Download Exercises in Applied Thermodynamics, First Year, Set 3

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
 of the extensive property is given by
The rate of change, 
n
 =   i

i =1
Applications
i) Conservation of mass
The T-elbow of hot and cold 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.
Warm water to
shower
Hot water in
Cold water in
Open system
(ii) Conservation of matter in a combustion chamber
In a combustion chamber, under steady flow conditions, net inflow of matter is zero.
Also conservation can be applied to each chemical element.
Fuel in
Products of combustion out
Air in
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 i1 + i2 + i3 + i4 = 0).
i1
i4
i2
i3
2. Properties common to two systems in equilibrium
i) Simple thermodynamic systems: temperature
ii) Electrically charged bodies: electric potential
iii) Fluids in separate vessels connected by a valve: pressure
3
pi = 300 kPa
Vi= 150 l
po=105 kPa
Given that theprocess is fully rtesisted,
2
W1 2   p dV
1
450
400
ABSOLUTE PRESSURE/kPa
350
300
250
200
150
100
50
0
0
100
200
300
400
500
600
700
VOLUME/ l
FIG.Q3:ABSOLUTE PRESSURE VS VOLUME
By graphical interpolation of Figure 1 to give 6 equal intervals of V at 100 litres each,
Table Q3a: Absolute pressure-volume relationship
p/ kPa
V/l
405
150
245
250
185
350
155
450
115
550
85
650
70
750
800
Using Simpson’s rule, for any even number (2n) of equal width strips width x, area A is
given by, A = x (y1+ 4y2+2y3+ . . . + 4y2n+ y2n+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
V/l
405
150
210
300
155
450
95
600
70
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 = Ws = W – po (V1– V2) = – 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 partly 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 piston.
(c)
5
The expression is not for open systems and therefore cannot be used for any flow
process.
For an ideal gas, pV = mRT
For the process, pVn = K
2
Therefore, work done in process 1-2 = W1 2    pdV
1
n
=   KV dV = (p2V2 – p1V1) / (n-1)
For a constant pressure process, p = constant. Therefore n = 0. p2 = p1= p
W1-2 = (p2V2 – p1V1) / (n-1) = p (V1-V2)
For a constant volume process, V = constant. Therefore n = .
W1-2 = 0
For constant temperature process, n = 1. The formula, W1-2 = (p2V2 – p1V1) / (n-1)
cannot be used, and the result is obtained by integration as
W1-2 = mRT ln (V1/V2)
6
Cyclic process: There is no net change in the system boundary during a cycle. So there 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
Qc
Ws
Qin = 68 kW
Wd
Qx
System boundary
The boundary interactions including the unaccounted heat interaction rate Qx are as
identified in the figure above. Ws and Ws are, respectively the work output rates to the
shaft and the dynamo, Qin is the heat supply rate, and Qc is the heat removal rate by the
cooling water.
Rate of work transfer through shaft = Ws
Power delivered to dynamo =
Total work interaction rate
Wd
Ratio of shaft power to heat input
= 
= 40*2*4000/60 W
= 16.8 kW
= 0.5 kW
(by data)
= 16.8 kW + 0.5 kW = 17.3 kW
= 17.3 / 68
= 0.254
(or 25.4%)
Applying the first law of thermodynamics to the system,
Qin – Qc – Qx – Ws- Wd
=0
Qx = 68 - 12.5 -17.3
= 38.2 kW
2
partition
system boundary
acid
water
Before breaking of partition
mixture
After breaking of partition
Consider the rigid insulated sealed vessel shown above. There is no boundary interaction
Therefore there is no change 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.
p
3
Data:
m = 0.2 kg
R = 287 J/kg K
 = 1.4
p1 = 800 kPa, T1 = 1000 K
Using ideal gas equation,
V1 = m R T1/ p1 = 0.0717 m3 = 71.7 litres
1
pV1.2 = constant
2
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).
Cv = R / (–1) = 717 J/kg K
Cp =  R / (–1) = 1004 J/kg K
V1 = m R T1/ p1 = 0.0717 m3 = 71.7 litres
According to the law of expansion,
p1V11.2 = p2V21.2
 V2 = (100/800) 1/1.2 V1 = 407 l
Using ideal gas equation,
T2 = p2V2 T1 / p1V1 = 709 K
Work done on the system, W1-2 = (p2V2 – p1V1) / (n –1) = m R (T2 – T1) / (n –1)
= 0.2*287* (709 - 1000) / 0.2
= – 83.3 kJ
Change in internal energy U = m Cv (T2 – T1) = 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
V
4
Data:
position 2
position 1
V = 1 litre
R = 287 J/kg K
 = 1.4
p1 = 100 kPa, T1 = 303 K
50 mm
p
Using ideal gas equation,
m = p1 V1 / R T1 = 1.15*10 -3 kg = 1.15 g
Cv = R / (–1) = 717 J/kg K
Cp =  R / (–1) = 1004 J/kg K
2
pV = constant
Table of properties of the air
State p/kPa V / l
T/K
1
1
100
1.0
303
2
201
0.607
370
(Numbers in red are directly obtained from the
data. Numbers in green are calculated values).
V1 – V2 = ( d 2 /4)* x, where x is the inward displacement of the piston.
V2 = 1 – (*0.1*0.1*0.05 / 4 * 103) litre
= 0.607 l
According to law of expansion,
p1V1 = p2V2

p2 = 100* (1/0.607) = 201 kPa
Using ideal gas equation,
T2 = p2V2 T1 / p1V1 = 367 K
V
Work done on the system, W1-2 = (p2V2 – p1V1) / (n –1) = m R (T2 – T1) / (n –1)
= 1.15*10 -3 *287* (370 - 303) / 0.4
= 55 J
Change in internal energy U2 – U1 = m Cv (T2 – T1) = 1.15*10 -3 * 717 (370-303) J = 55 J
By first law of thermodynamics, U2 – U1 = Q1-2 + W1-2
Heat interaction, Q1-2
= 0 kJ
(A small difference between W1-2 and U2 – U1 is possible because of rounding errors in the
arithmetic.)