1. PM3125 Content of Lectures 1 to 6: Heat transfer: Source of heat
Steam and electricity as heating media
Determination of requirement of amount of steam/electrical energy
Steam pressure
Mathematical problems on heat transfer
uploaded at
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2. What is Heat?
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3. Heat is energy in transit.
What is Heat?
Heat is energy in transit.
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4. Units of Heat The SI unit is the joule (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 the temperature 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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5. Conversion between different
units of heat:
1 J = cal = 0.239x10-3 kcal = Btu
1 cal = J = x 10-3 Btu
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6. Sensible heat is associated with a temperature change
What is 'sensible heat‘?
Sensible heat is associated with a temperature change
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7. Specific Heat Capacity
To raise the temperature by 1 K, different substances need different amount of energy because substances have different molecular configurations and bonding (eg: copper, water, wood)
The amount of energy needed to raise the temperature of 1 kg of a substance by 1 K is known as the specific heat capacity
Specific heat capacity is denoted by c
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8. Calculation of Sensible Heat
Q = m c dT ∫
Q is the heat lost or gained by a substance
m is the mass of substance
c is the specific heat of substance which changes with temperature
T is the temperature
When temperature changes causes negligible changes in c,
Q = m c dT ∫
= m c ∆T
where ΔT is the temperature change in the substance
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9. Calculation of Sensible Heat
When temperature changes causes significant changes in c,
Q = m c ∆T cannot be used.
Instead, we use the following equation:
Q = ∆H = m ∆h
where ΔH 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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10. Calculation of Sensible Heat
Calculate the amount of heat required to raise the temperature of 300 g Al from 25oC to 70oC.
Data: c = J/g oC for Al
Q = m c ΔT (since c is taken as a constant)
= (300 g) (0.896 J/g oC)( )oC
= 12,096 J
= 13.1 kJ
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11. Data: c = 0.452 J/g oC for iron and 4.186 J/g oC for water
Exchange of Heat
Calculate the final temperature (tf), when 100 g iron at 80oC is tossed into 53.5g of water at 25oC.
Data: c = J/g oC for iron and J/g oC for water
Heat lost by iron = Heat gained by water
(m c ΔT)iron = (m c ΔT)water
(100 g) (0.452 J/g oC)(80 - tf)oC
= (53.5 g) (4.186 J/g oC)(tf - 25)oC
80 - tf = (tf -25)
tf = 34.2oC
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12. Latent heat is associated with phase change of matter
What is ‘latent heat‘?
Latent heat is associated with phase change of matter
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13. Phases of Matter
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14. Phase Change Heat required for phase changes:
Melting: solid  liquid
Vaporization: liquid  vapour
Sublimation: solid  vapour
Heat released by phase changes:
Condensation: vapour  liquid
Fusion: liquid  solid
Deposition: vapour  solid
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15. Phase Diagram: Water
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16. Phase Diagram: Water Compressed liquid Saturated liquid Superheated
steam
Saturated steam
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17. Phase Diagram: Water
Explain why water is at liquid state at atm pressure
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18. Phase Diagram: Carbon Dioxide
Explain why CO2 is at gas state at atm pressure
Explain why CO2 cannot be made a liquid at atm pressure
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19. 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 removed to 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 of heat removed to condense a unit mass of steam.
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20. 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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21. Water: Specific Heat Capacities and Latent Heats
How to evaluate the sensible heat gained (or lost) by superheated steam?
Q = m c ∆T
cannot be used since changes in c with changing temperature is NOT negligible.
Instead, we use the following equation:
Q = ∆H = m ∆h
provided the system is at constant pressure and the associated volume change is negligible.
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Enthalpies could be referred from the Steam Table

22. Properties of Steam
Learnt to refer to Steam Table to find properties of steam such as saturated (or boiling point) temperature and latent heat of vapourization at give pressures, and enthalpies of superheated steam at various pressures and temperatures.
Reference:
Chapter 6 of “Thermodynamics for Beginners with worked examples” by R. Shanthini
(published by Science Education Unit, Faculty of Science,
University of Peradeniya)
(also uploaded at
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23. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
-20oC ice
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24. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
-20oC ice
0oC melting point of ice
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25. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
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.]
