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 gore@purdue.edu

Gatewood Wing 3166, 765 494 0061Office 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 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

6

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.

6

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

7

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

7

electricW I shaftWWQdt

dE

motor τ = 10.2 N-mω = 1000 RPM

I = 10 ampV = 110 V

+

-

8

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

11

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

13

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

12

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