine the lost work

(

)

2 env

lost 0 0 air env 0 v 0

1 0 2 lost 0 v out 1 out el. lost / ln ln / ln 0.2 0.2 1kW 223.44s 44.69kJ kJ 295.15 273.15K 62.5kg 0.715 ln 44.69kJ 166.56kJ+44.6 kg K 291.15 gen Q T V m W T S T S S T m c R T T V m T T W T mc Q T Q W W





∆

=

∆

+ ∆

=

_

+

_

+





∆

=

−

= − ⋅ ⋅ τ = − ⋅

⋅

= −

∆

=

⋅

+

=

⋅

ɺ

9kJ lost 211.25kJ W

∆

=

(9)

8 8 8 8

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 20O_C

and 100kPa and leaves at the same pressure and temperature of 25O_C.

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 20O_C.

(10)

9 9 9 9 ENERGY TRANSFER 2013 / 2014

PROBLEM 3 - SOLUTION

AD.1. P

ower of the electric heater.

The electric heater power is equal to the rate of heat delivered by the heater to the system:

N =Q

_heat_heat_heat_heat

ɺ

_heat_heat_heat_heat.

From the First Law of Thermodynamics:

, ( ) ( ) ( ) ∆ + = ∆ − = ɺ ɺ _ɺ ɺ ɺ ɺ ɺ ɺ ɺ _ɺ ɺ ɺ ɺ ɺ _ɺ ɺ ɺ p 2 1 fan p 2 1 fan p 2 1 fan p 2 1 fan heat. loss fan p 2 1 fan heat. loss fan p 2 1 fan heat. loss fan p 2 1 fan heat. loss fan p 2 1 fan heat. loss p 2 1 fan loss heat. loss p 2 1 fan loss heat. loss p 2 1 fan loss heat. loss p 2 1 fan loss H=mc T -T =Q+W H=mc T -T =Q+W H=mc T -T =Q+W H=mc T -T =Q+W Q=Q Q H W mc T -T -W Q=Q Q=Q Q Q H W mc T -T -WH W mc T -T -W Q=Q Q H W mc T -T -W Q =Q-Q =mc T -T -W -Q Q =Q-Q =mc T -T Q =Q-Q =mc T -T -W -Q-W -Q Q =Q-Q =mc T -T -W -Q  ρ           ɺ ɺ ɺ ɺ 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 5 55 5 3 33 3 ₂₂₂₂ p pp p p pp p m =V =V m =V =V m =V =V m =V =V RT RT RT RT N NN N 10 10 10 10 _kg_k_k_g_k m mm m _m_m_m_m m =0.5 =0.5946 m =0. m =0.5 5 =0.=0.59465946 m =0.5 =0.5946 J JJ J s s s s s s s _{287 ×293K}_{287 ×293K}_{287 ×293K}_{287 ×293K} s kgK kgK kgK kgK 7 J 7 J 7 J 7 J c = R=1004.5 c = R=1004. c = R=1004.55 c = R=1004.5 2 kgK 2 2 kgKkgK 2 kgK ( ) ( ) ɺ _ɺ ɺ ɺ ɺ ɺ p 2 1 fan p 2 1 fan p 2 1 fan p 2 1 fan heat loss heat loss heat loss heat loss kg kg kg kg J JJJ Q=mc T -T -W =0.5946 ×1004.5 25-20 K-300W =2686. Q=mc T -T -W =0. Q=mc T -T -W =0.5946 ×1004.5946 ×1004.5 5 25-20 K-300W =2686.25-20 K-300W =2686. Q=mc T -T -W =0.5946 ×1004.5 25-20 K-300W =2686. 4W 4W 4W4W s kgK s kgK s kgK s kgK Q =Q-Q =2686.4W -(-400)W Q =Q-Q =2686. Q =Q-Q =2686.4W -(4W -(--400)400)WW Q =Q-Q =2686.4W -(-400)W ɺ heat heat heat heat Q =3086.4W Q =3086.4W Q =3086.4W Q =3086.4W

(11)

10 1010 10

PROBLEM 3 - SOLUTION, cont.

AD.2 P

ower loss in the system.

