(A) Calculate the Power Transferred Through a 10 Cm Diameter$1 \mathrm {~m}$

Question

Examlex · Accepted Answer

The Answer is (a) The power transferred is

P = k A \Delta T / l

The cross sectional area is

A = ( 3.14 ) imes ( 0.05 \mathrm {~m} ) _ { 2 } = 0.00785 \mathrm {~m} ^ { 2 }

For iron the thermal conductivity is

\mathrm { k } = 4600 ( \mathrm {~W} \cdot \mathrm { cm } ) / \left( \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } ight)

. So

P = \left( 4600 ( \mathrm {~W} \cdot \mathrm { cm } ) / \left( \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } ight) ight) imes \left( 0.00785 \mathrm {~m} ^ { 2 } ight) imes \left( 20 ^ { \circ } \mathrm { C } ight) / ( 100 \mathrm {~cm} ) = 7.2 \mathrm {~W}

(b) Solving the above for area gives

A = P l / ( k \Delta T )

and the diameter as

d = [ 4 P l / ( \pi k \Delta T ) ] ^ { 1 / 2 }

Using

P = 4 \mathrm {~W}

, the diameter is found to be

d = \left[ 4 imes ( 4 \mathrm {~W} ) imes ( 100 \mathrm {~cm} ) / \left( ( 3.14 ) imes \left( 4600 ( \mathrm {~W} \cdot \mathrm { cm } ) / \left( \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } ight) ight) imes \left( 20 ^ { \circ } \mathrm { C } ight) ight] ^ { 1 / 2 } = 0.074 \mathrm {~m} = 7.4 \mathrm {~cm} ight.

(A) Calculate the Power Transferred Through a 10 Cm Diameter $1 \mathrm {~m}$