Download presentation
Presentation is loading. Please wait.
Published byRalf Melton Modified over 9 years ago
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 Prof. R. Shanthini & 12 March 2012
2
What is Heat? Prof. R. Shanthini & 12 March 2012
3
Heat is energy in transit.
What is Heat? Heat is energy in transit. Prof. R. Shanthini & 12 March 2012
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) Prof. R. Shanthini & 12 March 2012
5
Conversion between different
units of heat: 1 J = cal = 0.239x10-3 kcal = Btu 1 cal = J = x 10-3 Btu Prof. R. Shanthini & 12 March 2012
6
Sensible heat is associated with a temperature change
What is 'sensible heat‘? Sensible heat is associated with a temperature change Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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. Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
12
Latent heat is associated with phase change of matter
What is ‘latent heat‘? Latent heat is associated with phase change of matter Prof. R. Shanthini & 12 March 2012
13
Phases of Matter Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
15
Phase Diagram: Water Prof. R. Shanthini & 12 March 2012
16
Phase Diagram: Water Compressed liquid Saturated liquid Superheated
steam Saturated steam Prof. R. Shanthini & 12 March 2012
17
Phase Diagram: Water Explain why water is at liquid state at atm pressure Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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. Prof. R. Shanthini & 12 March 2012
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? Prof. R. Shanthini & 12 March 2012
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. Prof. R. Shanthini & 12 March 2012 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 Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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] Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
31
Application: Heat Exchanger
Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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. Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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). Prof. R. Shanthini & 12 March 2012
36
Solution to Worked Example 1 in Heat Exchanger
Prof. R. Shanthini & 12 March 2012
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 = kJ/kg [You could calculate the above by taking the difference in enthalpies at the 2 given states from tables available.] Prof. R. Shanthini & 12 March 2012
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) kJ/kg = 1951 kJ/kg Prof. R. Shanthini & 12 March 2012
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 kJ/kg Mass flow rate of steam / Mass flow rate of water = / 1951 = 0.30 kg stream / kg of water Prof. R. Shanthini & 12 March 2012
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 Prof. R. Shanthini & 12 March 2012
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. Prof. R. Shanthini & 12 March 2012
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 ( )/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 – ] kJ/kg = Mass flow rate of feed-water x [4.18 x ( ) ] kJ/kg Mass flow of steam / Mass flow of feed-water = / 2066 = kg stream / kg of water Prof. R. Shanthini & 12 March 2012
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 ( )/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 ( ) + ( )/2 + [( )/2 – ( )/2]} kJ/kg = { [ – ]} kJ/kg = 2433 kJ/kg of feed-water Prof. R. Shanthini & 12 March 2012
44
Heat Transfer is the means by which energy moves from
a hotter object to a colder object Prof. R. Shanthini & 12 March 2012
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) Prof. R. Shanthini & 12 March 2012