( ) ( ) ( ) ∆     ∆ _ _ _ _   ɺ ɺ ɺ ɺ ɺ ɺ ɺ ɺ _ɺ ɺ _ɺ ɺ lost t,rev. t lost t,rev. t lost t,rev. t lost t,rev. t t fan t fan t fan t fan 1 11 1 lost t,rev. t p 1 2 0 1 2 fun p 1 2 0 p fun lost t,rev. t p 1 2 0 1 2 fun p 1 2 0 p fun lost t,rev. t p 1 2 0 1 2 fun p 1 2 0 p fun lost t,rev. t p 1 2 0 1 2 fun p 1 2 0 p fun 2 22 2 W =W -W W =W -W W =W -W W =W -W W =-W W =- W =-WW W =-W T TT T W =W -W =m c T-T -T s-s +W =m c T-T -Tcln +W W =W -W =m W =W -W =m c T-T -T s-s +W =m c T-T -T s-s +W =m c T-T -Tclc T-T -Tcln +Wn +W W =W -W =m c T-T -T s-s +W =m c T-T -Tcln +W T TT T OR

Use the Guoy Stodola Law

(

)

  ∆ ∆ ∆ ∆ _ _   ɺ ɺ ɺ ɺ ɺ ɺ ( ( ( (m) m) m) m) QQQQ _ɺ 2 2 2 2 _ɺ 2222 lost 0 n 0 0 0 0 p p 0 l losost t 0 n 0 n 0 0 0 0 0 0 0 0 p p p 0p 0 lost 0 n 0 0 0 0 p p 0 1 0 1 1 1 0 0 11 1 0 1 T T T T T T T Q Q QQ T W =T S =T S + S =T mcln - =mcTln -Q W =T S =T S + S =T mcl W =T S =T S + S =T mcln - =mcTln - =mcTln -Qn -Q W =T S =T S + S =T mcln - =mcTln -Q T T T T T T T TT T T T ∆ ɺ ⋅ ⋅ ⋅ lost lost lost lost kg kg kg kg J J J J 298298298298 W =0.5946 1004.5 293K ln -2686.4W W =0. W =0.5946 1004.5946 1004.5 5 293K l293K ln -2686.n -2686.4W4W W =0.5946 1004.5 293K ln -2686.4W s kgK 293 s s kgK kgK 293293 s kgK 293 ∆ ɺ lost l losostt lost W =274.8W W =274.8W W =274.8W W =274.8W

(12)

11 1111 11

PROBLEM 4

Water of 38O_{C 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 10O_{C and the hot one at 80}O_C.

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).}

(13)

12 1212 12 ENERGY TRANSFER 2013 / 2014

PROBLEM 4 - SOLUTION

⋅

ρ

ɺ

3 3 3 3 3 3 3 3 -4 - -44 -4 3 33 3 3 333 -4 - -44 -4 3 3 3 3 3 3 3 3

l

i

t

r

l

i

t

r m

m

l

i

t

r

l

i

t

r m

m

l

i

t

r

l

i

t

r m

m

l

i

t

r

l

i

t

r m

m

=1,

6610

=1,

6610

=1,

6610

mi

1mi

n

1mi

n

1mi

n

1mi

n

V =10 =10

60s

kg

= V =

n mi

n 10 l

i

t

r

s

n

mi

n 10 l

i

t

r

s

n mi

n 10 l

i

t

r

s

n

mi

n 10 l

i

t

r

s

m

1,

6610

m

1,

6610

m

1,

6610

m

1000

1,

6610

=0.

1666

s

mass balance:

m

ɺ

₃₃₃₃

m

= m

=

m +

=

ɺ

₁₁₁₁

m

ɺ

m

₂₂₂₂

First Law of Thermodynamics for an open steady-flow

system (energy balance)

(

)

(

)

∆

ɺ

ɺ ɺ ɺ ɺ

ɺ

_ɺ

ɺ

1 2 3 1 2 3 1 2 3 1 2 3 1 1 2 2 3 1 1 1 1 2 2 2 2 33 1 1 2 2 3 1 1 3 1 2 3 1 1 1 1 3 3 1 2 1 2 33 1 1 3 1 2 3 1 2 1 3 1 2 1 1 2 1 33 1 2 1 3 3 33 3 3 33 3 3 33 3 2 3 2 3 2 3 2 3

U=Q-W +H H H

Q+m +m

- Q+m +m -

Q+m +m

- Q

Q

+

- 0=

h

h m h

0=

h

h m h

0=

h

h m h

0=

h

h m h

0=

h m -m h m h

0=

h m -m h m h

0=

h m -m h m h

0=

h m -m h m h

h -h

+m +

- +m +

- m

Q+

- m

Q+

- m

Q+

- m

= m h m h

Q+

-AD.1.

Mass flow rates of the cold and hot water streams.

ɺ

₂₂₂

=0.

₂

=0.

06834

kg

m

s

(

)

(

)

(

)

ɺ

3 2 3 w 2 3 2 3 w 2 3 2 3 w 2 3 2 3 w 2 1 11 1 2 1 w 2 1 2 1 2 1 w 2 1w 2 1 2 1 w 2 1 2 22 2 3 33 3 3 33 3 3 33 3 1 11 1

m h -h

m c T -T

m h -h

m c T -T

m h -h

m c T -T

=

h -h

c T -T

h -h

c T -T

h -h

c T -T

h -h

c T -T

kg

= - =(

0. 16666-0.

