Question #97470

1 Answer
Dec 21, 2016

It is clear that the slab of steel at 70^@C will loose heat via all three modes: Conduction, Convection and Radiation simultaneously and finally will be at ambient temperature given as 10^@C.

"Total heat lost by the steel slab"=mst
where m is mass of slab, s specific heat of steel and t is change in its temperature.

Assuming density rho of steel=7850kgm^-3, s=511Jkg^-1K^-1
m=rhoxx"volume"
m=7850xx(0.01xx1xx3)=235.5kg

:."Total heat lost by the steel slab"=235.5xx511xx(70-10)=7220430J
-.-.-.-.-.-.-.-.-.-.-.-.-.-.

A. Now, rate of heat loss due to conduction dotQ from slab to deck is calculated as below.

Lets assume that the 1.0 cm thick slab is lying flat on the deck.

Deck is at temperature of surroundings=10^@C

The heat loss from top to bottom of slab is found by taking
ΔT = 70^@C – 10^@C = 60^@C = 60 K, the temperature difference between the slab and deck.

Area A through which heat is lost=1xx3=3m^2

Thermal conductivity k("steel") = 50Wm^-1K^-1

We know that basic conduction problem equation is

dot Q=(kAΔT)/"thickness of slab"

Inserting above values we get
dot Q=(50xx3xx60)/0.01
=>dot Q=900000W

B. Similarly basic convection problem equation is
basic equation for convection,

dotQ=hAΔT
where heat transfer coefficient h=10Wm^-2K^-1.

Area A in contact with air=2xx0.01(3+1)+3xx1=3.08m^2

Inserting above values we get
dotQ=10xx3.08xx60=1848W

C. For radiation loss we have Stefan-Boltzmann Law.

dotQ=εAσ(T_1^4– T_2^4)
where sigma is the Stefan Boltzmann constant = 5.670367 xx 10^-8 Wm^-2 K^-4, T_1K is temperature of radiating body and T_2K is temperature of surroundings.

We are given the emissivity for steel, ε= 0.85. The radiating surface area as calculated above, A= 3.08 m^2
Inserting given values we get
dotQ=0.85xx3.08xx5.670367 xx 10^-8(343^4– 283^4)=1102W

D. Total heat loss rate =900000+1848+1102=902950W

Assuming rate of heat loss via each mode is constant throughout the time of loss.
Heat lost via conduction=7220430xx900000/902950approx7196840J