46
Mechanisms of Heat Transfer
Latent heat Conduction Convection Radiation Prof. R. Shanthini & 12 March 2012
47
Heat travels along the rod
Conduction HOT (lots of vibration) COLD (not much vibration) Heat travels along the rod Prof. R. Shanthini & 12 March 2012
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. Prof. R. Shanthini & 12 March 2012
49
Conduction: Fourier’s Law
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 ΔT Q = L k A t ( ) What is the unit of k? Prof. R. Shanthini & 12 March 2012
50
Thermal Conductivities
Substance Thermal Conductivity k [W/m.K] Syrofoam 0.010 Glass 0.80 Air 0.026 Concrete 1.1 Wool 0.040 Iron 79 Wood 0.15 Aluminum 240 Body fat 0.20 Silver 420 Water 0.60 Diamond 2450 Prof. R. Shanthini & 12 March 2012
51
Conduction through Single Wall
Use Fourier’s Law: T1 ΔT Q = L k A t ( ) Q . Q . Q . k A (T1 – T2) T2 T1 = x Δx Δx Prof. R. Shanthini & 12 March 2012
52
Conduction through Single Wall
Q . k A (T1 – T2) T1 = Δx Q . Q . T1 – T2 = Δx/(kA) T2 T1 x Δx Thermal resistance (in K/W) (opposing heat flow) Prof. R. Shanthini & 12 March 2012 52
53
Conduction through Composite Wall
B C T1 Q . Q . T2 T3 T4 kA kB kC x ΔxA ΔxB ΔxC Q . T1 – T2 T2 – T3 T3 – T4 = = = (Δx/kA)A (Δx/kA)B (Δx/kA)C Prof. R. Shanthini & 12 March 2012 53
54
Conduction through Composite Wall
T1 – T2 = (Δx/kA)A Q . T2 – T3 T3 – T4 (Δx/kA)C (Δx/kA)B + (Δx/kA)B (Δx/kA)A + (Δx/kA)C [ ] Q . = T1 – T2 + T2 – T3 + T3 – T4 Q . T1 – T4 = + (Δx/kA)B (Δx/kA)A + (Δx/kA)C 54 Prof. R. Shanthini & 12 March 2012
55
Example 1 . Tin – Tout Q = + (Δx/kA)insulation (Δx/kA)fireclay
An industrial furnace wall is constructed of 21 cm thick fireclay brick having k = 1.04 W/m.K. This is covered on the outer surface with 3 cm layer of insulating material having k = 0.07 W/m.K. The innermost surface is at 1000oC and the outermost surface is at 40oC. Calculate the steady state heat transfer per area. Solution: We start with the equation Q . Tin – Tout = (Δx/kA)fireclay + (Δx/kA)insulation Prof. R. Shanthini & 12 March 2012
56
Example 1 continued . . Q (1000 – 40) A = (0.21/1.04) + (0.03/0.07) Q
= W/m2 A Prof. R. Shanthini & 12 March 2012
57
Example 2 . . Q Tin – Tout = (Δx/kA)fireclay + (Δx/kA)insulation Q
We want to reduce the heat loss in Example 1 to 960 W/m2. What should be the insulation thickness? Solution: We start with the equation Q . Tin – Tout = (Δx/kA)fireclay + (Δx/kA)insulation Q . (1000 – 40) = 960 W/m2 = A (0.21/1.04) + (Δx)insulation /0.07) (Δx)insulation = 5.6 cm Prof. R. Shanthini & 12 March 2012
58
Conduction through hollow-cylinder
ri To L Q . Ti – To = [ln(ro/ri)] / 2πkL Prof. R. Shanthini & 12 March 2012
59
Conduction through the composite wall in a hollow-cylinder
To Material A Ti r1 Material B Q . Ti – To = [ln(r2/r1)] / 2πkAL + [ln(r3/r2)] / 2πkBL Prof. R. Shanthini & 12 March 2012
60
Example 3 . Ti – To Q = [ln(r2/r1)] / 2πkAL + [ln(r3/r2)] / 2πkBL
A thick walled tube of stainless steel ( k = 19 W/m.K) with 2-cm inner diameter and 4-cm outer diameter is covered with a 3-cm layer of asbestos insulation (k = 0.2 W/m.K). If the inside-wall temperature of the pipe is maintained at 600oC and the outside of the insulation at 100oC, calculate the heat loss per meter of length. Solution: We start with the equation Q . Ti – To = [ln(r2/r1)] / 2πkAL + [ln(r3/r2)] / 2πkBL Prof. R. Shanthini & 12 March 2012
61
Example 3 continued . . 2 π L ( 600 – 100) Q = + [ln(5/2)] / 0.2
= 680 W/m L Prof. R. Shanthini & 12 March 2012
62
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) Prof. R. Shanthini & 12 March 2012
63
Convection currents are set up when a pan of water is heated.
