Given:

The thickness of each radiator is 0.5 mm.

The temperature of engine fluid is 400 K.

The ambient air temperature is 300 K.

Formula used:

The expression heat flux of radiator is given by,

q=−kdTdx …… (I)

Here, q is the amount of heat flux, dTdx is the temperature gradient and k is the thermal conductivity.

The temperature gradient is given by,

dTdx=Ta−Tfdx …… (II)

Here, Ta is the ambient air temperature and Tf is the temperature of fluid and dx is the wall thickness.

Substitute (Ta−Tfdx) for dTdx in equation (I).

q=−k(Ta−Tfdx) …… (III)

Calculation:

The heat flux of aluminum radiator is calculated as,

Substitute 247 W/m⋅K for k, 300 K for Ta, 400 K for Tf and 0.5 mm for dx in equation (III).

qAl=−(247 W/m⋅K)( 300 K−400 K 0.5 mm× 10 −3  m 1 mm )=−(247 W/m⋅K)(−200,000 m⋅K)=49.4×106 W/m2

Here, qAl is the heat flux for aluminum.

The heat flux of brass radiator is calculated as,

Substitute 120W/m⋅K for k, 300 K for Ta, 400 K for Tf and 0.5 mm for dx in equation (III).

qBr=−(120 W/m⋅K)( 300 K−400 K 0.5 mm× 10 −3  m 1 mm )=−(120 W/m⋅K)(−200,000 m⋅K)=24×106 W/m2

Here, qBr is the heat flux for Brass.

Conclusion:

Therefore, the heat flux for the brass radiator is 24×106 W/m2 and for the aluminum radiator the heat flux is 49.4×106 W/m2.

Given:

The heat transfer rate is the same for brass and aluminum.

The brass is 30 weight percent zinc and 70 weight percent copper.

The density of brass is 8.4 g/cm3.

Formula used:

The relation for rate of heat transfer in aluminum is given by,

( dQ dt)Al=AAlqAl …… (IV)

Here, ( dQ dt)Al is the rate of heat transfer in aluminum and AAl is the area of aluminum radiator.

The relation for rate of heat transfer in brass is given by,

( dQ dt)Br=ABrqBr …… (V)

Here, ( dQ dt)Br is the rate of heat transfer in brass and ABr is the area of brass radiator.

The heat transfer rate is same for brass and aluminum.

Equate equation (IV) and (V) to obtain expression for AAl.

AAlqAl=ABrqBRAAl=A Brq BRq Al …… (VI)

The expression for weight is given by,

W=D×V …… (VII)

Here, W is the weight, D is the density and V is the volume.

Calculation:

The area of aluminum is calculated as,

Substitute 24×106 W/m2 for qBr and 49.4×106 W/m2 for qAl in equation (VI).

AAl=A Br( 24× 10 6  W/ m 2 )49.4× 106 W/ m 2=0.486ABr×100%=48.6% ABr≈49%ABr

The wall of thickness of both the radiators is same.

VAl=49% VBr

Here, VAl is the volume of aluminum and VBr is the volume of Brass.

The weight of brass is calculated as,

Substitute 8.4 g/cm3 for D and VBr for V in equation (VII).

wBr=8.4 g/cm3(VBr) …… (VIII)

Here, wBr is the weight of brass and VBr is the volume of

The weight of aluminum is calculated as,

Substitute 2.71 g/cm3 for D and VAl for V in equation (VII).

wAl=2.71 g/cm3(VAl) ……. (IX)

Here, wAl is the weight of aluminum.

Substitute 0.49VBr for VA in equation (IX).

wAl=2.71 g/cm3(0.49V Br)=(1.33 g/ cm 3)VBr  …… (X)

Divide equation (X) by (VIII).

w Brw Al=8.4 g/ cm 3( V Br ) ( 1.33 g/ cm 3 )V BrwBr=6.31wAl

Conclusion:

Therefore, the relative area of aluminum is 49% of the area of brass and the weight of brass is 6.31 times the weight of aluminum.