Question

Consider a 1.5-m-high and 2.4-m-wide glass window whose thickness is 6 mm and thermal conductivity is k = 0.78 W/m⋅K. Determine the steady rate of heat transfer through this glass window and the temperature of its inner surface for a day during which the room is maintained at 24°C while the temperature of the outdoors is −5°C. Take the convection heat transfer coefficients on the inner and outer surfaces of the window to be h1 = 10 W/m2⋅K and h2 = 25 W/m2⋅K, respectively, and disregard any heat transfer by radiation.

Answer 1

The steady rate of heat transfer through the glass window is 707.317 watts.

Step-by-step explanation:

A figure describing the problem is included below as attachment. From First Law of Thermodynamics we get that steady rate of heat transfer through the glass window is the sum of thermal conductive and convective heat rates, all measured in watts:

`\dot Q_(total) = \dot Q_(cond) + \dot Q_(conv, in) + \dot Q_(conv, out)` (Eq. 1)

Given that window is represented as a flat element, we can expand (Eq. 1) as follows:

`\dot Q_(total) = (T_(i)-T_(o))/(R)` (Eq. 2)

Where:

`T_(i)`,
`T_(o)` - Indoor and outdoor temperatures, measured in Celsius.

`R` - Overall thermal resistance, measured in Celsius per watt.

Now, we know that glass window is configurated in series and overall thermal resistance is:

`R = R_(cond) + R_(conv, in)+R_(conv, out)` (Eq. 3)

Where:

`R_(cond)` - Conductive thermal resistance, measured in Celsius per watt.

`R_(conv, in)`,
`R_(conv, out)` - Indoor and outdoor convective thermal resistances, measured in Celsius per watt.

And we expand the expression as follows:

`R = (l)/(k\cdot w\cdot d) + (1)/(h_(i)\cdot w\cdot d) + (1)/(h_(i)\cdot w\cdot d)`

`R = (1)/(w\cdot d)\cdot \left((l)/(k)+(1)/(h_(i))+(1)/(h_(o)) \right)` (Eq. 4)

Where:

`w` - Width of the glass window, measured in meters.

`d` - Length of the glass window, measured in meters.

`l` - Thickness of the glass window, measured in meters.

`k` - Thermal conductivity, measured in watts per meter-Celsius.

`h_(i)`,
`h_(o)` - Indoor and outdoor convection coefficients, measured in watts per square meter-Celsius.

If we know that
`w = 2.4\,m`,
`d = 1.5\,m`,
`l = 0.006\,m`,
`k = 0.78\,(W)/(m\cdot ^(\circ)C)`,
`h_(i) = 10\,(W)/(m^(2)\cdot ^(\circ)C)` and
`h_(o) = 25\,(W)/(m^(2)\cdot ^(\circ)C)`, the overall thermal resistance is:

`R = \left[(1)/((2.4\,m)\cdot (1.5\,m))\right] \cdot \left((0.006\,m)/(0.78\,(W)/(m\cdot ^(\circ)C) )+(1)/(10\,(W)/(m^(2)\cdot ^(\circ)C) )+(1)/(25\,(W)/(m^(2)\cdot ^(\circ)C) ) \right)`

`R = 0.041\,(^(\circ)C)/(W)`

Now, we obtain the steady rate of heat transfer from (Eq. 2): (
`R = 0.041\,(^(\circ)C)/(W)`,
`T_(i) = -5\,^(\circ)C`,
`T_(o) = 24\,^(\circ)C`)

`\dot Q_(total) = (24\,^(\circ)C-(-5\,^(\circ)C))/(0.041\,(^(\circ)C)/(W) )`

`\dot Q_(total) = 707.317\,W`

The steady rate of heat transfer through the glass window is 707.317 watts.