098 = - =(

= - =(

0.

0. 16666-0.

16666-0.

098

098 = - =(

0. 16666-0.

098 Q

Q

2

2 +

Q+

+

Q+

+

Q+

m

=

m

=

m

=

m

=

m m m

6)

s

1

ɺ

m =0.

0 m =0.

m =0.

0

0 m =0.

=

= 9826

=

9826

0 kg

kg

s

(14)

13 1313 13

PROBLEM 4 - SOLUTION, cont.

AD.2. T

otal entropy generation and the power loss.

∆

ɺ

_ɺ

ɺ

m1 m2 m1 m1 m2m2 m1 m2 3 3 3 3 3 3 3 3 m1 m2 1 w 2 w m1 m1 m2 m2 1 w 1 w 2 w2 w m1 m2 1 w 2 w 1 11 1 m1 m2 m1 m1 m2m2 m1 m2 m1 m2 m1 m1 m2m2 m1 m2 (Q) ( (Q)Q) (Q) gen 0 gen gen 00 gen 0 2 22 2

S S

S S m

+m

S S m

+m

S S m

+m

S S =0.

09826×4200×l

n

+0.

0683

S S =0.

09826×4200×l

n

+0.

0683

S S =0.

09826×4200×l

n

+0.

0683

S = + + S

T

+

= cl

n

cl

n

+

= cl

n

cl

n

+

= cl

n

cl

n

T

311.

15

3

311.

15

3

311.

15

3

311.

15

3 4×4

4×4

11.

15

15 +

+

283.

1

283.

1

283.

1

283.

5

1 200×l

200×l

n

353.

15

15 S S =2.

S S =2.

+

573

573 W

W

K

∆

_S

ɺ

(

Q)

_{= =-}

Q

ɺ

0

0 Q

Q

ɺ

_=-

-

500

500 W

W

T T 2

9

3.

9 =1.

=1.

7

0

7

0

0 ₁₅

₁₅

0

56

56 K

K

0

0 ∆

ɺ

∆

ɺ

∆

ɺ

m1 m2 m1 m2 m1 m2 m1 m2 (Q) ( (Q)Q) (Q) gen 0 gen 0 gen 0 gen 0

W

S = + + S =(

2. 573+1.

7056

S = + + S =(

2. 573+1.

7056

S = + + S =(

2. 573+1.

7056

S = + + S =(

S S

2. 573+1.

7056

)

=4.

279

279 K

K

The loss power from Gouy-Stodola Law: W W T S 293.15 W 0 K 4, 366 1254.3 lost

=

ɺ

gen.

=

⋅

_K

=

ɺ

(15)

14 1414 14 ENERGY TRANSFER 2013 / 2014 T S cooler comp. turbine 1 1 g m p , t ɺ 2 2r g m p , t ɺ 3 3 a m p , t ɺ 4 4 a m p , t ɺ 4 5 a m p , t ɺ 1 c c m , tɺ m , tɺc c2 m NtT η T S cooler comp. turbine 1 1 g m p , t ɺ 2 2r g m p , t ɺ 3 3 a m p , t ɺ 4 4 a m p , t ɺ 4 5 a m p , t ɺ 1 c c m , tɺ m , tɺc c2 m NtT η

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 400O_C

and leave at 350O_{C. 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 70O_{C and 95kPa and leaves at}

135kPa. For simplicity assume that the exhaust gases and the air are ideal gases of the same c_p=1kJ/(kgK) and κ=1.4. What is an isentropic efficiency of the

compressor?

To avoid the possibility of an engine knock (due to a side effect of the air temperature increase in the compressor), a cooler is placed between the

compressor and the engine suction manifold to decrease the air temperature to 80O_{C. Cold}

ambient air is used as a cooling fluid. Its

temperature rises from 20O_{C to 40}O_{C between the}

inlet and the outlet of the cooler. What is a total lost power in the turbocharger assuming that the pressure in the turbine decreases 1.7 times, and

(16)

PROBLEM 5 - SOLUTION

η

ɺ

_g_g_g_g

ɺ

_a_a_a_a O O O O O O O O O O O O O O O O 1 2r 3 5 1 2r 3 5 1 2r 3 5 1 2r 3 5 1 2 3 4 1 1 2 2 3 3 44 1 2 3 4 m mm m O O O O O O O O c1 c2 c1 c2 c1 c2 c1 c2

m =0.

02kg/s;m =0.

018kg/s;

m =0.

02kg/s;m =0.

018kg/s;

m =0.

02kg/s;m =0.

018kg/s;

m =0.

02kg/s;m =0.

