In a manufacturing facility, 2-in-diameter brass balls (k = 64.1 Btu/h·ft·°F, rho = 532 lbm/ft^3, and cp = 0.092 Btu/lbm·°F) initially at 360°F are quenched in a water bath at 120°F for a period of 2 min at a rate of 120 balls per minute. The convection heat transfer coefficient is 42 Btu/h·ft^2·°F. Take I to be the initial temperature. Determine:

a. The temperature of the balls after quenching.
b. The rate at which heat needs to be removed from the water in order to keep its temperature constant at 1200F.

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

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