Exercises in Applied Thermodynamics, First Year, Set 3

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Oct 27, 2013 (3 years and 11 months ago)

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Solutions to Exercises in Applied Thermodynamics


Set 3


1.

General equation of continuity for an extensive property

of an open system with
n

streams transporting

i

into the system is


n





=



i


i

=1


The rate of change
,

of the extensive property is given by



n




=


i


i
=1


Applications

i)

Conservation of mass


The T
-
elbow of hot and co
ld water shower serves as the mixing chamber for the hot
and cold water streams. Under steady flow conditions the net mass flow will be zero.




Hot water in






Cold water in







(ii) Conservation of matter in a combustion chamber

In a combustion cham
ber, under steady flow conditions, net inflow of matter is zero.
Also conservation can be applied to each chemical element.











iii)
Conservation of electric charge

The junction of an electric circuit cannot store electric charge and conservation of

electric charge is stated as Kirchoff’s law (which, in this case, is
i
1
+

i
2
+

i
3
+

i
4

= 0
).













i
4

i
3

i
2

i
1

Products of combustion out

Air in

Fuel in


Warm water to
shower

Open system



2.


Properties common to two systems in equilibrium

i) Simple thermodynamic systems: temperature

ii) Electrically charged bodies: electri
c potential

iii) Fluids in separate vessels connected by a valve: pressure



3



p
i
= 300 kPa




V
i
= 150 l

p
o
=105 kPa


Given that theprocess is fully rtesisted,



FIG.Q3:ABSOLUTE PRESSURE VS VOLUME


By graphical interpolation of F
igure 1 to give 6 equal intervals of
V

at 100 litres each,


Table Q3a: Absolute pressure
-
volume relationship

p
/ kPa

405

245

185

155

115

85

70

V
/
l

150

250

350

450

550

650

750



Using Simpson’s rule, for any even number (
2n)
of equal width strips width

x
, area
A

is
given by,
A

=

x

(
y
1
+ 4
y
2
+2
y
3
+ . . . + 4
y
2n
+
y
2n+1
) / 3


Work done,
W

= area under
p
-
V

curve in Figure Q3

=

100 [405+ 4(245+ 155+ 85) + 2(185+115) + 70] / 3

=

100 500 J =


100 kJ



The result could be obtained with the tabulated values
, ignoring
V

= 200 litres to give
Table 3b and very nearly the same result as above


Table Q3b: Absolute pressure
-
volume relationship

p
/ kPa

405

210

155

95

70

V
/
l

150

300

450

600

750


Work done,
W

= area under
p
-
V

curve in Figure Q3

=

150 [405+ 4(210+
95) + 2(155) + 70] / 3

=

100 250 J =


100 kJ


Shaft work =
W
s

=
W


p
o
(
V
1


V
2
) =


100 kJ


105 (150
-
750)/1000 =


37 kJ




4.

The expression is valid only for fully resisted expansion of a closed system.

(a)


i)

The expression cannot be used for a p
artly resisted expansion.

ii) The expression cannot be used for an un
-
resisted expansion.

iii) The expression can be used for a fully resisted expansion.




(b)



i)

Friction does not affect the process within the closed system comprising the gas.
Thus,
the expression can be used to find the work done on the system.

ii)

The expression cannot be used to find the work done on the system comprising
the surroundings, as friction occurs within the surroundings and affects the work
transmitted across the pist
on.



(c)

The expression is not for open systems and therefore cannot be used for any flow
process.



5

For an ideal gas,
pV

=
mRT


For the process,
pV
n

=
K

Therefore, work done in process 1
-
2 =


=

=
(
p
2
V
2



p
1
V
1
) / (
n
-
1)

For a constant pressure process,
p

= constant. Therefore
n
= 0.
p
2

=
p
1
=
p

W
1
-
2

=
(
p
2
V
2



p
1
V
1
) / (
n
-
1)


= p (V
1
-
V
2
)


For a constant volume process,
V

= constant. Therefore
n
=

.

W
1
-
2

= 0





For const
ant temperature process,
n

= 1. The formula,
W
1
-
2
=

(
p
2
V
2



p
1
V
1
) / (
n
-
1)
cannot be used, and the result is obtained by integration as

W
1
-
2

=
mRT

ln

(V
1
/V
2
)












6

Cyclic process
: There is no net change in the system boundary during a cycle. So th
ere is
no work done in displacing the matter in the surroundings. Thus the use of either gauge
pressure or absolute pressure will give the same result.


Non
-
cyclic process
: Here the use of gauge pressure will give the shaft work or useful work
and the use
of absolute pressure the work interaction across the system boundary.
































