A 2-m-internal-diameter spherical tank made of 0.5-cm-thick stainless steel (k =

a. 6.48 kW b. 1678.34 kg c. 777.92 W d. 201.42 kg

Explanation:

(a) the rate of heat transfer to the iced water in the tank

The rate of heat transfer to the outer surface of the spherical tank is P = P₁ + P₂ where P₁ = rate of heat transfer to the outer surface by radiation and P₂ = rate of heat transfer to the outer surface by convection through air

P₁ = εσAT⁴ where ε = emissivity of outer surface ε= 1, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m-K⁴, A = area of outer surface of spherical tank = 4πR² where R = outer radius of spherical tank = inner radius + thickness = inner diameter/2 + 5 cm = 2 m/2 + 0.05 m = 1 m + 0.05 m = 1.05 m and T = temperature of surroundings = 20 °C = 273 + 20 = 293 K.

P₁ = εσAT⁴

P₁ = 1 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.05 m)² × (293 K)⁴

P₁ = 1 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.1025 m²) × 7370050801 K⁴

P₁ = 184285909263.7647π × 10⁻⁸ W

P₁ = 578951258703.16 × 10⁻⁸ W

P₁ = 5789.51 W

P₂ = hA(T - T₁) where h = coefficient of thermal convection of air = 2.5 W/m²-K, A = outer surface area of spherical tank = 4πR², T = temperature of surroundings = 20 °C = 273 + 20 = 293 K and T₁ = temperature of outer surface of spherical tank = 0 °C = 273 + 0 = 273 K.  

P₂ = hA(T - T₁)

P₂ = 2.5 W/m²-K × 4π(1.05 m)² × (293 K - 273 K)

P₂ = 2.5 W/m²-K × 4π(1.1025 m²) × 20 K

P₂ = 220.5π W

P₂ = 692.72 W

So, P = P₁ + P₂ = 5789.51 W + 692.72 W = 6482.23 W

Since we are neglecting the thermal resistance of the spherical tank, the rate of heat absorption of the outer surface equals the rate of heat absorption in the inner surface. The rate of heat absorption at the inner surface equals the rate of heat transfer to the iced water.

So, rate of heat transfer to the iced water = P = 6482.23 W = 6.48223 kW 6.48 kW

(b) the amount of ice at 0°C that melts during a 24-h period. The heat of fusion of water is 333.7 kJ/kg.

Since the amount of heat, Q = Pt where P = heat transfer rate to iced water = 6482.23 W and t = time = 24 h = 24 h × 60 min/h × 60 s/min = 86400 s.

Also, Q = the latent heat required to melt the ice at 0 °C = mL where m = mass of ice melted and L = latent heat of fusion of ice = 333.7 kJ/kg

So, Pt = mL

m = Pt/L

= 6482.23 W × 86400 s/333.7 × 10³ J/kg

= 560064672/333.7 × 10³

= 1678.34 kg

(c) the rate of heat transfer to the iced water in the tank for this double-walled tank configuration

Since P is the rate of heat transfer to the outer surface, this is also the rate of heat transfer to the outer 0.5 cm thick wall = P₃ = 6482.23 W

P₃ = kA(T - T₃)/d where k = thermal conductivity of outer wall = 15 W/m²-K

A = surface area of outer wall = 4πR'² where R' = radius of outer wall = radius of inner wall + thickness of inner wall + thickness of vacuum + thickness of outer wall = 2.0 m/2 + 0.5 cm + 1.5 cm + 0.5 cm = 1 m + 2.5 cm = 1 m + 0.025 m = 1.025 m, T = temperature of surroundings = 20 °C = 273 + 20 = 293 K, T₃ = temperature of inner surface of outer wall of spherical tank and d = thickness of outer surface of tank = 0.5 cm = 0.05 m

P₃ = kA(T - T₃)/d

making T₃ subject of the formula, we have

P₃d = kA(T - T₃)

P₃d/kA = (T - T₃)

T₃ = T - P₃d/kA

substituting the values of the variables into the equation, we have

T₃ = 293 K - 6482.23 W × 0.05 m/[15 W/m-K × 4π(1.025 m)²]

T₃ = 293 K - 324.1115 Wm/[15 W/m-K × 4π(1.050625 m²)]

T₃ = 293 K - 324.1115 Wm/[63.0375π W/m-K)]

T₃ = 293 K - 324.1115 Wm/[198.0381 W/m-K)]

T₃ = 293 K - 1.64 K

T₃ = 291.36 K

Since the 1.5 cm thick air space is evacuated, all the heat gets to the inner 0.5 cm thick wall by radiation.

So P = εσAT₃⁴

P₄ = εσAT₃⁴ where ε = emissivity of outer surface ε = 0.15, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m-K⁴, A = area of inner surface of outer wall of spherical tank = 4πR"² where R" = outer radius of inner thick wall of spherical tank = inner radius + thickness of inner wall = inner diameter/2 + 0.5 cm = 2 m/2 + 0.005 m = 1 m + 0.005 m = 1.005 m and T = temperature of outer wall = 291.36 K.

P₄ = 0.15 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.005 m)² × (291.36 K)⁴

P₄ = 0.15 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.010025 m²) × 7206422389.51 K⁴

P₄ = 24762024365.028π × 10⁻⁸ W

P₄ = 77792193833.18 × 10⁻⁸ W

P₄ = 777.92 W

Now P₄ is the heat transfer rate to the inner surface which is at temperature T₄

Since T₄ = 0 °C, P₄ is the rate of heat transfer to the iced water

So, rate of heat transfer to the iced water P₄ = 777.92 W

(d) the amount of ice at 0°C that melts during a 24-h period for this double-walled tank configuration

Since the amount of heat, Q = P₄t where P₄ = heat transfer rate to iced water = 777.92 W and t = time = 24 h = 24 h × 60 min/h × 60 s/min = 86400 s.

Also, Q = the latent heat required to melt the ice at 0 °C = mL where m = mass of ice melted and L = latent heat of fusion of ice = 333.7 kJ/kg

So, P₄t = mL

m = P₄t/L

= 777.92 W × 86400 s/333.7 × 10³ J/kg

= 67212288/333.7 × 10³

= 201.42 kg

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Explanation: