Transcript C H A P T E R 13 The Transfer of Heat

PM3125
Content of Lectures 1 to 6:
Heat transfer:
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Source of heat
Heat transfer
Steam and electricity as heating media
Determination of requirement of amount of
steam/electrical energy
Steam pressure
Mathematical problems on heat transfer
uploaded at http://www.rshanthini.com/PM3125.htm
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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 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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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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Sensible Heat
• What is 'sensible heat‘?
Sensible heat is associated
with a temperature change
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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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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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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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Calculation of Sensible Heat
Calculate the amount of heat required to raise the temperature
of 300 g Al from 25oC to 70oC.
Data: c = 0.896 J/g oC for Al
Q = m c ΔT (since c is taken as a constant)
= (300 g) (0.896 J/g oC)(70 - 25)oC
= 12,096 J
= 13.1 kJ
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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 = 0.452 J/g oC for iron and 4.186 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 = 4.955 (tf -25)
tf = 34.2oC
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Latent Heat
• What is ‘latent heat‘?
Latent heat is associated with
phase change of matter
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Phases of Matter
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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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Phase Diagram: Water
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Phase Diagram: Water
Compressed liquid
Saturated liquid
Superheated
steam
Saturated steam
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Phase Diagram: Water
Explain why water is at liquid
state at atm pressure
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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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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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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) 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 21
Table
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 http://www.rshanthini.com/PM3125.htm)
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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
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ice
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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?
0oC
-20oC
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melting point of ice
ice
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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
-20oC
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melting point of ice
ice
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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
120.2oC
superheated steam
Specific heat
boiling point of water at 2 bar
Latent heat
Specific heat
0oC
-20oC
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melting point of ice
Specific heat
ice
Latent heat
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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) [120.2 - 0)]oC = 1004.9 kJ
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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) = 4404 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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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 + 668 kJ + 1004.9 kJ + 4404 kJ + 126 kJ
= 6285.3 kJ
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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,out
Hot fluid
at TH,in
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Cold fluid
at TC,in
Heat lost by the hot fluid
= Heat gained by the cold fluid
Hot fluid
at TH,out
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Application: Heat Exchanger
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Heat Exchanger
Heat lost by the hot fluid = Heat gained by the cold fluid
.m
hot
.
)=m
chot (TH,in – TH,out
mass flow rate
of hot fluid
cold
mass flow rate
of cold fluid
Specific heat
of hot fluid
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ccold (TC,out – TC,in)
Temperature
decrease in the
hot fluid
Specific heat
of cold fluid
Temperature
increase in the
cold fluid32
Heat Exchanger
Heat lost by the hot fluid = Heat gained by the cold fluid
.m
hot
.
)=m
chot (TH,in – TH,out
cold
ccold (TC,out – TC,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 heat
exchanger.
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Heat Exchanger
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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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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Worked Example in Heat Exchanger
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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
= 585.2 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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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) + 1947 kJ/kg = 1951 kJ/kg
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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 585.2 kJ/kg
Mass flow of steam / Mass flow of water
= 585.2 / 1951
= 0.30 kg stream / kg of water
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Group Assignment 1
Give the design of a heat exchanger
which has the most effective heat
transfer properties.
Learning objectives:
1) To be able to appreciate heat transfer applications in pharmaceutical
industry
2) To become familiar with the working principles of various heat
exchangers
3) To get a mental picture of different heat exchangers so that solving
heat transfer problems in class becomes more interesting
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Group Assignment 2
Steam enters a heat exchanger at 10 bar and 200oC and
leaves it as saturated water at the same pressure. Feedwater 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 feedwater.
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