A closed hot water tank is of cubical shape with all of its edges 1 metre long.

Question

A closed hot water tank is of cubical shape with all of its edges 1 metre long. It is made of sheet copper of thickness 1 mm and is not insulated. It stands on a concrete floor, so that you can assume that no heat escapes through the bottom of the tank, though heat can escape through the top and sides. It is filled to capacity with hot water. The thermal conductivity of sheet copper is 400 Wm-1 K-1. The convective heat transfer coefficient from hot water to copper is 100 Wm-2 K-1. while that from copper to air is 10 Wm-2 K-1. The specific heat capacity of water is 4190 J kg-1; K-1. Show that the steady-state rate of loss of heat from the water in the tank when the temperature of die water is theta degree C and the temperature of the air surrounding the tank is theta degree n C,. is 45.5 (theta - theta a) watts, where the coefficient is correct to three significant figures. (You may assume that theta > theta a.)

Explanation / Answer

let the temperature at the inner surface of copper be (T_1), that of outer surface be (T_2). Now,total area of copper inner surface neglecting the thickness of copper is (A=5*1^2=5m^2)

Thus, the heat transferred per unit time from water to copper is:

(Q_1=A*100*(T_1- heta)=500(T_1- heta))

The heat conducted from the inner surface of copper to outer surface of copper perunit time is:

(Q_2=A*400*(T_2-T_1)/d)

where d is thickness of copper sheet or d=1mm=0.001 m, thus, (Q_2 = 2000000(T_2-T_1))

the heat taken away by air per unit time is,

(Q_3 = A*10*( heta_a-T_2))

under steady state, (Q_1=Q_2=Q_3)

since the heat conducted away from water should be given off to air.

Thus, using (Q_1=Q_2)

(500(T_1- heta)=2000000(T_2-T_1))

thus, (-4000T_2+4001T_1= heta)

this is equation 1

From, (Q2=Q3),

(2000000(T_2-T_1)=50( heta_a-T_2))

thus, (-39999T_2+40000T_1=- heta_a)

Now, add equation 1 to it, so we get

(-43999T_2+43999T_1= heta- heta_a)

Thus, (( heta- heta_a)/43999=(T_1-T_2))

thus, (Q=Q_2=-2000000*( heta- heta_a)/43999=-45.5( heta- heta_a)Watts)

thus heat loss is (-Q=45.5( heta- heta_a)Watts)

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