Steam Tabels

8/10/2019 Steam Tabels

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What is Heat?


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What is Heat?

Heat is energy in transit.


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Units of Heat The SI unit is thejoule (J),

which is equal to Newton-metre (Nm).

Historically, heat was measured in terms of the ability

to raise the temperature of water.

The calorie (cal): amount of heat needed to raise thetemperature of 1 gramme of water by 1 C0 (from

14.50C to 15.50C)

In industry, the British thermal unit (Btu) is still used:amount of heat needed to raise the temperature of 1 lb

of water by 1 F0 (from 630F to 640F)


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Conversion between different

units of heat:

1 J = 0.2388 cal = 0.239x10-3 kcal = 60.189 Btu

1 cal = 4.186 J = 3.969 x 10-3 Btu


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Calculation of Sensible Heat

where T is the temperature change in the substance

Q = m c dTQ is the heat lost or gained by a substancem is the mass of substance

c is the specific heat of substance which changes with temperatureT is the temperature

When temperature changes causes negligible changes in c,

Q = m c dT = m c T


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When temperature changes causes significant changes in c,

Q = m c T cannot be used.

Q = H = m h

Instead, we use the following equation:

where H is the enthalpy change in the substance

and h is the specific enthalpy change in the substance.

To apply the above equation, the system should

remain at constant pressure and the associated

volume change must be negligibly small.


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Q = m c T (since c is taken as a constant)= (300 g) (0.896 J/g oC)(70 - 25)oC

= 12,096 J

= 13.1 kJ

Calculate the amount of heat required to raise the temperatureof 300 g Al from 25oC to 70oC.

Data: c = 0.896 J/g oC for Al


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Exchange of Heat

Heat lost by iron = Heat gained by water

(m c T)iron = (m c T)water

(100 g) (0.452 J/g oC)(80 - tf)oC

= (53.5 g) (4.186 J/go

C)(tf - 25)o

C80 - tf = 4.955 (tf-25)

tf = 34.2oC

Calculate the final temperature (tf), when 100 g iron at 80oC is

tossed into 53.5g of water at 25oC.

Data: c = 0.452 J/g oC for iron and 4.186 J/g oC for water


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Latent Heat

What is latent heat?

Latent heat is associated withphase change of matter


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Phases of Matter


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Phase Diagram: Water


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Phase Diagram: Water

Saturated steam

Superheated

steam

Saturated liquid

Compressed liquid


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Phase Diagram: WaterExplain why water is at liquid

state at atm pressure


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Phase Diagram: Carbon DioxideExplain why CO2 is at gas

state at atm pressure

Explain why CO2

cannot be made a

liquid at atm

pressure


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Latent Heat

Latent heat is the amount of heat added per unit mass of

substance during a phase change

Latent heat of fusion is the amount of heat added to melt

a unit mass of ice OR it is the amount of heat removedto freeze a unit mass of water.

Latent heat of vapourization is the amount of heat added

to vaporize a unit mass of water OR it is the amount ofheat removed to condense a unit mass of steam.


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Water:

Specific Heat Capacities and Latent Heats

Specific heat of ice 2.06 J/g K (assumed constant)

Heat of fusion for ice/water 334 J/g (assumed constant)

Specific heat of water 4.18 J/g K (assumed constant)

Latent heat of vaporization cannot be assumed a

constant since it changes significantly with the pressure,

and could be found from the Steam Table

How to evaluate the sensible heat gained (or lost) by

superheated steam?


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Water:

Specific Heat Capacities and Latent Heats

How to evaluate the sensible heat gained (or lost) bysuperheated steam?

Q = m c Tcannot be used since changes in c with changing

temperature is NOT negligible.

Q = H = m h

Instead, we use the following equation:

provided the system is at constant pressure and the

associated volume change is negligible.

Enthalpies could be referred from the Steam Table


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Properties of Steam

Learnt to refer to Steam Table to find properties of

steam such as saturated (or boiling point) temperatureand latent heat of vapourization at give pressures, and

enthalpies of superheated steam at various pressures and

temperatures.

)


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Warming curve for waterWhat is the amount of heat required to change 2 kg of ice

at -20o

C to steam at 150o

C at 2 bar pressure?

-20oC ice


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at -20o

C to steam at 150o


-20oC ice

0oC melting point of ice


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at -20o

C to steam at 150o


-20oC ice


120.2oC boiling point of water at 2 bar

Boiling point of water at 1 atm pressure is

100oC.

Boiling point of water at 2 bar is 120.2oC.