0oC melting point of ice
-20oC ice
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26. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
150oC superheated steam
Specific heat
120.2oC
boiling point of water at 2 bar
Latent heat
Specific heat
0oC melting point of ice
Latent heat
Specific heat
-20oC ice
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27. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
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 0oC to 120.2oC
= (2 kg) (4.18 kJ/kg oC) [ )]oC = kJ
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28. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
Latent heat required to turn water into steam at 120.2oC and at 2 bar
= (2 kg) (2202 kJ/kg) = kJ
[Latent heat of vapourization at 2 bar is 2202 kJ/kg as could be referred to from the Steam Table]
Specific heat required to raise the temperature of steam from 120.2oC to 150oC
= (2 kg) (2770 – 2707) 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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29. Warming curve for water
What is the amount of heat required to change 2 kg of ice
at -20oC to steam at 150oC at 2 bar pressure?
Total amount of heat required
= 82.4 kJ kJ kJ kJ kJ
= kJ
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30. Application: Heat Exchanger
It 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.
Cold fluid
at TC,in
Cold fluid
at TC,out
Hot fluid
at TH,in
Heat lost by the hot fluid
= Heat gained by the cold fluid
Hot fluid
at TH,out
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31. Application: Heat Exchanger
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32. mhot chot (TH,in – TH,out) = mcold ccold (TC,out – TC,in)
Heat Exchanger
mhot chot (TH,in – TH,out) = mcold ccold (TC,out – TC,in) .
Heat lost by the hot fluid = Heat gained by the cold fluid
mass flow rate of hot fluid
mass flow rate of cold fluid
Specific heat of hot fluid
Specific heat of cold fluid
Temperature increase in the cold fluid
Temperature decrease in the hot fluid
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33. mhot chot (TH,in – TH,out) = mcold ccold (TC,out – TC,in)
Heat Exchanger
mhot chot (TH,in – TH,out) = mcold ccold (TC,out – TC,in) .
Heat lost by the hot fluid = Heat gained by the cold fluid
The above is true only under the following conditions:
Heat exchanger is well insulated so that no heat is lost to the environment
There are no phase changes occurring within the heat exchanger.
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34. If the heat exchanger is NOT well insulated, then
Heat lost by the hot fluid = Heat gained by the cold fluid
+ Heat lost to the environment
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35. Worked Example in Heat Exchanger
High pressure liquid water at 10 MPa (100 bar) and 30oC enters a series of heating tubes. Superheated steam 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 is assumed to be well insulated (adiabatic).
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36. Worked Example in Heat Exchanger
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37. Worked Example in Heat Exchanger
Solution:
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
= kJ/kg
[You could calculate the above by taking the difference in enthalpies at the 2 given states from tables available.]
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38. Worked Example in Heat Exchanger
Solution continued:
Superheated steam 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.
Heat lost by steam
= heat lost by superheated steam to become saturated steam
+ latent heat of steam lost for saturated steam to turn into
saturated water
= Enthalpy at 15 bar and 200oC
– Enthalpy of saturated steam at 15 bar
+ Latent heat of vapourization at 15 bar
= (2796 kJ/kg – 2792 kJ/kg) kJ/kg = 1951 kJ/kg
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39. Worked Example in Heat Exchanger
Solution continued:
Since there is no heat loss from the heater,
Heat lost by steam = Heat gained by high pressure water
Mass flow of steam x 1951 kJ/kg
= Mass flow of water x kJ/kg
Mass flow of steam / Mass flow of water
= / 1951
= 0.30 kg stream / kg of water
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40. Group Assignment 1
Give the design of a heat exchanger which has the most effective heat transfer properties.
Learning objectives:
To be able to appreciate heat transfer applications in pharmaceutical industry
To become familiar with the working principles of various heat exchangers
To get a mental picture of different heat exchangers so that solving heat transfer problems in class becomes more interesting
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41. Group Assignment 2
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 the feed-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 and 300oC, find the heat required by the boiler per kg of feed-water.
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