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

Average convection heat transfer coefficient,
`\dot{Q_(ave) } = 832.42 W`

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

Step-by-step 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 = (T_1 + T_2)/(2) \\T_s = (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 = (VD)/(v)`

`Re = (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)) ((\mu_(\infty))/(\mu_s))^(0.25) \\Nu = 2 + (0.4*78358.21^(0.5) + 0.06*78358.21^(0.67) ) (0.7282^(0.4)) ((1.872*10^(-5))/(2.934*10^(-5)))^(0.25)`

Nu = 179.95

The heat transfer coefficient can be calculated using the formula:

`h = (k* Nu)/(D) \\h = (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 = (\pi * D^3 )/(6)`

`V = (\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`

Answer 2

Step-by-step explanation:

The complete detailed explanation which answer the question efficiently is shown in the attached files below.

I hope it helps a lot !