Consider a house whose walls are 12 ft high and 40 ft long. Two of the walls of the house have no windows, while each of the other two walls has four windows made of 0.25-in.- thick glass (k = 0.45 Btu/h · ft · °F), 3 ft × 5 ft in size. The walls are certified to have an R-value of 19 (i.e., an L/k

Question

Consider a house whose walls are 12 ft high and 40 ft long. Two of the walls of the house have no windows, while each of the other two walls has four windows made of 0.25-in.- thick glass [latex] \left(k = 0.45 Btu / h \cdot ft \cdot{ }^{\circ} F ight), 3 ft imes 5 ft ext { in }[/latex] size. The walls are certified to have an R-value of 19 (i.e., an L/k value of [latex]19 h \cdot ft ^{2} \cdot{ }^{\circ} F / Btu[/latex]). Disregarding any direct radiation gain or loss through the windows and taking the heat transfer coefficients at the inner and outer surfaces of the house to be 2 and [latex]4 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F[/latex], respectively, determine the ratio of the heat transfer through the walls with and without windows.

Accepted Answer

Two of the walls of a house have no windows while the other two walls have 4 windows each. The ratio of heat transfer through the walls with and without windows is to be determined.

Assumptions 1 Heat transfer through the walls and the windows is steady and one-dimensional. 2 Thermal conductivities are constant. 3 Any direct radiation gain or loss through the windows is negligible. 4 Heat transfer coefficients are constant and uniform over the entire surface.

Properties The thermal conductivity of the glass is given to be [latex] k_{ ext {glass }}=0.45 Btu / h \cdot ft \cdot{ }^{\circ} F[/latex]. The R-value of the wall is given to be [latex]19 h \cdot ft ^{2} \cdot{ }^{\circ} F / Btu[/latex].

Analysis The thermal resistances through the wall without windows are

[latex] A=(12 ft )(40 ft )=480 m ^{2}[/latex]

[latex] R_{i}=\frac{1}{h_{i} A}=\frac{1}{\left(2 \operatorname{Btu} / h \cdot ft ^{2} \cdot{ }^{\circ} F ight)\left(480 ft ^{2} ight)}=0.0010417 h \cdot{ }^{\circ} F / Btu[/latex]

[latex] R_{ ext {wall }}=\frac{L}{k A}=\frac{19 h \cdot ft ^{2} \cdot ^{\circ} F / Btu }{\left(480 m ^{2} ight)}=0.03958 h \cdot{ }^{\circ} F / Btu[/latex]

[latex] R_{o}=\frac{1}{h_{o} A}=\frac{1}{\left(4 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F ight)\left(480 ft ^{2} ight)}=0.00052 h \cdot{ }^{\circ} F / Btu [/latex]

[latex] R_{ ext {total }, 1}=R_{i}+R_{ ext {wall }}+R_{o} [/latex]

[latex] = 0.0010417 + 0.03958 + 0.00052 = 0.0411417 h \cdot{ }^{\circ} F / Btu [/latex]

The thermal resistances through the wall with windows are

[latex] A_{ ext {windows }}=4(3 imes 5)=60 ft ^{2}[/latex]

[latex] A_{ ext {wall }}=A_{ ext {total }} - A_{ ext {windows }} = 480 - 60 = 420 ft ^{2}[/latex]

[latex] R_{2}=R_{ ext {glass }}=\frac{L}{k A}=\frac{0.25 / 12 ft }{\left(0.45 Btu / h \cdot ft \cdot{ }^{\circ} F ight)\left(60 ft ^{2} ight)} = 0.0007716 h \cdot{ }^{\circ} F / Btu[/latex]

[latex] R_{4}=R_{w a l l}=\frac{L}{k A}=\frac{19 h \cdot ft ^{2} \cdot{ }^{\circ} F / Btu }{\left(420 ft ^{2} ight)}=0.04524 h \cdot{ }^{\circ} F / Btu [/latex]

[latex]\frac{1}{R_{e q v}}=\frac{1}{R_{ ext {glass }}}+\frac{1}{R_{ ext {wall }}}=\frac{1}{0.0007716}+\frac{1}{0.04524} \longrightarrow R_{e q v}=0.00076 h \cdot ^{\circ} F / Btu [/latex]

[latex] R_{ ext {total }, 2}=R_{i}+R_{e q v}+R_{o}=0.001047+0.00076+0.00052=0.002327 h \cdot{ }^{\circ} F / Btu [/latex]

Then the ratio of the heat transfer through the walls with and without windows becomes

[latex] \frac{\dot{Q}_{ ext {total }, 2}}{\dot{Q}_{ ext {total }, 1}}=\frac{\Delta T / R_{ ext {total }, 2}}{\Delta T / R_{ ext {total }, 1}}=\frac{R_{ ext {total }, 1}}{R_{ ext {total }, 2}}=\frac{0.0411417}{0.002327}=17.7[/latex]

Question 3.33.E: Consider a house whose walls are 12 ft high and 40 ft long. ...

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