Set 5



1










The boundary interactions including the unaccounted heat interaction rate
Q
x

are as
identified in the figure above. W
s
and W
s
are, resp
ectively the work output rates to the
shaft and the dynamo, Q
in
is

the heat supply rate, and Q
c

is the heat removal rate by the
cooling water.


Rate of work transfer through shaft = W
s


=






= 40*2

*4000/60 W

= 16.8 kW


Power delivered to dynamo =

W
d


= 0.5 kW (by data)

Total work interaction rate




= 16.8 kW + 0.5 kW =
17.3 kW




Ratio of shaft power to heat input



= 17.3 / 68

=
0.254

(or 2
5.4%)


Applying the first law of thermodynamics to the system,

Q
in



Q
c



Q
x



W
s
-

W
d

= 0

Q
x

= 68
-

12.5
-
17.3


=
38.2 kW



2













Consider the rigid insulated sealed vessel shown above. There is no boundary interaction
Therefore there is no chan
ge energy content.


When the system is cooled there is a heat interaction with heat flow to the surroundings,
but no work. Therefore the energy content will decrease.




Q
in

= 68 kW





System boundary

Q
c

Q
x

W
s




W
d

partition system boundary

acid water mix
ture

Before breaking of partition

After breaking of partition

3

Data:


m

= 0.2 kg


R

= 287 J/kg K




= 1.4


p
1

= 800 kPa,
T
1

= 1000 K

Using ideal gas equation,


V
1

=
m R T
1
/

p
1
= 0.0717 m
3
= 71.7 litres


Table of properties of the air


State

p
/kPa

V

/
l

T

/ K

1

800

71.7

1000

2

100

407

709

(Numbers in red are directly obtained from the data. Numbers
in green are calculated
values).



C
v
=
R

/ (


1) = 717 J/kg K


C
p
=


R

/ (


1) = 1004 J/kg K


V
1

=
m R T
1
/

p
1
= 0.0717 m
3
= 71.7 litres


According to the law of expansion,


p
1
V
1
1.2

=
p
2
V
2
1.2




V
2

= (100/800)

1/1.2
V
1
= 407
l



Using ideal gas e
quation,


T
2

=
p
2
V
2
T
1
/
p
1
V
1

= 709 K





Work done on the system,

W
1
-
2

= (
p
2
V
2


p
1
V
1
) / (
n


1) =
m R
(
T
2


T
1
) / (
n


1)






= 0.2*287* (709
-

1000) / 0.2






=


83.3 kJ









Change in internal energy

U

=
m C
v

(
T
2


T
1
) = 0.2*

717 (709
-
1000) J =


41.7 kJ






By first law of thermodynamics,

U

=
Q


W




Heat interaction,
Q


=


41.7


(

83.3) =
41.6 kJ










pV
1.2

= constant

V

p

1

2

4

Data:


V

= 1 litre


R

= 287 J/kg K




= 1.4


p
1

= 100 kPa,
T
1

= 303 K





Usi
ng ideal gas equation,


m
=

p
1
V
1

/
R T
1
= 1.15*10
-
3

kg = 1.15 g


C
v
=
R

/ (


1) = 717 J/kg K


C
p
=


R

/ (


1) = 1004 J/kg K


Table of properties of the air


State

p
/kPa

V

/
l

T

/ K

1

100

1.0

303

2

201

0.607

370

(Numbers in red are directly obtaine
d from the
data. Numbers in green are calculated values).


V
1


V
2
= (


d

2
/4)*

x
, where

x

is the inward displacement of the piston.


V
2
= 1


(

*0.1*0.1*0.05 / 4 * 10
3
) litre

= 0.607
l

According to law of expansion,


p
1
V
1


=
p
2
V
2




p
2

= 100*
(1/0.607)


= 201 kPa

Using ideal gas equation,


T
2

=
p
2
V
2
T
1
/
p
1
V
1

= 367 K





Work done on the system,

W
1
-
2

= (
p
2
V
2


p
1
V
1
) / (
n


1) =
m R
(
T
2


T
1
) / (
n


1)






= 1.15*10
-
3

*287* (370
-

303) / 0.4






=

55 J









Change in

internal energy
U
2


U
1

=
m C
v

(
T
2


T
1
) = 1.15*10
-
3

* 717 (370
-
303) J = 55 J






By first law of thermodynamics,


U
2


U
1

=
Q
1
-
2

+
W
1
-
2



Heat interaction,
Q
1
-
2


=
0 kJ


(A small difference between
W
1
-
2

and
U
2


U
1
is possible

because of rounding errors in the
arithmetic.)

50 mm

pV



=

constant

V

p

2

1

position 2 position 1