Convection is the process in which heat is carried from place to place by the bulk movement of a fluid (gas or liquid). Convection currents are set up when a pan of water is heated. Prof. R. Shanthini & 12 March 2012
64
Convection It explains why breezes come from the ocean in the day and from the land at night Prof. R. Shanthini & 12 March 2012
65
Convection: Newton’s Law of Cooling
Flowing fluid at Tfluid Heated surface at Tsurface Q . conv. = h A (Tsurface – Tfluid) Area exposed Heat transfer coefficient (in W/m2.K) Prof. R. Shanthini & 12 March 2012
66
Convection: Newton’s Law of Cooling
Flowing fluid at Tfluid Heated surface at Tsurface Q . conv. Tsurface – Tfluid = 1/(hA) Convective heat resistance (in K/W) Prof. R. Shanthini & 12 March 2012
67
Example 4 The convection heat transfer coefficient between a surface at 50oC and ambient air at 30oC is 20 W/m2.K. Calculate the heat flux leaving the surface by convection. Solution: Use Newton’s Law of cooling : Q . conv. = h A (Tsurface – Tfluid) Flowing fluid at Tfluid = 30oC = (20 W/m2.K) x A x (50-30)oC Heated surface at Tsurface = 50oC Heat flux leaving the surface: . Q conv. A = 20 x 20 = 400 W/m2 h = 20 W/m2.K Prof. R. Shanthini & 12 March 2012
68
Example 5 Air at 300°C flows over a flat plate of dimensions 0.50 m by 0.25 m. If the convection heat transfer coefficient is 250 W/m2.K, determine the heat transfer rate from the air to one side of the plate when the plate is maintained at 40°C. Solution: Use Newton’s Law of cooling : Q . conv. = h A (Tsurface – Tfluid) Flowing fluid at Tfluid = 300oC Heated surface at Tsurface = 40oC = 250 W/m2.K x m2 x ( )oC = W/m2 h = 250 W/m2.K A = 0.50x0.25 m2 Heat is transferred from the air to the plate. Prof. R. Shanthini & 12 March 2012
69
Forced Convection In forced convection over external surface:
In forced convection, a fluid is forced by external forces such as fans. In forced convection over external surface: Tfluid = the free stream temperature (T∞), or a temperature far removed from the surface In forced convection through a tube or channel: Tfluid = the bulk temperature Prof. R. Shanthini & 12 March 2012
70
Free Convection In free convection, a fluid is circulated due to buoyancy effects, in which less dense fluid near the heated surface rises and thereby setting up convection. In free (or partially forced) convection over external surface: Tfluid = (Tsurface + Tfree stream) / 2 In free or forced convection through a tube or channel: Tfluid = (Tinlet + Toutlet) / 2 Prof. R. Shanthini & 12 March 2012
71
Change of Phase Convection
Change-of-phase convection is observed with boiling or condensation . It is a very complicated mechanism and therefore will not be covered in this course. Prof. R. Shanthini & 12 March 2012
72
Overall Heat Transfer through a Plane Wall
Fluid A at TA > T1 T1 Q . Q . T2 Fluid B at TB < T2 x Δx Q . TA – T1 = 1/(hAA) T1 – T2 = Δx/(kA) T2– TB = 1/(hBA) Prof. R. Shanthini & 12 March 2012
73
Overall Heat Transfer through a Plane Wall
Q . TA – T1 = 1/(hAA) T1 – T2 = Δx/(kA) T2– TB = 1/(hBA) . TA – TB Q = 1/(hAA) + Δx/(kA) + 1/(hBA) Q . = U A (TA – TB) where U is the overall heat transfer coefficient given by 1/U = 1/hA + Δx/k + 1/hB Prof. R. Shanthini & 12 March 2012
74
Overall heat transfer through hollow-cylinder
ri ro Ti To Fluid A is inside the pipe Fluid B is outside the pipe TA > TB L Q . = U A (TA – TB) where 1/UA = 1/(hAAi) + ln(ro/ri) / 2πkL + 1/(hBAo) Prof. R. Shanthini & 12 March 2012
75
Example 6 . Q = U A (TA – TB) = U A (120 – 35) What is UA?
Steam at 120oC flows in an insulated pipe. The pipe is mild steel (k = 45 W/m K) and has an inside radius of 5 cm and an outside radius of 5.5 cm. The pipe is covered with a 2.5 cm layer of 85% magnesia (k = 0.07 W/m K). The inside heat transfer coefficient (hi) is 85 W/m2 K, and the outside coefficient (ho) is 12.5 W/m2 K. Determine the heat transfer rate from the steam per m of pipe length, if the surrounding air is at 35oC. Solution: Start with Q . = U A (TA – TB) = U A (120 – 35) What is UA? Prof. R. Shanthini & 12 March 2012
76
Example 6 continued 1/UA = 1/(hAAi) + ln(ro/ri) / 2πkL + … + 1/(hBAo)
1/UA = 1/(85Ain) + ln(5.5/5) / 2π(45)L + ln(8/5.5) / 2π(0.07)L + 1/(12.5Aout) Ain = 2π(0.05)L and Aout = 2π(0.08)L 1/UA = ( ) / 2πL Prof. R. Shanthini & 12 March 2012
77
Example 6 continued . . UA = 2πL / (0.235 + 0.0021 +5.35 + 1) Q = U A
(120 – 35) steel air = 2πL (120 – 35) / ( ) steam insulation = 81 L Q . / L = 81 W/m Prof. R. Shanthini & 12 March 2012
78
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) Prof. R. Shanthini & 12 March 2012
79
Radiation Radiation is the process in which energy is transferred by means of electromagnetic waves of wavelength band between 0.1 and 100 micrometers solely as a result of the temperature of a surface. Heat transfer by radiation can take place through vacuum. This is because electromagnetic waves can propagate through empty space. Prof. R. Shanthini & 12 March 2012
80
The Stefan–Boltzmann Law of Radiation
Q t = ε σ A T4 ε = emissivity, which takes a value between 0 (for an ideal reflector) and 1 (for a black body). σ = x 10-8 W/m2.K4 is the Stefan-Boltzmann constant A = surface area of the radiator T = temperature of the radiator in Kelvin. Prof. R. Shanthini & 12 March 2012
81
Why is the mother shielding her cub?
