ENGR 2213 ThermodynamicsENGR 2213 Thermodynamics

F. C. Lai

School of Aerospace and Mechanical

Engineering

University of Oklahoma

Energy Analysis for a Control VolumeEnergy Analysis for a Control Volume

CV i em m m CV

i edm

m mdt

AVm AV

v

Conservation of Mass

Net Change in Mass within CV

Total Mass Entering CV

Total Mass Leaving CV

= -

Steady State i em m

Example 1Example 1

1m 40 kg/ s

33(AV) 0.06 m / s

Feedwater Heater:Inlet 1 T1 = 200 ºC, p1 = 700 kPa, Inlet 2 T2 = 40 ºC, p2 = 700 kPa, A2 = 25 cm2

Exit sat. liquid, p3 = 700 kPa,Find 2 3 2m ?, m ? and V ?

Inlet 1 Inlet 2

Exit

Example 1 (continued)Example 1 (continued)

i em m 1 2 3m m m

AVm AV

v

Steady State

Inlet 2: compressed liquid Table A-4, v2 = 0.001008 m3/kg

Exit: saturated liquid Table A-5, v3 = 0.001108 m3/kg

33

3

(AV) 0.06m 54.15 kg/ s

v 0.001108

Example 1 (continued)Example 1 (continued)

2 3 1m m m

2 22

2

m v (14.15)(0.001008)V 5.7 m/ s

A 0.0025

= 54.15 – 40 = 14.15 kg/s

Energy Analysis for a Control VolumeEnergy Analysis for a Control Volume

Flow workEnergy that is necessary for maintaining a continuousflow through a control volume.

A cross-sectional areap fluid pressureL width of fluid element

F = pA

W = FL = pAL = pV

Energy Analysis for a Control VolumeEnergy Analysis for a Control Volume

2Ve u pv gz

2

2Ve u gz

2

Energy carried by a fluid element in a closed system

Energy carried by a fluid element in a control volume2V

h gz2

Energy Analysis for a Control VolumeEnergy Analysis for a Control Volume

22CV ei

i ei i e e CV CVdE VV

m h gz m h gz Q Wdt 2 2

Conservation of Energy

Net Change in Energy of CV

Total Energy Carried by MassEntering CV

Total Energy Carried by MassLeaving CV

= -

Total Energy Crossing Boundary as Heat and Work

+

Steady-Flow ProcessSteady-Flow Process

A process during which a fluid flows through a control volume steadily.

● No properties within the control volume change with time.

● No properties change at the boundaries of the control volume with time.● The heat and work interactions between a steady- flow system and its surroundings do not change with time.

Steady-Flow ProcessSteady-Flow Process

Conservation of mass

i em m

Conservation of energy 2 2e i

e iCV CV e e i iV V

Q W m h gz m h gz2 2

Steady-Flow ProcessSteady-Flow Process

i em m m

2 2e i

CV CV e i e i(V V )

Q W m (h h ) g(z z )2

Conservation of mass

Conservation of energy

For single-stream steady-flow process

2 2CV CV e i

e i e iQ W (V V )

(h h ) g(z z )m m 2

Steady-Flow DevicesSteady-Flow Devices

CVW 0, PE 0. 2 2

CV e ie i

Q (V V )(h h )

m 2

● Nozzles and DiffusersA nozzle is used to accelerate the velocity of a fluid in the direction of flow while a diffuser is used to decelerate the flow.

The cross-sectional area of a nozzle decreases in the direction of flow while it increases for a diffuser.

For nozzles and diffusers,

i em m m,

Example 2Example 2

iV 10 m/ s.

eV 665 m/ s.

iV 10 m/ s.

Steam enters an insulated nozzle at a flow rate of 2 kg/s with Ti = 400 ºC, pi = 4 MPa, and

Find the cross-sectional area at the exit.

Inlet Ti = 400 ºCpi = 4 MPa

Exit pe = 1.5 MPa

It exits at pe = 1.5 MPa with a velocity of

eV 665 m/ s.

Example 2 (continued)Example 2 (continued)

2 2i e

e i(V V )

h h2

2 2CV e i

e iQ (V V )

(h h )m 2

ee

e

mvA

V

Inlet, superheated vaporTable A-6, hi = 3213.6 kJ/kg

2 2

3

(10) (665) 13213.6

2 10

= 2992.5 kJ/kg

Example 2 (continued)Example 2 (continued)

ee

e

mvA

V

Table A-6, he = 2992.5 kJ/kg

1.4 MPa 1.5 MPa 1.6 MPa250 2927.2 2923.2 2919.2300 3040.4 3037.6 3034.8

T = 280 ºCv = 0.1627 m3/kg

(2)(0.1627)

665 = 0.000489 m2

Steady-Flow DevicesSteady-Flow Devices

PE 0.

