Thermodynamics

Chapter 3

Diploma in EngineeringMechanical Engineering Division, Ngee Ann Polytechnic

Chapter 3Steady Flow Processes with

Steam

• Introduction• Steam boiler• Steam turbine• Steam condenser• Mixing chamber

Introduction• This chapter deals with steady flow processes in

open systems.

• Limited to steam and water in the following devices: Steam boiler Steam turbine Condenser Mixing chamber

2 21 2

1 21 1 2 2( ) ( )2 2

in outin outc cQ W m h gZ Q W m h gZ

Steam boiler• Steam boiler is a device used in a steam power

plant. Its function is to generate steam at constant pressure.

Boiler

steam

feed water

furnaceinQ

lossQ

system boundary

1

2

Heat energy is supplied to convert the water into steam

Steam boiler• kg/s of feed water enter

the boiler at point 1 with specific enthalpy, h1, the velocity, c1 and vertical distance above the datum, Z1

• kg/s of steam leave the boiler at point 2 with specific enthalpy, h2, the velocity, c2 and vertical distance above the datum, Z2

• be the rate of heat supplied from the furnace into the boiler and be the rate of heat loss from the boiler to its surrounding.

Steam boiler• Apply the continuity flow equation

• Apply the steady flow energy

• Since

• hence

2 21 2

1 21 1 2 2( ) ( )2 2

in outin outc cQ W m h gZ Q W m h gZ

1 2

in outm m

m m m

0

0

in

out

W

W

2 21 2

1 21 1 2 2( ) ( )2 2in outc cQ m h gZ Q m h gZ

Steam boilerExample• A boiler operates at a constant pressure of 20 bar

conditions. Steam is produced at the rate of 0.5 kg/s with a dryness fraction of 0.98. Feed water enters the boiler at a temperature of 60°C. Assuming the heat lost to surrounding, the change in kinetic and the change in potential energy are negligible. Determine the rate of heat energy supplied to the boiler.

Steam boilerSolution:Applying the continuity flow equation

At point 1, the feed water in a compressed liquid,

1 2 0.5 /

in outm m

m m kg s

1 60251.1 /

sf at t t Ch h kJ kg

Steam boilerAt point 2, the steam is a wet steam with dryness

fraction, x2=0.98

At p2=20 bar,

Applying

2 20

2 20

909 /

2799 /f f at p bar

g g at p bar

h h kJ kg

h h kJ kg

2 2 2 2 2

(1 )

(1 ) 0.98 2799 (1 0.98) 909 2761.2 /x g f

g f

h xh x h

h x h x h kJ kg

Steam boilerApplying the steady flow energy equation

Since

Hence

2 21 2

1 21 1 2 2( ) ( )2 2

in outin outc cQ W m h gZ Q W m h gZ

1 21 2

2 21 21 2

0

0

0

1 12 2

in

out

out loss

W

W

Q Q

m gZ m gZ

m c m c

1 21 2

3 3 32 12 1 0.5 (2761.2 10 252.1 10 ) 1255.05 10 /

in

in

Q m h m h

Q m h m h J s

Steam turbine• Steam turbine is a device used in a steam power

plant. Its function is to produce work output.

lossQ

outW

(power output)

1

2

high pressure steam

low pressure steam

1

1

1

1

mchz

2

2

2

2

mchz

During the expansion process work is produced by the turbine and heat energy may be lost from the turbine to its surrounding at a steady rate.

Steam turbine• kg/s of steam enter the

turbine at point 1 with specific enthalpy, h1, the velocity, c1 and vertical distance above the datum, Z1

• kg/s of steam enter the boiler at point 2 with specific enthalpy, h2, the velocity, c2 and vertical distance above the datum, Z2

• be the power output of the turbine and be the rate of heat loss from the turbine to its surroundings.

outW

Steam turbine• Apply the continuity flow equation

• Apply the steady flow energy

• Since

• hence

2 21 2

1 21 1 2 2( ) ( )2 2

in outin outc cQ W m h gZ Q W m h gZ

1 2

in outm m

m m m

0

0

in

in

W

Q

2 21 2

1 21 1 2 2( ) ( )2 2outoutc cm h gZ Q W m h gZ

Steam turbineExample• In a steam power plant as shown, 3.5 kg/s of

superheated steam at pressure 20 bar and temperature 450 °C enters the turbine. It then expends and leaves the turbine at pressure 0.12 bar and dryness fraction 0.92. The heat loss, the change in kinetic energy and the change in potential energy are assumed to be negligible. Determine the power output of the turbine.

Steam turbineSolution:Applying the continuity flow equation

Refer to the superheated steam table

1 2 3.5 /

in outm m

m m kg s

1 20 , 4503357 /

at p bar t Ch h kJ kg

Steam turbineAt point 2, the steam is a wet steam with dryness

fraction, x2=0.92

At p2=0.12 bar,

Applying

2 0.12

2 0.12

207 /

2590 /f f at p bar

g g at p bar

h h kJ kg

h h kJ kg

2 2 2 2 2

(1 )

(1 ) 0.92 2590 (1 0.92) 207 2399.4 /x g f

g f

h xh x h

h x h x h kJ kg

Steam turbineApplying the steady flow energy equation

Since

Hence

2 21 2

1 21 1 2 2( ) ( )2 2

in outin outc cQ W m h gZ Q W m h gZ

1 21 2

2 21 21 2

0

0

0

1 12 2

in

in

out loss

W

Q

Q Q

m gZ m gZ

m c m c

1 21 2

3 3 31 21 2 3.5 (3357 10 2399.4 10 ) 3351.6 10 /

out

out

m h W m h

W m h m h J s

Steam condenser• Steam condenser is a device used in a steam

power plant. It normally has a low pressure and it condenses the steam into water by taking away heat from the steam.

