EGR 334 ThermodynamicsChapter 4: Section 4-5

Lecture 13: Control Volumes and Energy Balance Quiz Today?

TEST #1 NEXT MONDAY

Today’s main concepts:• Be able to draw a graphic representation of a control

volume with energy balance.• Explain what flow work.• Write the energy balance equation• Identify some commons assumptions that simplify the

energy balance equation.• Use mass balance and energy balance to solve

thermodynamic problems.

Reading Assignment:

Homework Assignment:

• Read Chap 4: Sections 6-9 (for next Wed.)

From Chap 4: 25, 28, 31, 34

TEST #1 NEXT

MONDAY

3

em

CVdm

dtim

Mass Rate Balance:

Energy Rate Balance:

e em e

CVdE

dti im e

Q W

For a Control Volume:

The rate at which mass accumulates in the control volume is the difference between the rate of mass flow in and flow out.

The rate at which energy accumulates in the control volume is the difference between the heat flow rate in and the power out of the system and the difference between the rate at which mass carries energy with it either into and out of the control volume.

4Sec 4.4: Conservation of Energy for a Control Volume

E withinthe system

net Qinput

net W output[ ]=[ ] +[ ]

Energy Balance

m withinthe system

net minput

net m output[ ]=[ ] +[ ]

Mass Balance

Net E from Mass transfer

+[ ]Add a term to the energy balance.

net netnetm e U PE KE

CVi ei e

dEQ W m e m e

dt

2 2net V V

( )2 2in exit

in exit in in in exit exit exit

dEm e e m u gz m u gz

dt

where then energy due to mass transfer is

mass transfer energy expressed as a rate equation

then the full 1st law of Thermodynamics is given as

5Sec 4.4: Conservation of Energy for a Control Volume

Two Types of WorkFlow Work: work done BY the flowControl Volume Work: work done BY the control volume (WCV)

- work done by PV, turning a shaft, electricity (types of work we have dealt with thus far)

CV e e e i i iW W m p v m p v

[ ]Time rate of Energytransfer from the control volume at exit due to mass transfer

e e e eW m p v

(-) Work Done ON system as mass enters

(+) Work Done On environment as mass exits

Define Flow Work

Work done by system due to rotating shafts, electrical effects, or boundary displacement (∫pdV)

6Sec 4.4: Conservation of Energy for a Control Volume

Putting all these terms together including the separate work terms, W = WCV + WPAv

2 2CV V V

2 2i e

CV CV i i i i i e e e e e

dEQ W m u p v gz m u p v gz

dt

2 2CV V V

2 2i e

CV CV i i i e e e

dEQ W m h gz m h gz

dt

Thus u “heat energy”h “heat energy + flow work”

hi he

Recall that enthalpy is defined

h = u + p v

then the form that will usually be your starting equation is

7Sec 4.5: Analyzing Control Volumes at Steady State

For Steady State Conditions:

i

exiti

in mm Mass Balance at steady state:

and 0CV dt

dE

e

ee

eei

ii

iiCVCV gz

vhmgz

vhmWQ

220

22

Often W = 0 if No change in system volume (fixed container) No turbines/pumps or electrical devices

CV 0dm

dt

Energy Balance at steady state:

Often Q = 0 if Small surface area System is insulated T between system & environment is small

Common Simplifying Assumptions:

8

Example 1: (Problem 4.27)Air at 600 kPa, 330 K enters a well-insulated, horizontal pipe having a diameter of 1.2 cm and exits at 120 kPa, 300 K. Applying the ideal gas model for air, determine at steady state (a) the inlet and exit velocities, each in m/s (b) the mass flow rate, in kg/s.

600 kPa330 K

120 kPa300 K

d =1.2 cm

Sec 4.5: Analyzing Control Volumes at Steady State

state Inlet Exit

P (kPa) 600 120

T (K) 330 300

h (kJ/kg) 330.34 300.19

Identify State Propertiesstate Inlet Exit

P (kPa) 600 120

T (K) 330 300

h (kJ/kg)

Look up values for h on Table A-22

state Inlet Exit

p [kPa]

T [K]

h [kJ/kg]

9

Assumptions: System at steady state No significant heat loss No work done by system No large change in elevation

Sec 4.5: Analyzing Control Volumes at Steady State

Energy balance:

state Inlet Exit

P (kPa) 600 120

T (K) 330 300

h (kJ/kg) 330.34 300.19

Table A-22

2 2CV V V

2 2i e

CV CV i i i e e e

dEQ W m h gz m h gz

dt

2 2V V0 0 0 0 0

2 2i e

i i e em h m h

2 2V V0

2 2i e

i em h m h

is reduced with assumptions to:

Steady state also implies that mass flow rate is equals mass flow rate out.

