quiz
Chapter Two
Basic Thermal Science
• The science behind heat flow is called heat transfer
• There are three ways by which heat can be transferred: conduction, convection, and radiation
• Heat flow depends on the temperature difference (sensible heat transfer)
98.6oF= 37oC is the body temperature Conduction Heat Transfer
Conduction – The form of heat transfer through a body that occurs without any movement of the body.
It is due to molecular or electron vibration or movement.
Fourier’s Law of Heat Conduction: The rate of heat transfer by conduction (Qcond) is proportional to the temperature difference (∆T) and the heat flow area (A) and is inversely proportional to the distance (∆x) through which conduction occurs.
Where: Qcond = conductive heat transfer rate, W K = thermal conductivity, W/(m.K) T1-T2 = temperature difference, K A = area through which conduction occurs, m
2
= thickness, m
Q: What are the other forms of the equation? Q: How the total thermal resistance of a composite structures is calculated?
24 o C 10
o C 38
o C
Q:What is meant by steady-state? A: Temperatures T1 and T2 do not change in time
Fig 2.13. Thermal conductivity varies with temperature.
However, for most HVAC applications, the range of operation in temperature
is so narrow that conductivity can be assumed constant.
Definitions: Thermal conductivity (k) is the heat transfer rate of a material per meter thickness when the temperature difference (∆T) across is equal to 1
o C. Units (W/m.K)
Resistance (R) is the temperature difference (∆T) needed to transfer 1 watt across the material. Units (K/W). Thermal resistance (Rth) is the temperature difference (∆T) needed to transfer 1 watt across the material “holding the area equal to 1 m
2 . Or, the area (A) needed to transfer 1 watt across the material “holding
the temperature difference equal to 1 m 2 . Units (m
2 .K)/W.
Thermal conductance (U) is the heat transfer rate of a material per meter squared of its area when the temperature difference (∆T) across is equal to 1
o C. Units (W/m
2 .K).
Example (1): Calculate the conductive heat transfer rate through a 200-mm think concrete block wall of area 10 m by 3 m if its k =0.35 W/(m.K) and the outside temperature is 39oC, and the inside air temperature is 24oC.
Example (2): Draw the thermal resistance network (TRN) for the wall structure shown and write its total resistance (Rtot) equation. Example (3): Draw the thermal resistance network (TRN) for the wall structure shown and write its total resistance (Rtot) equation.
ules when drawing TRN’s
1. when you split paths, you should split from the beginning not when the split occurs.
2. when layers have different areas in a structure, you should use R not Rth, because R is independent of area.
R
Example (4): A wall with an area of (14 × 3) m
2 is constructed from the layers listed in the table below and shown
graphically in the drawing. If the outside temperature (To) is 39 o C and the inside temperature (Ti) is 24
o C,
answer the following questions:
Layer Thickness (∆x) (cm)
Thermal Conductivity (k)
(W/(m.K)
Resistance (Rth) (m
2 .K)/W
1 Cement plaster- Sand Aggregate 1 - 0.013
2 Normal-weight concrete blocks filled with perlite
20 - 0.35
3 Air space 5 0.024 =0.05/0.024=2.08
4 Molded beads expanded polystyrene insulation 24 kg/m
3
8 0.037 =0.08/0.037=2.16
5 Cement board 1150 kg/m 3 1.1 0.25 =0.011/0.25=0.044
Total thermal resistance = 4.65
Heat flow (Qcond)
To =39oC Ti=24oC
1. Draw the thermal resistance network (TRN) for the wall.
2. Calculate the total thermal resistance of the wall.
Resistances are in series, then: Rth, total = Rth1 + Rth2 + Rth3 + Rth4 + Rth5 = 4.65 (m
2 .K)/W
3. Calculate the conductive heat transfer rate (Qcond)
Qcond = A×∆T/Rth,total = 42 m 2 × (39 – 24)K /4.65 (m
2 .K)/W = 135 W
4. What is the temperature of the insulation( Tins)?
To =39oC Ti=24oC
Cavities & Air Spaces
In principle, use of cavities is similar to use of a insulating material. If an air space is left between two layers making a wall or roof in any building, the air trapped between two layers being poor conductor of heat acts as a barrier to heat transfer.
Heat is transferred across an air space by a combination of conduction, convection and radiation. Heat transfer by conduction is inversely proportional to depth of the air space. Convection is mainly dependant on the height of the air space and its depth. Heat transfer by radiation is relatively independent of both thickness and height, but is greatly dependent on the reflectivity of the internal surfaces. All three mechanisms are dependent on the surface temperatures. The mathematical treatment of air cavity would be similar to that of insulation if natural convection in air is neglected. The thickness of air cavity is a very important design parameter that governs its effectiveness by controlling the heat transfer coefficient as in case of insulation.
It has been found that with gaps broader than 50 mm, movement of trapped air due to temperature gradient starts that in turn increases the coefficient of heat transfer. This increase in heat transfer takes place due to convective heat transfer taking place in addition to conductive heat transfer. Therefore, cavities broader than 50 mm are normally not preferred. However, if more thickness of air cavity is required for getting heavy insulation, by putting partitions in the main broad cavity multiple cavities can be used as an alternative.
Some typical values of thermal resistance for air cavities are given below:
Air cavity Placement Thickness of air layer (mm) Thermal resistance (m 2 K/W)
Vertical 10- 20 0.14
20- 50 0.17
Horizontal- heat flow from bottom to top 10- 50 0.17
Horizontal- heat flow from top to bottom 10- 50 0.21
If it is possible to ventilate the air gap between the roof and the ceiling, then we could expect a reduction of heat transfer especially by convection. If the ventilation is effective then the air in the void will remain close to the ambient temperature, thus reducing the convective heat transfer to zero. Ventilated air, however, does not reduce the radiative heat transfer from the roof to the ceiling. The radiative component of the heat transfer may be reduced by using low emissivity or high reflective coating (e.g. aluminum foil) on either surface facing the cavity.
In addition to application on walls and roofs, the concept of air cavities also finds very important place in development of insulating windows using double and triple glazing details.
Temperature Profile in a Construction
In a steady-state situation, because there is no heat-storage or heat-production in the construction, the heat flow through every layer in the building construction must be the same. The temperature change across each layer is linear and the rate of change depends on the thermal resistance. When the thermal resistance is small, the temperature difference across the layer is small, but when the thermal resistance is large, the temperature difference across the layer is also large.
The same calculation method is followed as in electrical theory. Ohm’s Law says that the voltage differences over resistances, connected in series, are proportional to the measure of the resistances