018kg/s;

t=400 C;t =350 C;t=70 C;

t=80 C

t=400 C;t =350 C;t =70 C;

t=80 C

t=400 C;t =350 C;t=70 C;

t=80 C

t=400 C;t =350 C;t =70 C;

t=80 C

p /p =1.

7;p =95kPa;

p =130kPa

p /p =1.

7;p =95kPa;

p =130kPa

p /p =1.

7;p =95kPa;

p =130kPa

p /p =1.

7;p =95kPa;

p =130kPa

=0.

95 =0.

=0.

95

95 =0.

95 t =20 C;

t =40 C

t =20 C;

t =40 C

t =20 C;

t =40 C

t =20 C;

t =40 C

Data 0 O p

c

=

1kJ/(kgK);

κ =

1.4; t

=

20 C

AD.1 – an isentropic efficiency of the compressor

κ κ





η

_

_

_

_

_



_

_

_

_













3 33 3 -1 -1 -1 -1 4 3 4 3 4 3 4 3 _0._0._0._0._4/1._4/1._4/1._4/1.₄₄₄₄ C, C, C, C, 4 44 4 4 3 4 4 33 4 3 ₄₄₄₄ ₃₃₃₃ 3 33 3 is. i is.s. is. r rr r

T =(

70+273.

15)

K=343.

15K

T =(

70+273.

15)

K=343.

15K

T =(

70+273.

15)

K=343.

15K

T =(

70+273.

15)

K=343.

15K

T -T

=

_p

₁₃₅

T -T

_{T =T}

_=343.

_15K

_=379.

_4K

p

95 p

95

(17)

PROBLEM 5 - SOLUTION, cont.

(

)

(

)

(

)

Cr

η

∆

⋅

ɺ

_ɺ

ɺ

_ɺ

ɺ

4r 4r 4r 4r C,r m T,r C, C,r r m T,m T,rr C,r m T,r C,r a p 4r 3 C, C,r r a p 4r 3a p 4r 3 C,r a p 4r 3 T,r g p 1 2r T,r g p 1 2r T,r g p 1 2r T,r g p 1 2r T,r T,r T,r T,r

from the First Law of Thermodynamics f frrom om tthe he FiFirrsst t Law Law of of TherThermodynamimodynamicscs from the First Law of Thermodynamics

T

W = W

W =

H =m c T -T

W =

H =m c T -T

W =

H =m c T -T

W =

H =m c T -T

W =m c T-T

kg

kg kJ

kJ

W =0.

02 1

400-350K=1

kW

W =0.

02 1

400-350K=1

kW

W =0.

02 1

400-350K=1

kW

W =0.

02 1

400-350K=1

kW

s kgK

(

)

(

)

(

)

η

ɺ

⇒

η

ɺ

g gg g m g p 1 2r a p 4r 3 4r 3 m 1 2r m g p 1 2r m g p 1 2r a p 4r 3 a p 4r 3 4r 4r 3 3 m m 1 2r1 2r m g p 1 2r a p 4r 3 4r 3 m 1 2r a aa a

m

m c

T-T =m c

T -T

m c

T-T =m c

T -T

m c

T-T =m c

T -T

T =T +

T-T =395.

93K

m

(

)

(

)

η

4 3 4 3 4 34 3 C, C, C, C, 4 3 4 3 4 3 4 3 is. is. is. is. r rr r

379. 4-343.

15K

379. 4-343.

15K

379. 4-343.

15K

379. 4-343.

15K

T -T

=

T -T 395.

93-343.

15K

T -T 395.

93-343.

15K

(18)

17 1717 17

AD.2. T

otal lost power in the turbocharger - calculations based on an isentropic process and the G.S. Law

PROBLEM 5 - SOLUTION, cont.

( ) ( ) ( )

∆

ɺ

∆

ɺ

∆

ɺ

∆

ɺ

lost lost lost lost

lost lost _T_T_T_T lost _C_C_C_C lost_cool_cool_cool_cool_._._._.

W = W

+ W

W = W

+ W

W = W

+ W

W = W

+ W

(

)

(

)

(

)

(

)

(

)





∆

_{⇒ }



∆

=



ɺ

_ɺ

ɺ

_ɺ

ɺ

_ɺ

T,r g p 1 2r T, T,r r g p 1 2rg p 1 2r T,r g p 1 2r lost T Tr T g p 1 2 l losost t T T Tr Tr T T g p 1 2g p 1 2 lost T Tr T g p 1 2 T TT T lost g p 2r 2 l losost t g p 2r 2g p 2r 2 lost g p 2r 2 T TT T

W =m c T-T

W

=W -W

W =m c T-T

W

=W -W

W =m c

T-T

W

=W -W

W =m c T-T

W

=W -W

W =m c

T-T

W

m c T -T

W

m c T -T

W

m c T -T

W

m c T -T

(

)

(

)

(

)

(

)

κ κ





_

_

















∆

ɺ

⋅

-1 -1 -1 -1 _0._0._0._0.₄₄₄₄ 1.4 1.4 1.4 1.4 _O OO O 2 22 2 2 1 2 1 2 1 2 1 1 11 1 lost l losostt lost_T_T_T_T

p

1

1 T =T

= 673.

