Lecture to Heat Transfer

R. Shanthini 16 July 2009

PM3125

Lectures: 1 to 3

Lecture Content of Lectures 1 to 6: Heat transfer: source of heat, heat transfer, steam and electricity as heating media, determination of requirement of amount of steam/electrical energy, steam pressure and mathematical problems on heat transfer


What is Heat?


What is Heat?

Heat is energy in transit.


Units of Heat• The SI unit is the joule (J), or Newton-metre (Nm).

• Historically, heat was measured in terms of the ability to raise the temperature of water.

• The kilocalorie (kcal): amount of heat needed to raise the temperature of 1 kilogramme of water by 1 C0 (from 14.50C to 15.50C)

• 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)


Units of Heat

• 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)


Conversion between different

units of heat:

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


Sensible Heat

• What is 'sensible heat‘?

Sensible heat is associated with a temperature change


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

• 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


Calculation of Sensible Heat

Q is the heat lost or gained by a substance

m is the mass of substance

ΔT is the temperature change of substance

c is the specific heat of substance

Q = m c ΔT


Calculation of Sensible Heat

Q = m c ΔT

= (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 temperature of 300 g Al from 25oC to 70oC.

Data: c = 0.896 J/g oC for Al


Exchange of Heat

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

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


Latent Heat

• What is ‘latent heat‘?

Latent heat is associated with phase change of matter


Phases of Matter


Phase Change• Heat required for phase changes:

» Vaporization: liquid vapour» Melting: liquid solid» Sublimation: solid vapour

• Heat released by phase changes:» Condensation: vapour liquid» Fusion: liquid solid» Deposition: vapour solid


Phase Diagram: WaterExplain why water is at liquid

state at atm pressure


Phase Diagram: Carbon DioxideExplain why CO2 is at gas state

at atm pressure

Explain why CO2 cannot be made a

liquid at atm pressure


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.


Water: Specific Heat Capacities and Latent Heats

Specific heat of ice ≈ 2.06 J/g K

Heat of fusion for ice/water ≈ 334 J/g

Specific heat of water ≈ 4.18 J/g K

Latent heat of vaporization depends on the pressure and could be found from the Steam Table

Sensible heat required by superheated steam could also be found from the Steam 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)


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

at -20oC to steam at 150oC at 2 bar pressure?

Heat required to raise the temperature of ice from -20oCto 0oC= (2 kg) (2.06 kJ/kg oC) [0 - (-20)]oC = 82.4 kJ

Heat required to turn ice into water at 0oC= (2 kg) (334 kJ/kg) = 668 kJ

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

[water boils at 120.2oC at 2 bar not at 100oC as could be referred to from the Steam Table]


Warming curve for water (contd)What is the amount of heat required to change 2 kg of ice


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/kgas could be referred to from the Steam Table]

Heat required to raise the temperature of steam from 120.2oC to 150oC= (2 kg) (2770 – 2707) kJ/kg = 126 kJ

[the enthalpy at 120.2oC and 2 bar is the saturatedsteam enthalpy is 2707 kJ/kg and the enthalpy at 150oC and at 2 bar is 2770 kJ/kg as could be

referred to from the Steam Table]


Warming curve for water (contd)What is the amount of heat required to change 2 kg of ice


Total amount of heat required 4404 kJ

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

= 6285.3 kJ


Heat ExchangerIt is an industrial equipment in which heat from a hot fluid is transferred to a cold fluid to heat it by not bringing them into direct contact with each other.

Hot fluid at TH,in Hot fluid

at TH,out

Cold fluid at TC,out

Cold fluid at TCC,in

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


Heat Exchanger

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


mass flow rate of hot fluid

Specific heat of hot fluid

mass flow rate of hot fluid

Specific heat of hot fluid

Temperature decrease in the hot fluid

Temperature increase in the cold fluid


Heat Exchanger

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


The above is true only when the heat exchanger is well insulated so as to not loose heat to the

environment, and when there are no phase changes.


Heat Exchanger

mhot chot (TH,in – TH,out) = mcold ccold (TC,out – TC,in)

+ Heat lost to the environment

. .



If the heat exchanger is NOT well insulated and when there are no phase changes, then


Heat Exchanger

mhot (hH,in – hH,out) = mcold (hC,out – hC,in)


. .



If the heat exchanger is NOT well insulated and when there are phase changes, then


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 198.3oC. What is the mass of steam required per unit mass of incoming liquid water? The heat is assumed to be well insulated (adiabatic).

Example in Heat Exchanger


Example in Heat Exchanger Solution: High pressure (100 bar) water enters at 30oC and leaves at 198.3oC. At 100 bar, water boils at 311.0oC. Therefore, no phase changes in the high pressure water that is getting heated up n the heater.

Heat gained by high pressure water

= ccold (TC,out – TC,in)

= (4.18 kJ/kg oC) x (198.3-30)oC

= 703.5 kJ/kg

[You could calculate the above by taking the difference in enthalpies at the 2 given states from tables available, and the answer is (858.54 – 134.86) = 723.7 kJ/kg, which is the accurate method. ]


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


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 703.5 kJ/kg

Mass flow of steam / mass flow of water

= 703.5 / 1951

= 0.36


Give the design of a heat exchanger which has the most

effective heat transfer properties.

Assignment 1


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.

Lecture to Heat Transfer

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