Question

A decorative plastic film on a copper sphere of 10 mm diameter in an oven at 750C. Upon removal from the oven, the sphere is subjected to an air stream at 1 atm. , and 230C having a velocity of 10 meter/sec. Estimate how long it will take to coal the sphere to 350C? Density of copper is 8933 kg/m3 , thermal conductivity k = 399 W/m.K and specific heat Cp = 387 J/kg.K. for Air at an average temperature of T[infinity] = 296 K has viscosity µ= 181x10-7 N.s/m2 , kinematic viscosity ν = 15.36x10-6 m2/s and thermal conductivity k = 0.0258 W/m.K and Prandtl No. Pr = 0.709 and air at surface temperature Ts = 328 K will have viscosity µs = 197x10-7 N.s/m2

Answer 1

Answer:

3.26 secs

Explanation:

Diameter of sphere ( D )= 10 mm

T1 = 75°C

P = 1 atm

T∞ = 23°C

T2 = 35°c

Velocity = 10 m/s

Determine how long it will take to cool the sphere to 35°C

Using the properties of copper and air given in the question

Nu = 2 + (Re)^0.8 (Pr)^0.33

hd / k = 2 + ( vd/v )^0.8 (Pr)^0.33

∴ h ≈  2594.7 W/m^2k

Given that :

(T2 - T∞) / ( T1 - T∞ )  = exp [ ( -hA / pv CP ) t ]  

( 35 - 23 ) / ( 75 - 23 ) = exp [  - 2594.7 * 6 * t / 8933 * 387 * 10 * 10^-3 ]

=  ln ( 12/52 ) = -1.466337069  =     - 0.45032919 *  t

∴ t ≈ 3.26 secs        ( -1.466337069 / -0.45032919 )