15K

=578.

45K 305.

3 C

T =T

= 673.

15K

=578.

45K 305.

3 C

T =T

= 673.

15K

=578.

45K 305.

3 C

T =T

= 673.

15K

=578.

45K 305.

3 C

p

1.

7 p

p

1.

7

7 p

1.

7 kg

kg

kg kJ

kJ

W

=0.

02 1 350-305.

3K=0.

894kW

W

=0.

02 1 350-305.

3K=0.

894kW

W

=0.

02 1 350-305.

3K=0.

894kW

W

=0.

02 1 350-305.

3K=0.

894kW

s kgK

Lost power in the turbine

T

turbine

1 1 g

m

p , t

ɺ

2 2r g

m

p , t

ɺ

T

turbine

1 1 g

m

p , t

ɺ

2 2r g

m

p , t

ɺ

(19)

18 1818 18

PROBLEM 5 - SOLUTION, cont.

Lost power in the compressor

(

)

(

)

(

)

(

)

(

)





∆

_



∆



ɺ

_ɺ

ɺ

_ɺ

ɺ

_ɺ

C,r a p 4r 3 C, C,r r a p 4r 3a p 4r 3 C,r a p 4r 3 lost C,r C C a p 4 3 l losost t C,C,r r C C C C a p 4 3a p 4 3 lost _C C,r C C a p 4 3 CC C lost a p 4r 4 lost a p 4r 4 lost a p 4r 4 lost _C a p 4r 4 CC C

W =m c

T -T

W =m c

T -T

W =m c

T -T

W =m c

T -T

W

=W -W

W

=W -W

W

=W -W

W

=W -W

W =m c

T -T

W

=m c T

-T

W

=m c T

-T

W

=m c T

-T

(

∆

ɺ

)

⋅

(

)

lost lost lost lost_C CC C

kg

kg kJ

kJ

W

=0.

018 1 395.

93-379.

4K=0.

298kW

W

=0.

018 1 395.

93-379.

4K=0.

298kW

W

=0.

018 1 395.

93-379.

4K=0.

298kW

W

=0.

018 1 395.

93-379.

4K=0.

298kW

s kgK

S

comp.

3 3 a

m

p , t

ɺ

4 4 a

m

p , t

ɺ

S

comp.

3 3 a

m

p , t

ɺ

4 4 a

m

p , t

ɺ

₌

_395.93K;

₌

_{379, 4K} 4r 4 4r 4 4r 4 4r 4

T

(20)

19 1919 19

PROBLEM 5 - SOLUTION, cont.

Power loss in the cooler

( )

5 4 5 cool. 4r 4 4r a a p a p T p T S m c ln R ln m c ln T p T



 



∆

=

_

−

_{ }

_

=

 





ɺ

_ɺ

cooler 4 5 a

m

p , t

ɺ

1 c c

m , t

ɺ

m , t

ɺ

c c2 cooler 4 5 a

m

p , t

ɺ

1 c c

m , t

ɺ

m , t

ɺ

c c2

ɺ

4 4 a m t ,p

(

∆

W

ɺ

lost

)

_cool.

= ∆ = ∆

T0 S

ɺ

cool. T0

(

( )

S

ɺ

a _cool.

+

∆

S

ɺ

c

)

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

(

)

cool. cool. cool. a c H Q W H H H Q W 0 0

∆ = −

∆ = ∆

∆

=



₊

₌



=

_

ɺ

( )

3 cool. a c2 c c p c1 kg kJ (80 273.15) kW S 0.018 1 ln 2.06 10 s kg K 395.93 K T S m c ln T −

+

∆

=

⋅

=

−

⋅

∆ =

ɺ

_ɺ

(21)

20 2020 20

PROBLEM 5 - SOLUTION, cont.

