Explanation Given information: The diameter of the tank is 6 m , the density of the oxygen is 1141 kg/m 3 , the specific heat of the oxygen is 1.71 kJ/kg ⋅ ° C , the temperature of the oxygen is − 184 ° C , the increased temperature of the oxygen is − 183 ° C and the time period is 144 h . Write the expression for the radius of the spherical tank. r = d 2 ....... (I) Here, the radius is r and the diameter of the tank is d . Write the expression for the volume of the spherical tank. V = 4 3 π r 3 ....... (II) Here, the volume is V and the radius of the spherical tank is r . Write the expression for the mass of the oxygen inside the spherical tank. m = ρ V ....... (III) Here, the mass is m , the density of the oxygen is ρ and the volume of the spherical tank is V . Write the expression for the heat loss. Q = m c p ( T o − T c ) ....... (IV) Here, the total heat lost is Q , the specific heat is c p , the temperature of the oxygen is T o and the changed temperature of the oxygen is T c . Write the expression for the rate of heat transfer is Q ′ = Q δ t ....... (V) Here, the rate of heat transfer is Q ′ , the total heat lost is Q and the time period is δ t . Calculation: Substitute 6 m for d , in Equation (I). r = 6 m 2 = 3 m Substitute 3 m for r in Equation (II). V = 4 3 π ( 3 m ) 3 = 113.097 m 3 Substitute 113.097 m 3 for V and 1141 kg/m 3 for ρ in Equation (III)

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ISBN: 9781307573060

Author: CENGEL

Publisher: MCG/CREATE

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Chapter 3, Problem 203P

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The average rate of heat transfer to the tank.

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Chapter 3 Solutions

HEAT+MASS TRANSFER-SI ED. EBOOK >I<

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