ME 200 L16:ME 200 L16: Transient & Steady State ProcessesRead 4.12

https://engineering.purdue.edu/ME200/ThermoMentor© Program Launched

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

2

In this Lecture …

• Example with significant simplification of the energy equation (note the role of enthalpy difference Δh, kJ/kg)

• Transient and Steady State Processes in Control Volumes

• Example of Transient Mass or/and Energy Balance.

3

Example

Steam with a specific enthalpy of 3000 kJ/kg and a mass flow rate of 0.5 kg/s enters a horizontal pipe. At the exit, the specific enthalpy is 1700 kJ/kg. If there is no significant change in kinetic energy, determine the rate of heat transfer between the pipe and its surroundings, in kW. Assume steady state.

• Find– Q = ? in kW

• Sketch

• Assumptions– The control volume is at steady

state.– ΔWcv = Δke = Δpe = 0

• Basic Equations

1 2

h2 = 1700 kJ/kgh1 = 3000 kJ/kgm1 = 0.5 kg/ssteam

e

ee

eei

ii

iicvcv

cv gzV

hmgzV

hmWQdt

dE

22

22

3

e

eii

cv mmdt

dm ;

4

Example

Steam with a specific enthalpy of 3000 kJ/kg and a mass flow rate of 0.5 kg/s enters a horizontal pipe. At the exit, the specific enthalpy is 1700 kJ/kg. If there is no significant change in kinetic energy, determine the rate of heat transfer between the pipe and its surroundings, in kW. Assume steady state.

Find– Q = ? in kW

System (mass flowing through pipe)

1 2

h2 = 1700 kJ/kgh1 = 3000 kJ/kgm1 = 0.5 kg/ssteam

4

cvQ

cvQ

5

Example

Assumptions

• Wcv = 0

• ΔKE = 0

• ΔPE = 0

• Steady State, Steady Flow Operation

Basic Equations

•Solution

e

ee

eei

ii

iicvcv

cv gzV

hmgzV

hmWQdt

dE

22

22

cv 2 1Q m h h

skJ

kW

kg

kJskgQcv 1

1300017005.0

kWQcv 6505

e

eii

cv mmdt

dm mmm 21

Conservation of Mass and Conservation of Energy (CV)

2 2( 2 ) ( 2 )CVe eCV CV i i

i e

dEQ W m u pv V gz m u pv V gz

dt

Exit Flow workInlet Flow work

( )CV CV CV CVi e CV

i e

dm d V dV dm m V

dt dt dt dt

1/v

/CV CV CVm V v V

CV CV CVCV

dE dm dee m

dt dt dt

Conservation of Mass (CM)

Inlet Flow work=0

=0=0

Control Mass

Mass

Note that Volume, SpecificVolume and Density of aControl Mass can change,While the “Mass,” remainsconstant. Example: Piston-Cylinder Device

( )0

0; ln( ) ln( ) ln( ) ln( .)

.

CM CM CM CM CMCM CM

CM CMCM CM CM CM

CM CM

CM CM CM

dm d V dV dV

dt dt dt dtdV d

V V ConstV

V m Const

Examples: Transient Mass Conservation

Problem: An initially empty tank is filled with a liquid (density= 1000 kg/m3, specific heat: 4.2 kJ/kg-K, T=10oC) at a rate of 2 kg/minute for 20 minutes and then with a rate that decreases linearly with time to 0 kg/minute over 10 minutes

Find :(1) The final mass in the tank,

(2) Time taken by a 2 kW heater to warm the liquid in the tank from

10oC to 20oC,

Given:

CVi e

i e

dmm m

dt

t = 0

40 Kg0 kg 50 kg

t = 20 mins. t = 30 mins.

Equations

2

2

( 2 )

( 2 )

CVCV CV i i

i

e ee

dEQ W m u pv V gz

dt

m u pv V gz

20 30

0 20

2 2

2 1

0

2 1

22 (2 ( 20))

10

2(20) 2(30 20) 0.2 / 2(30 20 )

4(30 20) 40 20 0.1(500) 4(30 20)

40 20 50 40 50

( 0) ( )

2( ) ( ) (50)(4.2)(20 10)

1050sec 17.5min

CV

t

CV CV CV

CV

m dt t dt

kg

U U W dt

W t mc T T

kW t s

t

•Filling time = 30 minutes•Heating time = 17.5 minutes•HW: Heating begins during filling