PM3125: Lectures 6 to 9

# PM3125: Lectures 6 to 9

## PM3125: Lectures 6 to 9

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1. PM3125: Lectures 6 to 9 • Content of Lectures 6 to 12: • Heat transfer: • -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

2. What is Heat?

3. What is Heat? Heat is energy in transit.

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)

5. 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

6. Sensible Heat • What is 'sensible heat‘? Sensible heat is associated with a temperature change

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

8. Q = m cdT ∫ Q = m cdT Calculation of Sensible Heat 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, = m c ∆T where ΔT is the temperature change in the substance

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.

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 = 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

11. 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

12. Latent Heat • What is ‘latent heat‘? Latent heat is associated with phase change of matter

13. Phases of Matter

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

15. Phase Diagram: Water

16. Phase Diagram: Water Compressed liquid Saturated liquid Superheated steam Saturated steam

17. Phase Diagram: Water Explain why water is at liquid state at atm pressure

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

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.

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?

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. 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 http://www.rshanthini.com/PM3125.htm)

23. -20oC ice 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?

24. 0oC melting point of ice -20oC ice 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?

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

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

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) [120.2 - 0)]oC = 1004.9 kJ

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) = 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]

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 + 668 kJ + 1004.9 kJ + 4404 kJ + 126 kJ • = 6285.3 kJ

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

31. Application: Heat Exchanger

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

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

34. 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

35. Worked Example 1 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 exchanger is assumed to be well insulated (adiabatic).

36. Solution to Worked Example 1 in Heat Exchanger

37. Solution to Worked Example 1 in Heat Exchanger contd. 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.]

38. Solution to Worked Example 1 in Heat Exchanger contd. 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 of superheated steam 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

39. Solution to Worked Example 1 in Heat Exchanger contd. 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

40. Assignment 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

41. Worked Example 2 in Heat Exchanger 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.

42. Solution to Worked Example 2 in Heat Exchanger • - 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 [2829 – 2778 + 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

43. Solution to Worked Example 1 in Heat Exchanger contd. • 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. • Temperature of feed-water leaving the heat exchanger is 159.9oC • Boiling point of water at 25 bar is (221.8+226.0)/2 = 223.9oC • The feed-water is converted to superheated steam at 300oC • Heat required by the boiler per kg of feed-water • = {4.18 x (223.9-159.9) + (1850+1831)/2 • + [(3138+3117)/2 – (2802+2803)/2]} kJ/kg • = {267.52 + 1840.5 + [3127.5 – 2802.5]} kJ/kg • = 2433 kJ/kg of feed-water

44. Heat Transfer is the means by which energy moves from a hotter object to a colder object

45. Mechanisms of Heat Transfer Conduction is the flow of heat by direct contact between a warmer and a cooler body. Convection is the flow of heat carried by moving gas or liquid. (warm air rises, gives up heat, cools, then falls) Radiation is the flow of heat without need of an intervening medium. (by infrared radiation, or light)

46. Mechanisms of Heat Transfer Latent heat Conduction Convection Radiation

47. Conduction HOT (lots of vibration) COLD (not much vibration) Heat travels along the rod

48. Conduction Conduction is the process whereby heat is transferred directly through a material, any bulk motion of the material playing no role in the transfer. Those materials that conduct heat well are called thermal conductors, while those that conduct heat poorly are known as thermal insulators. Most metals are excellent thermal conductors, while wood, glass, and most plastics are common thermal insulators. The free electrons in metals are responsible for the excellent thermal conductivity of metals.

49. Conduction: Fourier’s Law ( ) ΔT k A t Q = L Cross-sectional area A L Q = heat transferred k = thermal conductivity A = cross sectional area DT = temperature difference between two ends L = length t = duration of heat transfer What is the unit of k?

50. Thermal Conductivities