[Refer the Steam Table.]


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at -20o

C to steam at 150o


-20oC ice


120.2oC boiling point of water at 2 bar

150oC superheated steam

Specific heat

Specific heat

Specific heat

Latent heat

Latent heat


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at -20o

C to steam at 150o


Specific heat required to raise the temperature of ice from -20oCto 0oC

= (2 kg) (2.06 kJ/kg oC) [0 - (-20)]oC = 82.4 kJ

Latent heat required to turn ice into water at 0oC

= (2 kg) (334 kJ/kg) = 668 kJ

Specific heat required to raise the temperature of water from 0o

C to120.2oC

= (2 kg) (4.18 kJ/kg oC) [120.2 - 0)]oC = 1004.9 kJ


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at -20o

C to steam at 150o


Latent heat required to turn water into steam at 120.2oC and at 2 bar

= (2 kg) (2202 kJ/kg) = 4404 kJ

[Latent heat of vapourization at 2 bar is 2202 kJ/kg as could bereferred to from the Steam Table]

Specific heat required to raise the temperature of steam from 120.2oC

to 150oC

= (2 kg) (27702707) kJ/kg = 126 kJ

[Enthalpy at 120.2oC and 2 bar is the saturated steam enthalpy of

2707 kJ/kg and the enthalpy at 150oC and 2 bar is 2770 kJ/kg as

could be referred to from the Steam Table]


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at -20o

C to steam at 150o


Total amount of heat required

= 82.4 kJ + 668 kJ + 1004.9 kJ + 4404 kJ + 126 kJ

= 6285.3 kJ


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Application: Heat ExchangerIt is an industrial equipment in which heat is transferred from a hot

fluid (a liquid or a gas) to a cold fluid (another liquid or gas) without

the two fluids having to mix together or come into direct contact.

Hot fluid

at TH,in Hot fluid

at TH,out

Cold fluid

at TC,out

Cold fluid

at TC,in

Heat lost by the hot fluid

= Heat gained by the cold fluid


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Application: Heat Exchanger


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Heat Exchanger

mhot chot (TH,inTH,out) = mcold ccold (TC,outTC,in). .

Heat lost by the hot fluid = Heat gained by the cold fluid

mass flow rate

of hot fluid

Specific heatof hot fluid

mass flow rate

of cold fluid

Specific heatof cold fluid

Temperature

decrease in the

hot fluid

Temperature

increase in the

cold fluid


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Heat Exchanger

mhot chot (TH,inTH,out) = mcold ccold (TC,outTC,in). .


The above is true only under the following conditions:

(1) Heat exchanger is well insulated so that no heat is lost to the

environment

(2) There are no phase changes occurring within the heatexchanger.


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Heat Exchanger


+ Heat lost to the environment

If the heat exchanger is NOT well insulated, then

W k d E l 1 i H E h


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High pressure liquid water at 10 MPa (100 bar) and

30oC enters a series of heating tubes. Superheatedsteam at 1.5 MPa (15 bar) and 200oC is sprayed over

the tubes and allowed to condense. The condensed

steam turns into saturated water which leaves the

heat exchanger. The high pressure water is to be

heated up to 170oC. What is the mass of steam

required per unit mass of incoming liquid water?

The heat exchanger is assumed to be well insulated(adiabatic).

Worked Example 1 in Heat Exchanger

S l i W k d E l 1 i H E h


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Solution to Worked Example 1 in Heat Exchanger

S l i W k d E l 1 i H E h d


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High pressure (100 bar) water enters at 30oC and leaves at 198.3oC.

Boiling point of water at 100 bar is 311.0oC. Therefore, no phase

changes in the high pressure water that is getting heated up in the

heater.

Heat gained by high pressure water

= ccold (TC,out

TC,in)= (4.18 kJ/kg oC) x (170-30)oC

= 585.2 kJ/kg

[You could calculate the above by taking the difference in enthalpies at

the 2 given states from tables available.]

Solution to Worked Example 1 in Heat Exchanger contd.


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S l ti t W k d E l 1 i H t E h td


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Since there is no heat loss from the heater,

Heat lost by steam = Heat gained by high pressure water

Mass flow rate of steam x 1951 kJ/kg

= Mass flow rate of water x 585.2 kJ/kg

Mass flow rate of steam / Mass flow rate of water

= 585.2 / 1951= 0.30 kg stream / kg of water

Solution to Worked Example 1 in Heat Exchanger contd.