(

)

(

)

(

)

(

)

(

)

(

)

(

)

cool. _4r ₅ a a p 5 4r c a c2 c1 c c p c2 c1 c H m c T T _T _T m m T T H m c T T 395.93 80 273.15 kg kg m 0.018 0.0385 s 40 273.15 20 273.15 s

∆

=

−



₋

 ⇒ =



−

∆ =

−

_

−

+

=

+

−

+

ɺ

_ɺ

ɺ

_ɺ

ɺ

(

)

(

)

3 c2 c c p c1 40 273.15 T kg kJ kW S m c ln 0.0385 1 ln 2.54 10 T s kg K 20 273.15 K −

+

∆

=

⋅

=

⋅

+

ɺ

_ɺ

(

)

(

( )

)

(

)

(

)

3 3

lost _cool. 0 _cool. lost cool. a c kW W T S S 293.15K 2.06 10 2.54 10 K W 0.1407kW − −

∆

= ∆

+ ∆ =

⋅

− ⋅ + ⋅

∆

=

ɺ

( ) ( ) ( )

lost lost _T lost _C lost _cool.

W W W W 0.894kW 0.298kW 0.1407kW

∆ =

ɺ

∆

ɺ

+

∆

ɺ

+

∆

ɺ

=

+

lost

W

1.3327kW

(22)

21 2121 21

PROBLEM 6

Liquid water of 20O_{C and the mass flow rate of 2.5kg/s is heated to 60}O_C

by mixing it with superheated steam of 150O_{C 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 25O_{C. 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 150O_C

and 200kPa, are 2769kJ/kg and 7.28kJ/(kgK), respectively.

Specific enthalpy and specific entropy of a liquid water can be approximated as h_w=c_wt_w and s_w=c_wln(T_w /273K), respectively.

(23)

22 2222 22 ENERGY TRANSFER 2013 / 2014

PROBLEM 6 - SOLUTION

(

)

(

)

0 w v w w w m w v v w m v H H H Q W W H m c t t H m c t h



∆ = ∆ + ∆ = −



=

_



∆

=

−

_



∆

=

−

_

ɺ

_ɺ

ɺ

_ɺ

(

)

(

)

(

)

(

)

(

)

kg kJ 20kW 2 5 4 22 60 20 K s kgK 0 16kg/s kJ kJ 4 22 60K 2769 kgK kgK w w m w v w m v w w m w v w m v v m c t t m c t h Q Q m c t t m c t h . . m . .

− +

− =

−

=

−

⋅

−

=





−









ɺ

_ɺ

ɺ

(

)

0 0 lost gen gen m w v W T S S S S S S



∆

=

_



= − + + ∆ 

ɺ

ɺ ɺ ɺ

ɺ

(

)

0 0 273K 273K m m w v w w w w w v v v T S m m c ln T S m c ln S m s Q S T

=

+

=

∆ = −

ɺ

_ɺ

ɺ

_ɺ

ɺ

_ɺ

ɺ

(24)

23 2323 23

PROBLEM 6 - SOLUTION, cont.

(

)

(

)

(

)

0 kg kJ 60 273 K kW 2 5 0 16 4 22 2 23 s kgK 273K K kg kJ 20 273 K kW 2 5 4 22 0 746 s kgK 273K K kg kJ kW 0 16 7 28 1 165 s kgK K 20kW kW 0 067 25 273 K K m w v ( ) S . . . ln . ( ) S . . ln . S . . . S .

+

=

+

⋅

=

+

=

⋅

=

⋅

=

− −

∆ =

=

+

ɺ

(

)

0 kW kW kW kW kW 2 23 0 746 1 165 0 067 0 386 K K K K K kW 25 273 K 0 386 K gen lost gen S . . . . . W T S .





=

−

_

+

_

+

=





∆ =

= +

⋅

ɺ

115kW lost W

∆

ɺ

=

(25)

24 2424 24

PROBLEM 7

Superheated steam enters the turbine of 4MW power at 2.1MPa and

temperature of 475O_{C. 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 30O_{C by the stream of cooling water,}

which enters the condenser at 15O_{C and leaves at 25}O_C.

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 15O_{C. Specific heat of liquid water is 4.19kJ/(kgK)}

and its specific enthalpy and specific entropy can be approximated as h_w=c_wt_w and s_w=c_wln(T_w /273K ), respectively.

From the steam tables the following data are given:

for 2.1MPa & 475O_{C 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.

(26)

25 2525 25 ENERGY TRANSFER 2013 / 2014

PROBLEM 7 - SOLUTION

(

)

{

(

1

)

2

(

)

2 1 4000kW 0 4 835kg/s 3411 3 2584 kJ/kg t t v v H Q W W Q m . h h . H m h h



∆ = −



=

_

→

=

−



∆ =

−

_

ɺ

_ɺ

ɺ

_ɺ

Mass flow rate of the steam from the First Law of Thnermodynamics for the turbine

(

)

(

)

(

)

12 0 12 0 2 1 0 2 1 kg kJ 288 15K 4 835 8 15 7 34 s kgK " t , v , v v W T S T m s s T m s s . . . .

