a piece of glass is 6mm thick and the coefficient of thermal conductivity of this glass is K=0.75Wm^-1K^-1.
if we add a 2nd identical piece of glass separated by an airgap of 5cm, the total thermal conduction will be reduced.
what is the ratio of heat transfer by conduction for the double pane to the single case.
K of air =0.023W m^-1K^-1
Add the thermal resistances of the two glass slabs and the air gap. The conductance is the reciprocal of the resistance. Are you sure the air gap is 6 cm and not 6 mm? With a gap that large, you are likely to get convection currents in the gap.
Thermal Resistance of one slab of glass:
R1 = W1/K1
Thermal Resistance of two slabs of glass with gap = Wair:
R2 = 2W1/K1+ Wair/Kair
where Wair is the air gap and kair is the termal conductivity of glass.
What you want is the ratio R1/R2
what is W1 and K1?
so is Wair=thermal conductivity of air gap
and Kair=thermal conductivity of glass?
W1 is the width of the glass slabs. They say it is 0.6 cm.
K1 is the thermal conductivity of the glass. Yes to your other questions.
Both W's should be in meters since the units of K contain m^-1, but the answer to your question will be a dimensionless ratio anyway
To find the ratio of heat transfer by conduction for the double pane to the single case, we'll need to calculate the heat transfer for each case separately and then compare them.
Let's start by calculating the heat transfer for the single case, where the glass is 6mm thick.
The formula to calculate heat transfer by conduction is:
Q = (k × A × ΔT) / d
Where:
Q is the heat transfer (in watts),
k is the thermal conductivity (in W m^-1 K^-1),
A is the surface area (in square meters),
ΔT is the temperature difference (in Kelvin),
d is the thickness of the material (in meters).
Given:
k (for single glass) = 0.75 W m^-1 K^-1,
d (for single glass) = 6 mm = 6 × 10^-3 m.
Let's assume ΔT = 1 K and a common surface area of 1 square meter for both cases.
For the single case, the heat transfer can be calculated as:
Q_single = (0.75 × 1 × 1) / 6 × 10^-3
= 0.75 / 6 × 10^-3
= 125 W
Now, let's calculate the heat transfer for the double-pane case, where we have two identical pieces of glass separated by an airgap of 5 cm.
Given:
k (for air) = 0.023 W m^-1 K^-1,
d (for air) = 5 cm = 5 × 10^-2 m.
For the double-pane case, the heat transfer can be calculated as the sum of the heat transfers through each glass and the airgap:
Q_double = (k × A × ΔT) / d_glass1 + (k_air × A × ΔT) / d_airgap + (k × A × ΔT) / d_glass2
Since we have identical glass pieces and assuming the same temperature difference, this simplifies to:
Q_double = 2 × [(k × A × ΔT) / d_glass] + (k_air × A × ΔT) / d_airgap
Substituting the given values:
Q_double = 2 × [(0.75 × 1 × 1) / 6 × 10^-3] + (0.023 × 1 × 1) / 5 × 10^-2
= 250 + 0.046
= 250.046 W
Finally, we can calculate the ratio of heat transfer by conduction for the double pane to the single case:
Ratio = Q_double / Q_single
= 250.046 / 125
= 2.000368
Therefore, the ratio of heat transfer by conduction for the double pane to the single case is approximately 2.000368.