2 2CV CV e i

e iQ W (V V )

(h h )m m 2

● TurbinesA turbine is a device from which work is produced asa result of the expansion of a gas or superheatedsteam through a set of blades attached to a shaft freeto rotate.

For turbines,

i em m m,

Example 3Example 3

iV 10 m/ s.

eV 50 m/ s.

iV 10 m/ s.

Steam enters a turbine at a flow rate of 4600 kg/h.At the inlet, Ti = 400 ºC, pi = 6 MPa, and

If the turbine produces a power of 1 MW, find the heatloss from the turbine.

Inlet Ti = 400 ºCpi = 6 MPa Exit

xe = 0.9pe = 10 kPa

At the exit, xe = 0.9, pe = 10 kPa and

eV 50 m/ s.

Example 3 (continued)Example 3 (continued)2 2e i

CV CV e i(V V )

Q W m (h h )2

Inlet: superheated vapor at 6 MPa and 400 ºC Table A-6, hi = 3177.2 kJ/kg

Exit: x = 0.9, saturated mixture at 10 kPa Table A-5, hf = 191.83 kJ/kg, hfg = 2392.8 kJ/kg

he = hf + xehfg = 191.83 + 0.9 (2392.8)= 2345.4 kJ/kg

Example 3 (continued)Example 3 (continued)

2 2 2 2e i

3

V V (50) (10) 11.2 kJ/kg

2 2 10

2 2e i

CV CV e i(V V )

Q W m (h h )2

he - hi = 2345.4 – 3177.2 = - 831.8 kJ/kg

CV4600

Q 1000 ( 831.8 1.2)3600

= - 63.1 kW

Steady-Flow DevicesSteady-Flow Devices

PE 0.

2 2CV CV e i

e i e iQ W (V V )

(h h ) g(z z )m m 2

i em m m,

● Compressors and Pumps

Compressors and pumps are devices to which work is provided to raise the pressure of a fluid.

For compressors,

Compressors → gasesPumps → liquids

For pumps, CVQ 0, PE 0.

Example 4Example 4

iV 6 m/ s.

eV 2 m/ s.

iV 6 m/ s.

Air enters a compressor.At the inlet, Ti = 290 K, pi = 100 kPa, and

If given that Ai = 0.1 m2 and heat loss at a rate of 3 kW, find the work required for the compressor.

Inlet Ti = 290 Kpi = 100 kPa Exit

Te = 450 Kpe = 700 kPa

At the exit, Te = 450 K, pe = 700 kPa and

eV 2 m/ s.

CVQ 3kW

Example 4 (continued)Example 4 (continued)

i i

i

i

A V

RTp

2 2e i

CV CV e i(V V )

Q W m (h h )2

i i

i

A Vm

v

Table A-17, at 290 K, hi = 290.16 kJ/kg, at 450 K, he = 451.8 kJ/kg.

(0.1)(6)(100)

(0.287)(290) = 0.72 kg/s

Example 4 (continued)Example 4 (continued)

2 2i e

CV CV i e(V V )

W Q m (h h )2

2 2

CV 3

6 2 1W 3 (0.72) (290.16 451.8)

2 10

= - 119.4 kW

Example 5Example 5

iV 10 m/ s.

eV 40 m/ s.

iV 10 m/ s.

A pump steadily draws water at a flow rate of 10 kg/s.At the inlet, Ti = 25 ºC, pi = 100 kPa, and

If the exit is located 50 m above the inlet, find the workrequired for the pump.

Inlet Ti = 25 ºCpi = 100 kPa

Exit Te = 25 ºCpe = 200 kPa

At the exit, Te = 25 ºC, pe = 200 kPa and

eV 40 m/ s.

Example 5 (continued)Example 5 (continued)

Table A-4, at 25 ºC

vf = 0.001003 m3/kg

2 2CV CV e i

e i e iQ W (V V )

(h h ) g(z z )m m 2

he – hi ~ [hf + vf (p – psat)]e - [hf + vf (p – psat)]i

= vf (pe – pi)

= 0.001003 (200 – 100) = 0.1 kJ/kg2 2 2 2e i

3

V V (40) (10) 10.75 kJ/kg

2 2 10

g(ze – zi) = 9.8(50)/103 = 0.49 kJ/kg

Example 5 (continued)Example 5 (continued)2 2e i

CV e i e i(V V )

W m (h h ) g(z z )2

= 20 (0.1 + 0.75 + 0.49)

= 13.4 kW