me 200 l6: me 200 l6: energy rate balance, transient operation, cyclic repetitive operation, cycle...
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ME 200 L6: 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 [email protected]
Gatewood Wing 3166, 765 494 0061Office Hours: MWF 1130-1230
TAs: Robert Kapaku [email protected] Dong Han [email protected]
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 rate form expressed in words is
rate of change of energy
in the system at time t
net rate of transferin by heatat time t
net rate of transfer out by work
at time t
►The time rate form of the closed system energy balance is
(Eq. 2.37)dE
Q Wdt
2 2 2
1 1 1
dE Qdt Wdt
► Just as in calculus, separate variables and integrate
Change in Energy of a System
►The changes in energy of a system from state 1 to state 2 consist of internal, kinetic and potential energy changes.
E = U + KE + PE (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.
(Eq. 2.27a)
2 2 2 2 2 2
1 2 1 2
1 1 1 1 1 1
dE dU d(KE) d(PE) Qdt Wdt Q W
2
2 1 2 1 2 1 2 1 1 2 1 2
1
dE E E U U KE KE PE PE Q W
Home Work ProblemImagine a party at a college location as sketched below. Bob goes to the refrigerator door to get a soda…
TV
Refrigerator (fridge) door open
Door locked
Music speakers Well
insulated party room
A/C Vent
Electrical supply cable
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Example 1An 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.Find: ΔE in kJ?System
GivenW = -15 kWQ = -1.8 kWΔt = 8 h
AssumptionsThe battery is a closed system.The work and heat transfer rates are constant.
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storage battery
Q = ?1.8 kWΔt = 8 h
W = ?15 kW
dEQ W
dtIntegrating :
E Q W
Basic Equation 51kJ s 3600sW W t 15kW 8h 4.32 10 kJ
1kW 1h
1kJ s 3600sQ Q t 1.8kW 8h 51,800kJ
1kW 1h
5 54.32 10 3.851 0 10,8 0 E kJ
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Example 2Example 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.
• Find– Welectric in kW?– Wshaft in kW?– Q in kW?
Sketch
GivenI = 10 ampV = 110 Vτ = 10.2 N-mω = 1000 RPM
• Assumptions– The motor is a closed system.– The system is at steady state.
• Basic Equations
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electricW I shaftWWQdt
dE
motor τ = 10.2 N-mω = 1000 RPM
I = 10 ampV = 110 V
+
-
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Example 2Example 2
• Given– I = 10 amp– V = 110 V– τ = 10.2 N-m– ω = 1000 RPM
• Find– Welectric in kW?– Wshaft in kW?– Q in kW?
• Sketch
• Basic Equations
• Solution
8
VIWelectric shaftWWQdt
dE
8
motor τ = 10.2 N-mω = 1000 RPM
I = 10 ampV = 110 V
+
-
W
kW
volt
ampWattampVWelectric 310
1
1
110110
kWWelectric 1.1
smN
kW
srev
radrevmNWshaft
310
1
60
min12
min10002.10
kWWshaft 07.1
0
WQ
shaftelectric WWQ
kWkWQ 07.11.1
kWQ 03.0
WQdt
dE
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Example 3Example 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. Sketch the cycle on a p-V diagram.2. Calculate the net work for the cycle, in kJ.3. Calculate the heat transfer for process 2-3, in kJ.4. Calculate the heat transfer for process 3-1, in kJ.
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ExampleExample 33
• Find– p-V diagram– Wnet = ? in kJ– Q23 = ? in kJ– Q31 = ? in kJ
• System
• Given– 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
• Assumptions– The gas is the closed system.– For the system, ΔKE = ΔPE
= 0.– Volume change is the only
work mode.
• Basic Equations
UPEKEE
WQE gas
K
J K JW pdV
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Example 3Example 3
• Solution
312312 WWWWcycle 012 W 2
112
VV pdVW
0
3
223
VV pdVW V
cp 3 3 3
22 2
V V V 323 3 3VV V
2
Vc dVW dV c c ln V p V ln
V V V
3131 1
3VVppdVW V
V p
WVV 31
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3
21-2: Constant Volume Heat Addition
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
1
P, atm
V, m3
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Example 3Example 3
312312 WWWWcycle cycleW 0 18.78 10.5 kJ kJWcycle 28.8
2323 WQUPEKE 0 0
0
2323 WQ kJQ 78.1823
3131 WQUPEKE 0 0 313131 WUUQ
0312312 UUUUUU0
1231 UUUU kJUU 4.2631
31Q 26.4kJ 10.5kJ 36.9kJ
5 2 3
3323 3 3 3 3
2
V 10 N m 0.103m 1kJW p V ln 1.4bar 0.103m ln 18.78kJ
V 1bar 0.028m 10 N m
kJ
mN
mN
bar
bar
kJmV
1
10
10
1
4.1
5.10028.0
3
253
3
33 103.0 mV
Cycle Analysis, Efficiency and Cycle Analysis, Efficiency and Coefficient of PerformanceCoefficient 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.
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Return to Example 3Return to Example 3
cycle 12 23 31
cycle 12 23 31
W W W W
Q Q Q Q
3
2
1-2: Constant Volume Heat addition
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
1
P, atm
V, m3
cycleW 0 18.78 10.5 8.28kJ cycleQ 26.4 18.78 ( 36.9) 8.28kJ
Cycle CycleE Q W 0
First Law of Thermodynamics or Conservation of Energy is satisfied.
For this cycle 1-2 and 2-3 are the heat addition processes and the customer Pays for the fuel that leads to this heat.
cycle in out out
in in in
W Q Q Q 8.28(100)1 18.33%
Q Q Q (26.4 18.78)
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Flip the Engine to make it a Heating/Cooling DeviceFlip the Engine to make it a Heating/Cooling Device
cycle 12 23 31
cycle 12 23 31
W W W W
Q Q Q Q
3
2
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.
1
P, atm
V, m3
cycleW 0 18.78 10.5 8.28kJ cycleQ 26.4 18.78 36.9 8.28kJ
Cycle CycleE Q W 0
First Law of Thermodynamics or Conservation of Energy is satisfied.
LT cycleHTHeating
cycle cycle
HT cycleLTCooling
cycle cycle
Q WQ (26.4 18.78)COP 5.46
W W 8.28
Q WQ 36.9COP 4.46
W W 8.28