A room at 20^∘ C air temperature is losing heat to the...

A room at $20^{\circ} \mathrm{C}$ air temperature is losing heat to the outdoor air at $0^{\circ} \mathrm{C}$ at a rate of $1000 \mathrm{~W}$ through a $2.5$-m-high and 4-m-long wall. Now the wall is insulated with $2-\mathrm{cm}$-thick insulation with a conductivity of $0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. Determine the rate of heat loss from the room through this wall after insulation. Assume the heat transfer coefficients on the inner and outer surfaces of the wall, the room air temperature, and the outdoor air temperature remain unchanged. Also, disregard radiation. (a) $20 \mathrm{~W}$ (b) $561 \mathrm{~W}$ (c) $388 \mathrm{~W}$ (d) $167 \mathrm{~W}$ (e) $200 \mathrm{~W}$

A room at $20^{\circ} \mathrm{C}$ air temperature is losing heat to the outdoor air at $0^{\circ} \mathrm{C}$ at a rate of $1000 \mathrm{~W}$ through a $2.5$-m-high and 4-m-long wall. Now the wall is insulated with $2-\mathrm{cm}$-thick insulation with a conductivity of $0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. Determine the rate of heat loss from the room through this wall after insulation. Assume the heat transfer coefficients on the inner and outer surfaces of the wall, the room air temperature, and the outdoor air temperature remain unchanged. Also, disregard radiation.
(a) $20 \mathrm{~W}$
(b) $561 \mathrm{~W}$
(c) $388 \mathrm{~W}$
(d) $167 \mathrm{~W}$
(e) $200 \mathrm{~W}$

Key Concepts

Fourier's Law

Fourier's Law describes the conduction of heat, stating that the rate at which heat is transferred through a material is proportional to the temperature difference and the area perpendicular to the heat flow, and inversely proportional to the thickness of the material. It forms the basis for analyzing how heat moves through solids.

Thermal Resistance

Thermal resistance is a measure of a material's opposition to heat flow, analogous to electrical resistance in circuits. It is calculated based on the material's thickness, area, and thermal conductivity, and is used to quantify the resistance offered by each layer in a composite system.

Composite Wall Analysis

Composite wall analysis involves evaluating heat transfer through a wall made up of multiple layers, each with its own thermal resistance. The overall resistance is obtained by summing the individual resistances, and this allows for the calculation of the overall heat transfer rate when a temperature difference is applied across the wall.

Heat Transfer Coefficient

The heat transfer coefficient quantifies the convective heat transfer at the surfaces of a solid. Even when conduction through materials is considered, the additional resistance due to convection from the wall surfaces to the surrounding air must be taken into account to determine the total heat transfer rate.

Insulation Impact

Insulation involves adding materials with low thermal conductivity to increase the thermal resistance of a wall or other structure. This reduces the overall rate of heat loss by increasing the total resistance to heat flow, a key concept in energy conservation and efficiency in building design.

Transcript

00:01 By using the radiation heat transfer coefficient, radiation hit transfer and the stephen bolgman's law, we can write the heat transfer due to radiation, that is q, is equal to p, a, sigma, t2d power of 4 multiplied by time p.

00:23 Or from here, we can write, we can write, emissivity e, emissivity, e is equal to 2 divided by x multiplied by 1 divided by sigma t 2d power 4 so now we can further write e is equal to the value of q can be written as k a t multiplied by t divided by l multiplied by a t whole term multiplied by 1 divided by sigma t 2d power of 4 here k a del t divided by l is the heat transfer due to conductions so from here we get we get e is equal to emissivety is equals to k del t multiplied by 1 divided by l system multiplied by 1 divided by sigma b e2d power of 4 here a a will cancel out each other this will cancel out each other aa will cancel out each other t t t will cancel out each other so let's put the values here we will get amissivity is equal to the value of k we can write the value of k here in this equation is 1 .1 jule per second per meter degree centigrade multiplied by change in temperature.

02:34 Change in temperature is 20 degrees centigrade minus 0 degree centigrade divided where the length is 0 .10 meter multiplied where 1 divided by the value of bulgement constant sigma is 5 .67 multiplied by 10 to the power of minus 8 multiplied by the value of t the temperature here is given 0 degree centigrade that is 0 plus 273 kelvin, 273 kelvin to the power of 4.

03:08 So after solving we get emissivity value e is equals to 220, 220.

03:18 Now directly we will get emissivity is equal to 0 .696, 0 .0 .6.

03:30 So this is the value of emissivity...

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