Ratio of the surface area of a cub to its volume is much larger than for its mother. Prof. R. Shanthini & 12 March 2012
82
What is the Sun’s surface temperature?
The sun provides about 1000 W/m2 at the Earth's surface. Assume the Sun's emissivity ε = 1 Distance from Sun to Earth = R = 1.5 x 1011 m Radius of the Sun = r = 6.9 x 108 m Prof. R. Shanthini & 12 March 2012
83
What is the Sun’s surface temperature?
Q t = ε σ A T4 (4 π 6.92 x 1016 m2) = x 1018 m2 (4 π 1.52 x 1022 m2)(1000 W/m2) = x 1026 W 2.83 x 1026 W T4 = (1) (5.67 x 10-8 W/m2.K4) (5.98 x 1018 m2) ε σ T = 5375 K Prof. R. Shanthini & 12 March 2012
84
If object at temperature T is surrounded by an environment at temperature T0, the net radioactive heat flow is: Q t = ε σ A (T4 - To4 ) Temperature of the radiating surface Temperature of the environment Prof. R. Shanthini & 12 March 2012
85
Example 7 What is the rate at which radiation is emitted by a surface of area 0.5 m2, emissivity 0.8, and temperature 150°C? Solution: [( ) K]4 Q t = ε σ A T4 0.5 m2 0.8 5.67 x 10-8 W/m2.K4 Q = (0.8) (5.67 x 10-8 W/m2.K4) (0.5 m2) (423 K)4 t = 726 W Prof. R. Shanthini & 12 March 2012
86
Example 8 Q = ε σ A (T4 - To4 ) t Q t Solution: [(273+25) K]4
If the surface of Example 7 is placed in a large, evacuated chamber whose walls are maintained at 25°C, what is the net rate at which radiation is exchanged between the surface and the chamber walls? Solution: Q t = ε σ A (T4 - To4 ) [(273+25) K]4 [( ) K]4 Q = (0.8) x (5.67 x 10-8 W/m2.K4) x (0.5 m2) x [(423 K)4 -(298 K)4 ] t = 547 W Prof. R. Shanthini & 12 March 2012
87
Example 8 continued Note that 547 W of heat loss from the surface occurs at the instant the surface is placed in the chamber. That is, when the surface is at 150oC and the chamber wall is at 25oC. With increasing time, the surface would cool due to the heat loss. Therefore its temperature, as well as the heat loss, would decrease with increasing time. Steady-state conditions would eventually be achieved when the temperature of the surface reached that of the surroundings. Prof. R. Shanthini & 12 March 2012
88
Example 9 Q = ε σ A (T4 - To4 ) t Q t Solution: [(273+25) K]4
Under steady state operation, a 50 W incandescent light bulb has a surface temperature of 135°C when the room air is at a temperature of 25°C. If the bulb may be approximated as a 60 mm diameter sphere with a diffuse, gray surface of emissivity 0.8, what is the radiant heat transfer from the bulb surface to its surroundings? Q t = ε σ A (T4 - To4 ) Solution: [(273+25) K]4 [( ) K]4 Q = (0.8) x (5.67 x 10-8 J/s.m2.K4) x [π x (0.06) m2] x [(408 K)4 -(298 K)4 ] t = 10.2 W (about 20% of the power is dissipated by radiation) Prof. R. Shanthini & 12 March 2012
Similar presentations
© 2024 SlidePlayer.com. Inc.
All rights reserved.