Condenser

1

2

steam

condensate

rejectQ

Steam condenser• kg/s of steam enter the

boiler at point 1 with specific enthalpy, h1, the velocity, c1 and vertical distance above the datum, Z1

• kg/s of condensate leave the condenser at point 2 with specific enthalpy, h2, the velocity, c2 and vertical distance above the datum, Z2

• be the rate of heat rejected from the condenser to its surroundings.

rejectedQ

Steam condenserApply the continuity flow equation

Apply the steady flow energy

Since

Hence

2 21 2

1 21 1 2 2( ) ( )2 2

in outin outc cQ W m h gZ Q W m h gZ

1 2

in outm m

m m m

0

0

0

in

out

in

out rejected

W

W

Q

Q Q

2 21 2

1 21 1 2 2( ) ( )2 2rejectedc cm h gZ Q m h gZ

Steam condenserExample• In a steam plant as shown, 3.5 kg/s of steam at

pressure 0.12 bar and dryness fraction 0.92 enters the condenser. It then condenses and leaves the condenser as saturated water. Assuming the changes in kinetic energy and potential energy are negligible. Determine the rate of heat rejected from the condenser to its surroundings.

Steam condenser

Solution:Applying the continuity flow equation

1 2 0.35 /

in outm m

m m kg s

Condenser

1

2

steam

condensate

rejectQ

1

1

1

1

3.5 /0.12 10.92

m kg sP barxh

2

2

h

m

Steam condenserAt point 1, the steam is a wet steam with dryness

fraction, x1=0.92At p1=0.12 bar,

Applying

At point 2, the condensate is a saturated water at p2=0.12 bar

2 0.12

2 0.12

207 /

2590 /f f at p bar

g g at p bar

h h kJ kg

h h kJ kg

1 1 1 1 1

(1 )

(1 ) 0.92 2590 (1 0.92) 207 2399.4 /x g f

g f

h xh x h

h x h x h kJ kg

2 0.12 207 /f at p barh h kJ kg

Steam condenserApplying the steady flow energy equation

Since

Hence

2 21 2

1 21 1 2 2( ) ( )2 2

in outin outc cQ W m h gZ Q W m h gZ

1 21 2

2 21 21 2

0

0

0

1 12 2

in

out

in

out rejected

W

W

Q

Q Q

m gZ m gZ

m c m c

1 21 2

3 3 31 21 2 3.5 (3399.4 10 207 10 ) 7673.4 10 /

rejected

rejected

m h Q m h

Q m h m h J s

Mixing chamber• Mixing chamber is a device used to mix high

temperature fluid and low temperature fluid, and produce fluid at the required temperature. Hot fluid and cold fluid enter the mixing chamber at point 1 and 2 respectively, Fluid at the required temperature leaves the mixing chamber at point 3. During the mixing heat energy may be lost from the chamber to its surrounding.

Mixing chamber

1

2

3

High temp. fluid

Low temp. fluid

Fluid at the required temperature

lossQ

Mixing chamber• kg/s of hot fluid enter the chamber at point 1

with specific enthalpy, h1, the velocity, c1 and vertical distance above the datum, Z1

• kg/s of cold fluid enter the chamber at point 2 with specific enthalpy, h2, the velocity, c2 and vertical distance above the datum, Z2

• kg/s of warm fluid leave the chamber at point 3 with specific enthalpy, h3, the velocity, c3 and vertical distance above the datum, Z3

• be the rate of heat loss from the chamber to its surroundings.

Mixing chamberApply the continuity flow equation

Apply the steady flow energy

Since

Hence

22 231 2

1 2 31 1 2 2 3 3( ) ( ) ( )2 2 2

in outin outcc cQ W m h gZ m h gZ Q W m h gZ

1 2 3

in outm m

m m m

0

0

0

in

out

in

out loss

W

W

Q

Q Q

22 231 2

1 2 31 1 2 2 3 3( ) ( ) ( )2 2 2loss

cc cm h gZ m h gZ Q m h gZ

Mixing chamberExample• In a food processing industry, steam and water

are mixed to produce continuous supply of hot water. 2.5 kg/s of dry saturated steam at pressure 1.2 bar and 60 kg/s of water at temperature 30 °C enter the mixing chamber. If the heat loss from the chamber to its surrounding is 500 W and the changes in kinetic energy and potential energy are negligible. Determine the mass flow rate and the specific enthalpy of the hot water.

Mixing chamberSolution:Applying the continuity flow equation

At point 1, the dry saturated steam at p1=1.2 bar

At point 2, the compressed water at t=30 °C

1 2 3

3 2.5 60 62.5 /

in outm m

m m m

m kg s

1 1.2 2683 /g at p barh h kJ kg

2 30125.7 /

sf at t Ch h kJ kg

Mixing chamberApplying the steady flow energy equation

Since

1 2 31 2 3

2 2 21 2 31 2 3

0

0

0

1 1 12 2 2

in

out

in

W

W

Q

m gZ m gZ m gZ

m c m c m c

22 231 2

1 2 31 1 2 2 3 3( ) ( ) ( )2 2 2

in outin outcc cQ W m h gZ m h gZ Q W m h gZ

Mixing chamberHence

1 2 31 2 3

3 1 23 1 2

1 21 23

3

3 3

3

2.5 2683 10 60 125.7 10 50062.5

227.98 10 /

loss

loss

loss

m h m h Q m h

m h m h m h Q

m h m h Qh

m

J kg

Thank youQ & A