∆

ɺ

=

∆

ɺ

=

_ɺ

−

=

_ɺ

−

=

⋅

−

Lost power in the turbine from the Guy Stodola law

1 2 1128 5kW t ,

W .

∆

ɺ

=

Power loss in the turbine

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

0 c c tc tc c vc cw H Q W Q W H H H



∆

=

−



=

_



∆ = ∆ + ∆

_

ɺ

(

)

(

)

3 2 2 1 vc v cw cw w w w H m h h H m c t t

∆

=

−





∆ = ⋅

−



ɺ

_ɺ

ɺ

_ɺ

where

(27)

26 2626 26

PROBLEM 7 - SOLUTION, cont.

(

)

(

)

(

)

(

)

(

)

(

)

2 3 3 2 2 1 2 1 2 3 3 0 kJ kJ kJ 2584 4 19 30K 125 7 kg kgK kg kg kJ 4 835 2584 125 7 kg s kg 283 67 kJ _s 4 19 25 15 K kgK v v cw w w w cw w w w w cw m h h m h h m c t t m c t t h ; h c t . . . . m . .

−

− + ⋅

− = → =

−

=

⋅

=

−

=

−

ɺ

Lost power in the condenser from the Guy Stodola law

(

)

(

)

(

)

2 0 0 23 0 3 2 1 3 3 kJ 273 30 K kJ 4 19 0 437 273K kgK 273K kgK kg kJ kg kJ 298K 288K 4 835 0 437 8 15 283 67 4 19 s kgK s kgK 288K W tcond cond. v , cw v cw w w w tcond T W T S T S S T m s s m c ln T T ( ) s c ln . ln . W . . . . . ln





∆

=

= ∆ + ∆ =

_

− +

_





+

=





∆

=

_

−

+

⋅

_





ɺ

_ɺ

ɺ

943 88kW tcond W .

∆

ɺ

=

(28)

27 2727 27

PROBLEM 8

Pure nitrogen at 0.1MPa and 25O_{C 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 25O_{C are, respectively, 1.5}_·_10-10m2_{/s and}

0.00156 kmol/(m3_bar).

(29)

28 2828 28 ENERGY TRANSFER 2013 / 2014

PROBLEM 8 - SOLUTION

2 2

0

A A C

= ℜ

p

=

Molecular diffusion of nitrogen through a cylindrical wall

r A C A,1 C A,2 C L B A + A 2 r 1 r B r A C A,1 C A,2 C L B A + BB A A + A A 2 r 1 r B B

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

3 3 2 2 kmol kmol 0 79 1bar 0 001232 m bar m

0 00156

A A C

= ℜ

p

=

.

⋅

.

⋅

=

.

(

)

(

)

2 3 10 12 kmol 0 00156 0 001232 m m kmol 2 10[m] 1 5 10 2 47 10 s 0 017 0 015 s A . . . . ln . / .

J

− −





−

_

_





_

_

_

_

=

π⋅

⋅

_

_

⋅

=

⋅

_

_









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

(

)

(

)

2 3 10 11 kmol 0 00156 0 m m kmol 2 10[m] 1 5 10 1 17 10 s 0 017 0 015 s A . . . ln . / .

J

− −





− 







_

_

_

_

=

π ⋅

⋅

_

_

⋅

=

⋅

_

_









1 2 3 3 1 1 10m 0 015m 0 017m kmol kmol 0 00156 ×1bar 0 00156 m bar m A A where L ; r . ; r . C

p

. .

=

= ℜ

=

(

)

1 2 2 1 kmol 2 s A A A A AB C C n LD const. ln r / r

J

= π

−



_



_

=





≡

ɺ

(30)

29 2929 29

PROBLEM 9

A 230 mm diameter pan of water at 22O_{C has a mass loss rate}

1.5·10-5 _{kg/s when the ambient air is dry and at 22}O_C.

Determine the convection mass transfer coefficient.

Estimate the evaporation mass loss rate when ambient air has a relative humidity of 50% and remains at 22O_C.

Water vapour saturation pressure at 22O_{C is equal to 2.617kPa.}

(31)

30 3030 30 ENERGY TRANSFER 2013 / 2014

PROBLEM 9 - SOLUTION

Data

A – water vapour B - dry air 2 O 1 5 1 2 0.23m 22 C 297K; 1 5 10 kg/h 22 2 617kPa 0 0 0 5 18kg/kmol 8315J/(kmol K) ( ) A A O s A H O d = ; t = = J m . ; p ( C) . ; . ; . ; M M ; B −

= ∆ = ⋅

=

ϕ

= ϕ =

=

⋅

ɺ

Solution

(

)

(

)

(

)

1 1 1 1 1 1 1 ( ) ( ) ( ) A A m AS A A m _{( )} ( ) ( ) ( ) AS A A A m AS A J m A k _{m / A} k j m / A k ∞ ∞ ∞



= ∆

= ⋅ ρ − ρ

_∆

 ⇒

=



ρ − ρ

= ∆

= ρ − ρ

_

ɺ

_ɺ

ɺ

2 2 2 1 3 3 0 23m 0 0415m 4 4 22 C 22 C kg 0 0 19 2 10 m o o ( ) s s A AS A A where d ( . ) A . p ( ) p ( ) . ; M . R T BT − ∞

=

π =

π

=

ρ

=

ρ

=

⋅

(32)

31 3131 31 ENERGY TRANSFER 2013 / 2014

PROBLEM 9 - SOLUTION,cont.

