timmonskids2681
27.07.2021 •
Physics
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
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Ответ:
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 )
Ответ: