Question

A stainless steel ball (rho = 8055 kg/m3, cp = 480 J/kg·K) of diameter D = 0.21 m is removed from the oven at a uniform temperature of 350°C. The ball is then subjected to the flow of air at 1 atm pressure and 30°C with a velocity of 6 m/s. The surface temperature of the ball eventually drops to 250°C. Determine the average convection heat transfer coefficient during this cooling process and estimate how long this process has taken. The average surface temperature is 300°C, and the properties of air at 1 atm pressure and the free stream temperature of 30°C are Pr = 0.7282, μs at 300°C = 2.934 × 10–5 kg/m·s, μ[infinity] = 1.872 × 10–5 kg/m·s, v = 1.608 × 10–5 m2/s, and k = 0.02588 W/m·°C.

Answer 1

Answer:

Explanation:

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Answer 2

Answer:

Average convection heat transfer coefficient, $\dot{Q_{ave} } = 832.42 W$

time taken for the process, $\triangle t = 37.9 min$

Explanation:

The average convection heat transfer rate is calculated using the formula:

$\dot{Q_{ave} } = h A_{s} (T_s - T_{\infty} )$

The surface area of the steel ball is given by :

$A_{s} = \pi D^{2} \\A_{s} = \pi * 0.21^{2} \\A_{s} = 0.139 m^2$

Free stream temperature, $T_{\infty} = 30^{0} C$

Initial temperature of the ball, T₁ = 350°C

Final temperature of the ball, T₂ = 250°C

Average surface temperature of the ball:

$T_s = \frac{T_1 + T_2}{2} \\T_s = \frac{350 + 250}{2}\\T_s = 300^{0} C$

Velocity of air, V = 6 m/s

Diameter of the ball, D = 0.21 m

Viscosity, v = 1.608 * 10⁻⁵ m²/s

Reynold number Re can be calculated by using the formula: $Re = \frac{VD}{v}$

$Re = \frac{6 * 0.21}{1.608 * 10^{-5} }$

Re = 78358.21

The Nusselt number can be calculated by using the equation:

$Nu = 2 + (0.4Re^{0.5} + 0.06Re^{0.67} ) (Pr^{0.4}) (\frac{\mu_{\infty}}{\mu_s})^{0.25} \\Nu = 2 + (0.4*78358.21^{0.5} + 0.06*78358.21^{0.67} ) (0.7282^{0.4}) (\frac{1.872*10^{-5}}{2.934*10^{-5}})^{0.25}$

Nu = 179.95

The heat transfer coefficient can be calculated using the formula:

$h = \frac{k* Nu}{D} \\h = \frac{0.02588* 179.95}{0.21}\\h = 22.18 W/m^2 k$

$\dot{Q_{ave} } = h A_{s} (T_s - T_{\infty} )$

$\dot{Q_{ave} } = 22.18 * 0.139 (300 -30 )\\\dot{Q_{ave} } = 832.42 W$

The time taken for the process, $\triangle t = \frac{Q_{total} }{\dot{Q_{ave} }}$

$Q_{total} = mc_{p} (T_1 - T_2)$

Volume of the steel ball, $V = \frac{\pi * D^3 }{6}$

$V = \frac{\pi * 0.21^3 }{6}$

V = 0.0049 m³

Density of steel, $\rho = 8055 kg/m^{3}$

Mass of the steel, $m = \rho V$

m = 8055*0.0049

m = 39.47 kg

Total rate of heat transfer: $Q_{total} = mc_{p} (T_1 - T_2)$

Specific heat capacity of steel ball, $c_p$ = 480 J/kg

$Q_{total} = 39.47 *480 (350 - 250)$

$Q_{total} = 1894560 J$

$\triangle t = \frac{Q_{total} }{\dot{Q_{ave} }}$

$\triangle t = 1894560/832.42\\\triangle t = 2275.97s\\\triangle t = 2275.97/60\\\triangle t = 37.9 min$