82.9k views
4 votes
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.

User Rrrfusco
by
6.1k points

2 Answers

4 votes

Answer:

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

User Avasin
by
6.1k points
3 votes

Answer:

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 !

A stainless steel ball (rho = 8055 kg/m3, cp = 480 J/kg·K) of diameter D = 0.21 m-example-1
A stainless steel ball (rho = 8055 kg/m3, cp = 480 J/kg·K) of diameter D = 0.21 m-example-2
A stainless steel ball (rho = 8055 kg/m3, cp = 480 J/kg·K) of diameter D = 0.21 m-example-3
A stainless steel ball (rho = 8055 kg/m3, cp = 480 J/kg·K) of diameter D = 0.21 m-example-4
User Baf
by
7.2k points