Question 11.9: Construction and Thermal Insulation GOAL Calculate the R-val...

(a) Find the energy transferred in 1.00 h by conduction through a concrete wall.
Calculate the R-value of concrete plus two air layers:

\sum R=\frac{L}{k}+2 R_{\text {air layer }}=\frac{0.20  m }{0.80  J / s  \cdot  m  \cdot { }^{\circ} C }+2\left(0.030 \frac{ m ^2}{ J / s  \cdot { }^{\circ} C }\right)

=0.31 \frac{ m ^2}{ J / s  \cdot { }^{\circ} C }

Write the thermal conduction equation:

P=\frac{A\left(T_h  –  T_c\right)}{\sum R}

P=\frac{\left(7.3  m ^2\right)\left(20.0^{\circ} C  –  5.00^{\circ} C \right)}{0.31  m ^2  \cdot  s  \cdot { }^{\circ} C / J }=353  W \rightarrow 350  W

Multiply the power in watts times the seconds in an hour:

Q=P \Delta t=(350  W )(3600  s )=1.3 \times 10^6  J

(b) Calculate the R-factor of the newly insulated wall.
Refer to Table 11.4 and sum the appropriate quantities after converting them to SI units:

R_{\text {total }}=R_{\text {outside air layer }}+R_{\text {concrete }}+R_{\text {sheath }}

+R_{\text {fiberglass }}+R_{\text {drywall }}+R_{\text {inside air layer }}

= (0.030 + 0.25 + 0.233 + 1.92 + 0.079 + 0.030)

= 2.5 m² · °C · s/J

(c) Calculate the energy transferred in 1.00 h by conduction.
Write the thermal conduction equation:

P=\frac{A\left(T_h-T_c\right)}{\sum R}

Substitute values:

P=\frac{\left(7.3  m ^2\right)\left(20.0^{\circ} C  –  5.00^{\circ} C \right)}{2.5  m ^2  \cdot  s \cdot { }^{\circ} C / J }=44  W

Multiply the power in watts times the seconds in an hour:

Q=P \Delta t=(44  W )(3600  s )=1.6 \times 10^5  J

(d) Calculate the temperature between the concrete and the sheathing.
Write the thermal conduction equation:

P=\frac{A\left(T_h  –  T_c\right)}{\sum R}

Solve algebraically for T_h by multiplying both sides by ΣR and dividing both sides by area A:

P \sum R=A\left(T_h-T_c\right) \rightarrow\left(T_h-T_c\right)=\frac{P \sum R}{A}

Add T_c to both sides:

T_h=\frac{P \sum R}{A}+T_c

Substitute the R-value for the concrete wall from part (a), but subtract the R-value of one air layer from that calculated in part (a):

T_h=\frac{(44  W )\left(0.31  m ^2  \cdot  s  \cdot { }^{\circ} C / J  –  0.03  m ^2  \cdot  s  \cdot { }^{\circ} C / J \right)}{7.3  m ^2}+5.00^{\circ} C

= 6.7°C

REMARKS Notice the enormous energy savings that can be realized with good insulation!