MECH 390
Assignment 1
Fatima Alsoufi
1. Reversible Process, Closed System
For a closed system, the first and second laws in rate form are (see Chapters 3 and 4 of the
textbook for details):
dU  
= Q −W
dt
Q
dS
= ∑ i + Sgen
dt
i Ti
(1)
In a reversible process, the system is always in equilibrium which implies homogeneous
temperature and pressure in the system and vanishing entropy generation; under these
conditions, the power is:
dV
Wrev = p
dt
(2)
Manipulate the above equations to determine the expressions for specific heat (𝑞𝑞12) and work
(𝑤𝑤12) for a process between two states, state 1 and state 2, that is:
1. isochoric (constant volume),
2. isobaric (constant pressure),
3. isentropic (adiabatic and reversible),
4. isothermal (constant temperature).
See Ch. 7 of the textbook
2. Dual Cycle Engine
Consider an engine that operates on a dual cycle (Text Ch. 8.9), comprised of the following
processes:
o 1-2: adiabatic reversible compression from 100 kPa and 300 K with a
compression ratio of 14.34,
o 2-3’: isochoric heating with a pressure ratio of 1.494,
o 3’-3: isobaric heating to 2200 K,
o 3-4: adiabatic reversible expansion,
o 4-1: isochoric cooling.
Assume the working fluid is an ideal gas (air) with variable specific heats and R = 0.287
kJ/kg·K.
a) Draw the process curve on a p-v diagram and a T-s diagram.
b) Make a table of pressure p, temperature T, specific volume v, internal energy u, enthalpy
h, and entropy s at all points (keep in mind that the provided property tables give the
temperature part of entropy).
c) Determine the net-work of the cycle.
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© A. Rowe
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MECH 390
Assignment 1
d) Determine the thermal efficiency of the cycle.
e) If the engine is four-stroke and delivers 22 kW of power at 1600 rpm, what is the
total displacement volume of the cylinders?
3. Balance expressions
Assume the human body has a specific energy of e kJ/kg and the average human has a mass
mh kg. Provide an energy balance describing the rate of change in embodied human energy in
Canada. The number of people in Canada at some instance in time is N(t).
4. Open System Devices
Beginning with the full 1st and 2nd law for open systems, determine heat and work for
compressors, turbines, pumps, heating and cooling of flows, and throttling devices. Clearly
state all steps and assumptions. Textbook Secs. 9.5, 9.6, 9.8-9.10
5. Open System Entropy Generation
Ambient air enters a device operating at steady state at 1 bar, 22C and exits at 4 bar, 177 C (1
bar = 100 kPa). The rate of work input is 3.5 kW and the mass flow rate is 0.02 kg/s. Assume
air is an ideal gas (R=0.287 kJ/kgK), kinetic and potential energy effects are negligible. What
is the rate of entropy generation?
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© A. Rowe
2/2
9
closed
qui
Sys-gen
= o
Aft
SQ
=
wis
Sw-dr
=
=> u -u,
q
=
SQ-Su
:
M
-
Reverse work:
V Mu
Wor
:
pr-
pu
Sw =
>
=> Wi
Spor
=
Second law :
D
=> =Th Sq
=
=> de
=
Tts
plu
-
a) isochoric (constant volume
w
S Pfr , Gu
=o
=
=>
W
o
=
a
4 4
=
-
6) isobaric (constant pressure
-p = o
wa
Pau P(V-vi)
=
=
uz
u
-
uz + Pr
q
=>
92
=
,
-
P(V VI)
-
-
(u , + PV)
=
9, 2
-hi
=
c) I sentropic Cadiabatics reversable)
59
: 0
=
u
W
=
Tas
>
=
Asto
-9
,
=
,
S
=
constant
0
6) Isothermal (Constant T St 0)
=
,
T
Sts
4 U
,
Win
=
Wiz
f
=
=
12
=
T(x 5)
-
-
T(S2- S )
,
<U
=
u
w
q
q w
=
-
=
(sq
=
-
,
Tsi)
Ts
-
=
fi
-
-
f
u + 42
,
-
[Uz-Tse)
q2
:
5
-
a)
T
-
x3
30a
P
3
v
20
&
2
Y
·
Y
·
M
&
*
I
↓
2
>S
>
V
is
12 :
·
t
d
0
first Law 1 - U
s(t P) S(T)- Rho
:
=
,
=
W
+ 3
2
·
Ve
:
=
-W
u
=
-
,
4z
V
=
,
W2z
0
=
& =
=
,
- 3 + 3 :
P Ps
Waz
=
=
Ps (Vg-Ve)
=
9
*
,
-U
=
+ W
,
3- 4 :
S
-Ye
Sy ,y
=
1 :
V
Y
=
Wzy
0,
=
W
,
=
U
=
0
V
-
=-
,
Colour cobe :
Prepa
State
TCR0
1100
300k
0
.
861
23970 17830k 0 06
↓
35931 431740 0 06
.
.
u(ang)h(k5/g)S(k5/gh)
S
214 9300 997 130
7 150
.
