A House Has$$155 \mathrm {~m} ^ { 2 } \text { of exterior walls consisting of } 2.9 \mathrm {~cm} \text { of exterior }$$

Question

Examlex · Accepted Answer

The Answer is (a) The power loss is given by

P = A \Delta T / R

The total

R

-value is given by the sum of the values for the exterior wall, the insulation and the interior wall as

R = R _ { e x t } + R _ { i n s } + R _ { i n t }

where

R

is defined as

R = l / k

Using SI units the thermal conductivity of wood is

k = 13 ( \mathrm {~W} \cdot \mathrm { cm } ) / \left( \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } ight)

. For fibreglass wool

k = 3.8 ( \mathrm {~W} \cdot \mathrm { cm } ) / \left( \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } ight)

. So for walls with fibreglass wool the total

R

-value will be

R = \frac { 2.9 \mathrm {~cm} } { 13 \frac { \mathrm { W } \cdot \mathrm { cm } } { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } } } + \frac { 15 \mathrm {~cm} } { 3.8 \frac { \mathrm { W } \cdot \mathrm { cm } } { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } } } + \frac { 2.0 \mathrm {~cm} } { 13 \frac { \mathrm { W } \cdot \mathrm { cm } } { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } } } = 4.32 \frac { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } } { \mathrm { W } }

The power loss is given by

P = \left( 155 \mathrm {~m} ^ { 2 } ight) imes \left( 20 ^ { \circ } \mathrm { C } + 2 ^ { \circ } \mathrm { C } ight) / \left[ 4.32 \left( \mathrm {~m} ^ { 2 } { } ^ { \circ } \mathrm { C } ight) / \mathrm { W } ight] = 789 \mathrm {~W}

(b) If polystyrene foam is used with a thermal conductivity

k = 2.84 ( \mathrm {~W} \cdot \mathrm { cm } ) / \left( \mathrm { m } ^ { 2 \cdot } { } ^ { \circ } \mathrm { C } ight)

then the total

R

-value is

R = \frac { 2.9 \mathrm {~cm} } { 13 \frac { \mathrm { W } \cdot \mathrm { cm } } { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } } } + \frac { 15 \mathrm {~cm} } { 2.84 \frac { \mathrm { W } \cdot \mathrm { cm } } { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } } } + \frac { 2.0 \mathrm {~cm} } { 13 \frac { \mathrm { W } \cdot \mathrm { cm } } { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } } } = 5.66 \frac { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } } { \mathrm { W } }

The power loss is

P = \left( 155 \mathrm {~m} ^ { 2 } ight) imes \left( 20 ^ { \circ } \mathrm { C } + 2 ^ { \circ } \mathrm { C } ight) / \left[ 5.66 \left( \mathrm {~m} ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } ight) / \mathrm { W } ight] = 602 \mathrm {~W}

or

100 imes ( 789 \mathrm {~W} - 602 \mathrm {~W} ) / ( 789 \mathrm {~W} ) = 23.7 \% ext { reduction. }

A House Has $155 \mathrm {~m} ^ { 2 } \text { of exterior walls consisting of } 2.9 \mathrm {~cm} \text { of exterior }$