A 10-in.-thick, 30-ft-long, and 10-ft-high wall is to be constructed using 9-in.-long solid bricks ( k = 0.40 Btu / h \cdot ft \cdot{ }^{\circ} F) of cross section 7 \text { in. } \times 7 \text { in. }, or identical size bricks with nine square air holes (k = 0.015 Btu / h \cdot ft \cdot{ }^{\circ} F) that are 9 in. long and have a cross section of 1.5 in. \times 1.5 in. There is a 0.5-in.-thick plaster layer (k = 0.10 Btu / h \cdot ft \cdot{ }^{\circ} F) between two adjacent bricks on all four sides and on both sides of the wall. The house is maintained at 80°F and the ambient temperature outside is 30°F. Taking the heat transfer coefficients at the inner and outer surfaces of the wall to be 1.5 and 4 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F, respectively, determine the rate of heat transfer through the wall constructed of (a) solid bricks and (b) bricks with air holes.
Question 3.56.E: A 10-in.-thick, 30-ft-long, and 10-ft-high wall is to be con...
A wall is to be constructed using solid bricks or identical size bricks with 9 square air holes. There is a 0.5 in thick sheetrock layer between two adjacent bricks on all four sides, and on both sides of the wall. The rates of heat transfer through the wall constructed of solid bricks and of bricks with air holes are to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of change with time. 2 Heat transfer through the wall is one-dimensional. 3 Thermal conductivities are constant. 4 Heat transfer coefficients account for the radiation heat transfer.
Properties The thermal conductivities are given to be k = 0.40 Btu / h \cdot ft \cdot{ }^{\circ} F for bricks, k = 0.015 Btu / h \cdot ft \cdot{ }^{\circ} F for air, and k = 0.10 Btu / h \cdot ft \cdot{ }^{\circ} F for sheetrock.
Analysis (a) The representative surface area is A=(7.5 / 12)(7.5 / 12)=0.3906 ft ^{2}. The thermal resistance network and the individual thermal resistances if the wall is constructed of solid bricks are
R_{i}=\frac{1}{h_{i} A}=\frac{1}{\left(1.5 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F \right)\left(0.3906 ft ^{2}\right)}=1.7068 h ^{\circ} F / BtuR_{1}=R_{5}=R_{\text {plaster }}=\frac{L}{k A}=\frac{0.5 / 12 ft }{\left(0.10 Btu / h \cdot ft \cdot ^{\circ} F \right)\left(0.3906 ft ^{2}\right)}=1.0667 h ^{\circ} F / Btu
R_{2}=R_{\text {plaster }}=\frac{L}{k A}=\frac{9 / 12 ft }{\left(0.10 Btu / h \cdot ft \cdot { }^{\circ} F \right)[(7.5 / 12) \times(0.5 / 12)] ft ^{2}}=288 h ^{\circ} F / Btu
R_{3}=R_{\text {plaster }}=\frac{L}{k A}=\frac{9 / 12 ft }{\left(0.10 Btu / h \cdot ft \cdot ^{\circ} F \right)[(7 / 12) \times(0.5 / 12)] ft ^{2}}=308.57 h ^{\circ} F / Btu
R_{4}=R_{\text {brick }}=\frac{L}{k A}=\frac{9 / 12 ft }{\left(0.40 Btu / h \cdot ft \cdot{ }^{\circ} F \right)[(7 / 12) \times(7 / 12)] ft ^{2}}=5.51 h ^{\circ} F / Btu
R_{o}=\frac{1}{h_{o} A}=\frac{1}{\left(4 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F \right)\left(0.3906 ft ^{2}\right)}=0.64 h ^{\circ} F / Btu
\frac{1}{R_{\text {mid }}}=\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}=\frac{1}{288}+\frac{1}{308.57}+\frac{1}{5.51} \longrightarrow R_{\text {mid }}=5.3135 h ^{\circ} F / Btu
R_{\text {total }}=R_{i}+R_{1}+R_{\text {mid }}+R_{5}+R_{o}
= 1.7068 + 1.0667 + 5.3135 + 1.0667 + 0.64 = 9.7937 h ^{\circ} F / Btu
\dot{Q}=\frac{T_{\infty 1}-T_{\infty 2}}{R_{\text {total }}}=\frac{(80-30)^{\circ} F }{9.7937 h ^{\circ} F / Btu }=5.1053 Btu / h
Then steady rate of heat transfer through entire wall becomes
\dot{Q}_{\text {total }}=(5.1053 Btu / h ) \frac{(30 ft )(10 ft )}{0.3906 m ^{2}}=3921 Btu / h
(b) The thermal resistance network and the individual thermal resistances if the wall is constructed of bricks with air holes are
A_{\text {airholes }}=9(1.25 / 12) \times(1.25 / 12)=0.0977 ft ^{2}A_{\text {bricks }}=(7 / 12 ft )^{2}-0.0977=0.2426 ft ^{2}
R_{4}=R_{\text {airholes }}=\frac{L}{k A}=\frac{9 / 12 ft }{\left(0.015 Btu / h \cdot ft \cdot{ }^{\circ} F \right)\left(0.0977 ft ^{2}\right)}=511.77 h ^{\circ} F / Btu
R_{5}=R_{\text {brick }}=\frac{L}{k A}=\frac{9 / 12 ft }{\left(0.40 Btu / h \cdot ft \cdot { }^{\circ} F \right)\left(0.2426 ft ^{2}\right)}=7.729 h ^{\circ} F / Btu
\frac{1}{R_{m i d}}=\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}=\frac{1}{288}+\frac{1}{308.57}+\frac{1}{511.77}+\frac{1}{7.729} \longrightarrow R_{m i d}=7.244 h ^{\circ} F / Btu
R_{\text {total }}=R_{i}+R_{1}+R_{\text {mid }}+R_{6}+R_{o}
=1.7068+1.0667+7.244+1.0677+0.64=11.7252 h ^{\circ} F / Btu
\dot{Q}=\frac{T_{\infty 1}-T_{\infty 2}}{R_{\text {total }}}=\frac{(80-30)^{\circ} F }{11.7252 h ^{\circ} F / Btu }=4.2643 Btu / h
Then steady rate of heat transfer through entire wall becomes
\dot{Q}_{\text {total }}=(4.2643 Btu / h ) \frac{(30 ft )(10 ft )}{0.3906 ft ^{2}}= 3 2 7 5 \text { Btu/h }
