ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of...

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