2 2V V

2 2i e

e ih h

10

Next find mass flow rate

Sec 4.5: Analyzing Control Volumes at Steady State

Apply continuity equation and mass balance:

in exitm m

state Inlet Exit

P (kPa) 600 120

T (K) 330 300

h (kJ/kg) 330.34 300.19

600 kPa330 K

120 kPa300 K

d =1.2 cm

VV Am

v v

therefore:

pv RT

V Vi i e e

i e

A A

v v

V

Vi i

e e

v

v

applying Ideal Gas Equation:

V / 330 1200.22

V / 300 600i i i i e

e e e e i

RT p T p

RT p T p

iV 0.22V e

11

Finally combine what we know

Sec 4.5: Analyzing Control Volumes at Steady State

from Energy Balance: from Mass balance:

state Inlet Exit

P (kPa) 600 120

T (K) 330 300

h (kJ/kg) 330.34 300.19

600 kPa330 K

120 kPa300 K

d =1.2 cm

therefore:

solving for velocities:

iV 0.22V e2 2V V

2 2i e

e ih h

2 2V -V 2( )i e e ih h 2 2(0.22V ) -V 2( )e e e ih h 2-0.9516V 2(300.19 330.34) /e kJ kg

2 2 21000 /V 63.37 / 63367 /e

N m kg m skJ kg m s

kJ N

2 2V 63367 / 251.7 /e m s m s iV 0.22V 55.37 /e m s

12Sec 4.5: Analyzing Control Volumes at Steady State

to find the mass flow rate: from table 3.1: R = 0.2870 kJ/kg-K

state Inlet Exit

P (kPa) 600 120

T (K) 330 300

h (kJ/kg) 330.34 300.19

600 kPa330 K

120 kPa300 K

d =1.2 cm

2

2

(0.012 )120 251.7 /

4 1000 /0.0397 /

(0.2870 / )(300 ) 1000

mkPa m s

N m kJm kg s

kJ kg K K kPa N m

e eV V V

e

e

e e

e eRT

p

A A p Am

v RT

13

Example 2: (Problem 4.29)Refrigerant 134a flows at a steady state through horizontal pipe with an inside diameter of 4cm. It enters as -8 deg C saturated vapor with a mass flow rate of 17 kg/min. It exits at a pressure of 2 bar. If the heat transfer rate to the refrigerant is 3.4 kW, determinea) the exit temperatureb) the inlet and outlet velocities

Sec 4.5: Analyzing Control Volumes at Steady State

statec Inlet(sat. vapor)

Exit

p [bar] 2.1704 2

T [oC] -8

v [m3/kg] 0.0919

h [kJ/kg] 242.54

Identify State Properties

Look up values for h on Table A-10

-8 oC17 kg/min

2 bar

d= 4 cm

.Q = 3.4 kW

statec Inlet(sat. vapor)

Exit

p [bar]

T [oC]

v [m3/kg]

h [kJ/kg]

14Sec 4.5: Analyzing Control Volumes at Steady State

state Inlet(sat. vapor)

Exit

p [bar] 2.1704 2

T [oC] -8

v [m3/kg] 0.0919

h [kJ/kg] 242.54

Using continuity equation at the inlet

Find the volumetric flow rate

-8 oC17 kg/min

2 bar

d= 4 cm

.Q = 3.4 kW

Find the inlet velocity

VV Am

v v

3 31min(0.0919 / )(17 / min) 1.562 / min

60i i iV v m m kg kg ms

3

i 2

(0.0919 / )(17 / min) 1minV 20.7 /

(0.04 ) / 4 60i iv m m kg kg

m sA m s

15Sec 4.5: Analyzing Control Volumes at Steady State

state Inlet(sat. vapor)

Exit

p [bar] 2.1704 2

T [oC] -8

v [m3/kg] 0.0919

h [kJ/kg] 242.54

for steady state: mass balance is given as

For steady state the energy balance is given as:

-8 oC17 kg/min

2 bar

d= 4 cm

.Q = 3.4 kW

for steady state, no work done, and a horizontal pipe:

i em m ei VV

i e

AA

v v e iV Ve

i

v

v

2 2V V0

2 2i e

i i i e e eQ W m h gz m h gz

2 2V V

2 2e i

e i

Qh h

m

16Sec 4.5: Analyzing Control Volumes at Steady State

state Inlet(sat. vapor)

Exit

p [bar] 2.1704 2

T [oC] -8

v [m3/kg] 0.0919

h [kJ/kg] 242.54

combining the mass balance and energy balance equations:

or

-8 oC17 kg/min

2 bar

d= 4 cm

.Q = 3.4 kW

gives

e iV Ve

i

v

v 2 2V V

2 2e i

e i

Qh h

m

2

i 2V

V

2 2

e

i e ie i

v

vQh h

m

22V1

2i e

e ii

vQh h

m v

17Sec 4.5: Analyzing Control Volumes at Steady State

state Inlet(sat. vapor)

Exit

p [bar] 2.1704 2

T [oC] -8

v [m3/kg] 0.0919

h [kJ/kg] 242.54

This equation contains two unknowns, he and ve.

Note that both are properties of the exit state which can be both considered functions of an unknown temperature, TB and a known pressure, pB = 2 bar. One way to solve this would be to guess a temperature of the outlet….then look up the corresponding values of h and v….put them back in the equation and see if they give the correct value.

This iterative approach is exactly the type of solution that IT is very good at….so anther way to solve this is to use IT to define the h and v in terms of an unknown TB and let it iteratively find the temperature for us.

-8 oC17 kg/min

2 bar

d= 4 cm

.Q = 3.4 kW

22V1

2i e

e ii

vQh h

m v

18

Outlet temperature Te = -10.09 oC

Inlet velocity Vi = 20.7 m/s

Outlet velocity

e iV Ve

i

v

v

Results:

0.0889820.7

0.09186 20.05 /m s

c

19

TEST #1 NEXT MONDAY

end of Lecture 13 Slides