ME 200 L6: Energy Rate Balance,
Transient Operation, Cyclic Repetitive Operation,
Cycle Analysis,
Efficiency & Coefficient of Performance
Spring 2014 MWF 1030-1120 AM
J. P. Gore
gore@purdue.edu
Gatewood Wing 3166, 765 494 0061
Office Hours: MWF 1130-1230
TAs: Robert Kapaku rkapaku@purdue.edu
Dong Han han193@purdue.edu
Closed System Energy Balance
►Energy is an extensive property that
includes the internal energy, the kinetic energy
and the gravitational potential energy.
►For closed systems, energy is transferred in
and out across the system boundary by two
means only: by work and by heat.
►Energy is conserved. This is the first law of
thermodynamics.
Closed System Transient Energy Balance
►The time rate form of the closed system energy
balance is
dE
QW
dt
(Eq. 2.37)
►The rate form expressed in words is
rate of change
of energy
in the system
at time t
net rate of transfer
in by heat
at time t
net rate of transfer
out by work
at time t
► Just as in calculus, separate variables and integrate
2
2
2
1
1
1
 dE   Qdt   Wdt
Change in Energy of a System
2
2
2
2
2
2
 dE   dU   d(KE)   d(PE)   Qdt   Wdt 
1
1
1
1
1
1
Q 2  1W2
1
►The changes in energy of a system from state 1 to state
2 consist of internal, kinetic and potential energy changes.
2
 dE  E
2
 E1  U 2  U1  KE 2  KE1  PE 2  PE1  1 Q2  1W2
(Eq. 2.27a)
1
DE = DU + DKE + DPE
(Eq. 2.27b)
►Energy at state 1 or state 2 or any other state is defined in
reference to a standard state.
►Definition of energy at all states must have identical
standard base state.
►Changes in the energy of a system between states, defined
with identical standard state have significance.
Home Work Problem
Imagine a party at a college location as sketched below. Bob goes to the refrigerator
door to get a soda…
Music
speakers
A/C
Vent
Well
insulated
party
room
Electrical
supply
cable
Refrigerator
(fridge) door open
Door
locked
Example 1
An electric generator coupled to a windmill produces an average power
of 15 kW. The power is used to charge a storage battery. Heat transfer
from the battery to the surroundings occurs at a constant rate of 1.8 kW.
For 8 h of operation, determine the total amount of energy stored in the
battery, in kJ.
Given
Find: ΔE in kJ?
W = -15 kW
System
Q = -1.8 kW
Δt = 8 h
W = ?15 kW
storage
battery
Basic Equation
dE
QW
dt
Integrating :
DE  Q  W
Assumptions
The battery is a closed system.
The work and heat transfer rates
are constant.
1kJ s 3600s
W  WDt   15kW  8h 
 4.32 105 kJ
1kW 1h
Q = ?1.8 kW
Δt = 8 h
Q  QDt   1.8kW  8h 
1kJ s 3600s
 51,800kJ
1kW 1h
DE  51,800   4.32 105   3.8 105 kJ
6
Example 2
An electric motor draws a current of
10 amp with a voltage of 110 V.
The output shaft develops a
torque of 10.2 N-m and a
rotational speed of 1000 RPM.
For operation at steady state,
determine for the motor, each in
kW.
the electric power required.
the power developed by the output
shaft.
the rate of heat transfer.
Sketch
I = 10
amp
V = 110 V
motor
-
τ = 10.2 N-m
ω = 1000 RPM
Given
I = 10 amp
V = 110 V
τ = 10.2 N-m
ω = 1000 RPM
• Assumptions
– The motor is a closed system.
– The system is at steady state.
• Find
– Welectric in kW?
– Wshaft in kW?
– Q in kW?
+
• Basic Equations
dE  
 Q W
dt
Welectric  I
W shaft  
7
Example 2
• Given
–
–
–
–
• Basic Equations
I = 10 amp
V = 110 V
τ = 10.2 N-m
ω = 1000 RPM
dE  
 Q W
dt
Welectric  VI
W shaft  
• Solution
1Watt am p 1kW
Welectric  110V 10am p
1volt
103W

W
 1.1kW
• Find
electric
– Welectric in kW?
– Wshaft in kW?
– Q in kW?
rev  2 rad 1 min
1kW