.
.
·
.
617 59855 798 1937 136
.
.
.
.
·
given
from table
·
calculate)
·
Same Value
=
33931 43
2200K 0 1065
Y
1144
381 37
.
Sn
S1
(T p)
So(T)
=
,
=
7 130
Si
=
7 136
Sy
=
8 17
-
,
.
331
.
.
=
.
6577 4 %
.
2504 I
9 354
. Id
8
889 78
1213
6 53
8
.
.
861
.
.
08
.
.
.
17
:
1) V
=
RT1
S
,
.
0 18
=
.
.
86/m/kg
.
v
,
=
2=
4) P
=
0
=
.
=
14 . 3
830
287x85039701
pressurratio
5)
.
0
=
T
State1t 2 :
state 2-3 :
7 130
7 48
=
8
1872 7
R(n)
R(u(1)
-
.
0
1315 8
.
.
.
=
969 90
.
.
↓=
stat I
1 494
=
.
=
=
-o
=
Pyl
State 3-3 :
2) Whet
Wi + W + W
=
Wiz
2 14
=
Wes
=
Wylz
=
9
660
.
.
.
V (Ty)
=
0
.
95)
=> (Vr(ts)
o
593/(0 1067
.
-
.
1872 7
=
Wy,
275
.
0
.
06)
Ty
=
=
o
Un
2
-
884 782
.
.
State 3-4 :
617 59 =02 69
-
=
20040287=0
45
(7 677
.
Vru (Tu) = (0 95)
.
=
-
8 006
.
=
Wzy
.
=
=
987 92
.
=
7 590
.
-
884 78
.
Py =
=
7 590
(20
+ 1140
=
7 69 - interparta
.
-
=>
Ty
,
Sy
=
11444
.
,
hu
=
381 33
.
1213 83
.
kpa
=
6
.
357
=
5931.
&in
=
&y 9
+
&
969 9
=
.
zz1
9
.
2)
3)
1
-
1330
=
617 59
.
=
6145/Rg
352 31
7504 1 -1325 6
=
.
.
=
.
.
1178 3
.
13
Whet
860 4
.
:
=
Ein
0
=
.
56
36 %
=
1530 61
.
=
w
=
find
Imin
in
Whet
I cycle
.
zrev
60s
1600rpm
200
I
.
=
=
.
13 33 cys
.
16505kyde
13 33
.
Work
m
=
860 43
.
860
>
-
=
5
V
=
m(v,
=
0
.
-
=
.
4305/kg (
.
000kg
V)
0019 (0 86)
.
-
0
.
06)
=
0
.
00153m2
=
1 54 liters
.
93
En mue
=
e
Fro N(t) Eh
=
.
N(H
=
X
mu
me
-
w
enter
e
= -Nort
+
Nirth-Ndie
=
94 open sys Devices :
.
a
·
a
wi
lit
1
w
-
Q
&
&
>
·
&
-
E
M
-
-
p
1
M
&
&
A
Compressor of Turbine
firstlaw
: Em(h (yz) Em(h +y/z) 0
+
+
(Wor-hin)
in
=>
h-h
=>
,
=
+
-
Q
=
-
+
-
+
=
-
w
winch ,
adiabtic
w = m (n
2)
in
9-w
-
if me
have
one
mass
flow rate
m(S-S.) -
=
(2)
fre
w
Sen
-
ss
pumps
second Law :
,
O
=
:
mV(Pz
-
P)
,
·gen m(Sn-si)
=
if
inten
mith = Q-Swi
Min sout
,
Sh
mits
th-Tts
=
-Sw-TSSgen
-(th- Tbs)
=
Sq-sw
Pipe flow
1)
2)
Sosi
Sw
=
-
TSSgen
= See
Q
TGs
=>
Equ
Ch-ufp
=
Sw
.
=
-
utp-TS Sgen
S
>
-
=>
W
=
-Sulp-[TSsgen
=
,
:
Sc
=
S
,
heating cooling
,
m(hz hi)
eno
-
work
gen m
=
Sen m(s
=>
=
-S
.
(h)
-
Throttling Device
⑨
1)
2)
Gibbs
rev
oo
=y wi(h , m)
Ogen
+
=
= h = hz
m)S2-5) Syeninis -(i)
,
ov
95
i device
Ign
To
in (hn-hi)
1st Law :
aut law : mi(S-S)
=
RT
,
h(T)
=
iSen
=
m(s
-
,
S)
=
Q-W
- Sen
To
+(4) +
+
P
1)
=
lookpa
Pz
,
Enrgybalance
0
ha hi
<
-
Q
=
-
=
=
3
.
=
To
298K
=
2984
=
Yookpa
:
Q - Wont + mi(k, hz)
-
p(Tz -Til
3 - 0
=
1
0091450-299)
02(- 155 8)
.
.
.
=
-
0
,
agenMoses
.
=
153
.
8K5/kg
335kh
4) -S
3)
=>
Sen-1
.
0
W/k
124 =
0
.
0265245/8
ooon a