(

)

₍

₎

5 1 2 3 3 kg 1 5 10 s kg 0 0415m 19 2 10 0 0 m A m _{( )} AS A . m / A k . . . − − ∞

⋅

∆

=

ρ − ρ

_⋅

₋

ɺ

o o 2 2 2 1 3 3 2 3 3 22 C 22 C kg kg 0 5 19 2 10 9 6 10 m m ( ) s s ( ) A A A A p ( ) p ( ) M . , . R T BT − − ∞ ∞

ϕ ⋅

ρ =

=

ϕ

⋅ρ = ⋅ ⋅

= ⋅

(

)

(

)

2 2 2 2 3 3 3 m kg 0 0415m 1 883 10 19 2 9 6 10 s m ( ) ( ) ( ) A A m AS A J

=

m

∆ = ⋅ ρ − ρ =

ɺ

A k _∞ .

⋅

.

⋅

−

⋅

.

− ⋅

. − 2

m

1 883 10

s

m

k

=

.

⋅

− 2 2 6

kg

7 5 10

s

( ) ( ) A A

J

= ∆

m

ɺ

= ⋅

.

−

(33)

32 3232 32

PROBLEM 10

A cylindrical jug of 8cm internal diameter and 30cm height is half filled with water and left in a dry air at 15O_{C and pressure of 87kPa.}

Its top is open. Saturated pressure at 15O_{C 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).

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33 3333 33

It is Stefan flow.

Assumption of constant water level in the pitcher means a steady state with constant mass flux

PROBLEM 10 - SOLUTION

A A A m m J

∆ = τ = τ

_ɺ

where A A A A J

= ⋅ = ⋅ ⋅

A j A M j 0 1 1 A,L AB A A , x C D j ln L x

−

⋅

=

−

Stefan law 3 0 3 2 3 3 0 15 C 1 705kPa 19 6 10 87kPa N 87 10 kmol m _{36 33 10} J _m 8315 288K kmol×K A,L O s A, x p ( ) . x . p p C . BT − −

=

⋅

=

⋅

Assumptions

Solution

where A – water vapour

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34 3434 34 ENERGY TRANSFER 2013 / 2014 2 3 5 3 6 3 2 kmol m 36 33 10 2 6 10 1 0 kmol m s _{12 47 10} 0 15m 1 19 6 10 m s A . . j ln . . . − − − −

⋅

₋

=

⋅

−

⋅

(

2

)

2 2 3 2 8 10 m 5 027 10 m 2 4 D A . − −

π ⋅

π

=

⋅

A A A A A m m J A M j

∆ = τ = τ = ⋅ ⋅ ⋅ τ

ɺ

3 2 6 2 kg kmol 5 027 10 m 18 12 47 10 24 3600 s kmol m ×s A m . − . − ( )

∆

=

⋅

PROBLEM 10 - SOLUTION, cont.

2

9 75 10 kg

A

m

.

−

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35 3535 35

PROBLEM 11

To keep a can of beverage at 6O_{C in hot dry ambient air at temperature of}

27O_{C, 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 6O_{C is equal to 5kPa. Thermophysical properties of dry}

air at 27O_{C are as follows: thermal conductivity k=0.026W/mK; density}

ρ=1.16kg/m3; specific heat c_p=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).

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36 3636 36

PROBLEM 11 - SOLUTION

A

- vapour of the liquid agent;

B

- dry ambient air

(

)

(

)

(

)

S A A S m A,S A, A A m A,S A , S A m A ,S A, h(T T ) j r h(T T ) h r j h (T T ) h r h ∞ ∞ ∞ ∞ ∞ ∞

−

=

_

⇒

− = ρ − ρ ⋅



= ρ − ρ

_

−

=

⋅

ρ − ρ

Energy balance

(

)

2 3 2 3 5 3 2 3 3 3 3 W 0 026 mK _{1 121} kg J m 1 16 10 2 10 m kgK s kg J J 1 16 10 1 121 1251 78 m kgK m K / p m AB p AB / m h c Le h . k Le . D c D . h . . . h −