W shaft  10.2 N  m1000

min  rev
60s 103 N  m s

W shaft  1.07kW
0
• Sketch
I = 10 amp
V = 110 V
+
motor
-
τ = 10.2 N-m
ω = 1000 RPM
dE  
Q  W
 Q W
dt
Q  Welectric  W shaft
Q  1.1kW  1.07kW
Q  0.03kW
8
Example 3
A gas within a piston-cylinder assembly (undergoes a
thermodynamic cycle consisting of) three processes:
– Process 1-2: Constant volume, V = 0.028 m3, U2 – U1 = 26.4 kJ.
– Process 2-3: Expansion with pV = constant, U3 = U2.
– Process 3-1: Constant pressure, p = 1.4 bar, W31 = -10.5 kJ.
There are no significant changes in kinetic or potential
energy.
1.
2.
3.
4.
Sketch the cycle on a p-V diagram.
Calculate the net work for the cycle, in kJ.
Calculate the heat transfer for process 2-3, in kJ.
Calculate the heat transfer for process 3-1, in kJ.
9
Example 3
• Assumptions
• Find
–
–
–
–
p-V diagram
Wnet = ? in kJ
Q23 = ? in kJ
Q31 = ? in kJ
• System
• Given
– The gas is the closed system.
– For the system, ΔKE = ΔPE
= 0.
– Volume change is the only
work mode.
• Basic Equations
gas
– 1-2: V = 0.028 m3, U2 – U1 =
26.4 kJ
– 2-3: pV = constant, U3 = U2
– 3-1: p = 1.4 bar, W31 = -10.5 kJ
DE  Q  W
DE  DKE  DPE  DU
K
WK   pdV
J
J
10
Example 3
•
Solution
1-2: Constant Volume Heat Addition
P, atm
2
2-3: Isothermal Expansion, Heat added to
maintain T in spite of Expansion.
3-1: Constant Pressure Heat Rejection and
“exhaust,” leading to volume
reduction work is put into the system
3
1
V, m3
0
W12  VV2 pdV
Wcycle  W12  W23  W31
W23  VV3 pdV
2
p

c
V
W31  VV1 pdV  p V1  V3
3
W12  0
1
W23  
V3
V2

V3 dV
V
c
V
dV  c 
 c ln V V3  p3V3 ln 3
2
V2 V
V
V2
V3  V1 
W31
p
11
Example 3
 10.5kJ
1bar
103 N  m
V3  0.028m 
1.4bar 105 N m2
1kJ
3
V3  0.103m3
V3
105 N m2
0.103m3 1kJ
3
W23  p3V3 ln
 1.4bar
 0.103m  ln 0.028m3 103 N  m  18.78kJ
V2
1bar
Wcycle  W12  W23  W31
0
Wcycle  8.28kJ
0
0
DKE  DPE  DU  Q23  W23
0
Wcycle  0  18.78   10.5   kJ
0
DKE  DPE  DU  Q31  W31
0
Q23  W23
Q23  18.78kJ
Q31  U1  U3   W31
U2  U1  U3  U2   U1  U3   0 U1  U3   U2  U1
U1  U3  26.4kJ
Q31  26.4kJ   10.5kJ   36.9kJ
12
Cycle Analysis, Efficiency and
Coefficient of Performance
►When a working substance returns to the original state in a
cyclic manner while accepting and rejecting heat from two
reservoirs and delivering net work in the process, we have an
engine cycle.
►When a working substance returns to the original state in a
cyclic manner while accepting heat from a low temperature
reservoir and delivering heat to a high temperature reservoir
we have a refrigerator or a heat pump cycle.
►If the cold reservoir substance is the useful substance then
it is a refrigerator if the hot reservoir contains the useful
substance then we have a heat pump.
Return to Example 3
1-2: Constant Volume Heat addition
P, atm
2-3: Isothermal Expansion, Heat added to
maintain T in spite of Expansion.
3-1: Constant Pressure Heat Rejection and
“exhaust,” leading to volume
reduction work is put into the system
2
Wcycle  W12  W23  W31
1
3
Qcycle  Q12  Q23  Q31
V, m3 First Law of Thermodynamics or
Conservation of Energy is satisfied.
Wcycle  0  18.78   10.5    8.28 kJ
Qcycle   26.4  18.78  (36.9)  8.28kJ For this cycle 1-2 and 2-3 are the heat
DE  QCycle  WCycle  0

Wcycle
Qin
addition processes and the customer
Pays for the fuel that leads to this heat.
Qin  Qout
Qout
8.28(100)

 1

 18.33%
Qin
Qin (26.4  18.78)
14
Flip the Engine to make it a Heating/Cooling Device
3-1: Constant Volume Heat Rejection
2-3: Isothermal Compression, Heat removed to
maintain T in spite of Compression.
1-2: Constant Pressure Heat Extraction from
cold space leading to expansion of
working substance.
3
P, atm
Wcycle  W12  W23  W31
1
Wcycle  0  18.78   10.5    8.28 kJ
Qcycle   26.4 18.78  36.9  8.28kJ
DE  QCycle  WCycle  0
Qcycle  Q12  Q23  Q31
2
COPHeating
V, m3 First Law of Thermodynamics or
Conservation of Energy is satisfied.
Q LT  Wcycle (26.4  18.78)
Q HT



 5.46
Wcycle
Wcycle
8.28
COPCooling 
Q HT  Wcycle 36.9
Q LT


 4.46
Wcycle
Wcycle
8.28
15