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Give the design of a heat exchanger

which has the most effective heat

transfer properties.

Assignment

Learning objectives:

1) To be able to appreciate heat transfer applications in pharmaceutical

industry

2) To become familiar with the working principles of various heatexchangers

3) To get a mental picture of different heat exchangers so that solving

heat transfer problems in class becomes more interesting

W k d E l 2 i H t E h


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Steam enters a heat exchanger at 10 bar and 200oC and

leaves it as saturated water at the same pressure. Feed-

water enters the heat exchanger at 25 bar and 80oC and

leaves at the same pressure and at a temperature 20oC

less than the exit temperature of the steam. Determine the

ratio of the mass flow rate of the steam to that of thefeed-water, neglecting heat losses from the heat

exchanger.

If the feed-water leaving the heat exchanger is fed

directly to a boiler to be converted to steam at 25 bar and300oC, find the heat required by the boiler per kg of feed-

water.

Worked Example 2 in Heat Exchanger

S l ti t W k d E l 2 i H t E h


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- Steam enters at 10 bar and 200oC and leaves it as saturated water at

the same pressure.

- Saturation temperature of water at 10 bar is 179.9oC.

- Feed-water enters the heat exchanger at 25 bar and 80oC and leaves

at the same pressure and at a temperature 20oC less than the exit

temperature of the steam, which is 179.9oC.

- Boiling point of water at 25 bar is (221.8+226.0)/2 = 223.9oC.- Therefore, no phase changes in the feed-water that is being heated.

Heat lost by steam = Heat gained by feed-water (with no heat losses)

Mass flow rate of steam x [28292778 + 2015] kJ/kg

= Mass flow rate of feed-water x [4.18 x (179.9-20-80) ] kJ/kg

Mass flow of steam / Mass flow of feed-water

= 333.98 / 2066 = 0.1617 kg stream / kg of water

Solution to Worked Example 2 in Heat Exchanger


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Use of Steam Tables

Saturated Vapor or Liquid

Liquid at Bubble-Point

Vapor at Dew-Point

Liquid and Vapor Co-existing

Superheated Vapor or Sub-Cooled Liquid

Vapor Above Dew-Point Temperature or Below Dew-

Point Pressure Liquid Below Bubble-Point Temperature or Above

Bubble-Point Pressure


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Vaporize 1kg Water at 20 C and Bubble Point Pressure

U Q W Q P V

m U Q mP V

Q m U P V


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Q U P V


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3 5

2

2403.0 83.91

57.8 0.0010021 0.0234 1 10

1,000

2454.3

Q m U P V

kJkgQ

kgmkg bar N kJ

kg m bar N m

Q kJ

H U PVH U P V


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Definitions

Degrees of Superheat

Difference between actual temperature and

saturation temperature at the same pressure Degrees of Sub-Cooling

Difference between actual temperature and the

saturation temperature at the same pressure.


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Vapor

Liquid


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Water Vapor Heating

1 kg from 100

o

C, 1 Bar to 200

o

C, 1 Bar

U Q W Q P V m U Q mP V

Q m U P V m H


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Water Vapor Heating

1 kg from 100

o

C, 1 Bar to 200

o

C, 1 Bar

2875 26761199

Q m U P V m H

kJkgQ kJ

kg


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Constant Volume Water Vapor Heating

1 kg from 100

o

C, 1 Bar to 200

o

C, P=?

0

V

U Q

m U Q


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Interpolation for Pressure

0

1

2

34

5

6

0.3 0.5 0.7 0.9 1.1 1.3 1.5 1.7 1.9 2.1 2.3

Specific Volume

Pressure


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Interpolation for Internal Energy

2640

2645

2650

2655

2660

0.3 0.5 0.7 0.9 1.1 1.3 1.5 1.7 1.9 2.1 2.3

Specific Volume

InternalEnergy

Linear Interpolation


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Linear Interpolation

1 1 2 2

2 1 2 1

2 1

2 1

1 1

1 1

1 1

2 1

1 1

2 1

y mx b

y mx b y mx by y m x x

y ym

x x

b y mx

y mx y mx

y y m x x

y yy y x x

x x

2 1 P P

P P V V


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2 1

1 1

2 1

2 1

1 1

2 1

5 11 1.69 2.17 2.10.425 2.17

2643 2658 2658 1.69 2.17 2654

0.425 2.17

1 2654 2507 147

Desired

Desired

P P V VV V

P bar

U U

U U V V V V

kJU

kg

kJQ m U kg kJ

kg

Steam Tabels

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