Respuesta :
Answer:
Explanation:
First we compute the characteristic length and the Biot number to see if the lumped parameter
analysis is applicable.
Since the Biot number is less than 0.1, we can use the lumped parameter analysis. In such an
analysis, the time to reach a certain temperature is given by the following
From the data in the problem we can compute the parameter, b, and then compute the time for
the ratio (T – T)/(Ti
– T)
Answer:
A) 205.872°F
B) 2172 Btu/min
Explanation:
A) First of all, we will compute the characteristic length and the Biot number to see if the lumped parameter analysis is applicable.
Thus;
Lc = V/A = (πD³/6)/(πD²)
Simplifying; L = D/6
Since D = 2 inches or in ft as D= 0.1667
Lc = 2/6 = 1/3 inches
Now we have to convert to feet because K is in ft.
Thus Lc = 1/3 x 0.0833 = 0.0278 ft
Formula for Biot number is;
Bi = hLc/k
Thus, Bi = (42 x 0.0278)/641 = 0.018
Since the biot number is less than 0.1,we will make use of the lumped parameter analysis which is given by;
T = T∞ + (Ti - T∞)e^(-bt) and
b = h/(ρ(Cp)(Lc))
Let's find b first before the temperature T.
Thus; b = 42/(532 x 0.092 x 0.0278) = 30.9 per hr
Now lets find the temperature after 2 minutes. For sake of ease of calculation, let's convert 2 minutes to hours to give; 2/60 = 0.033 hr
From the question, T∞ = 120°F and Ti = 360°F
So, T = 120 + (360 - 120)e^(-30.9 x 0.033)
T = 120 + 240e^(-1.02897)
T = 120 + (240 x 0.3578) = 205.872°F
B) The heat transfer to each ball is the product of mass, heat capacity and difference between
the initial and final temperature.
Thus;
Q = M(Cp)(Ti - T∞)
We know that Density = Mass/Volume and thus mass(M) =
Density x Volume = ρV
So;
Q = ρV(Cp)(Ti - T∞)
= 532 x ((π x 0.1667³)/6) x (0.092) (360 - 205.872)
= 532 x 0.0024 x 0.092 x 154.128 = 18.1 Btu
So for 120 balls per minute the total heat removal ;
18.1 Btu x (120 per minutes) = 2172 Btu per minutes
