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andrey2020 [161]

3 years ago

7

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) ini

tially 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.

2 answers:

noname [10]3 years ago

5 0

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

bekas [8.4K]3 years ago

4 0

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)

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Explanation:

The optimum cutting speed for the minimum cost

$V_{opt}= \frac{C}{\left[\left(T_c+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]^n}\;\cdots(i)$

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C,n = Taylor equation parameters

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$C_e$ =Cost per grinding per edge

$C_m$ = Machine and operator cost per minute

On comparing with the Taylor equation $VT^n=C$ ,

Tool life,

$T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)$

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Cost of operator and machine time $=\$40/hr=\$0.667/min$

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Part handling time: $T_h=2.5$ min

Part diameter: $D=73 mm$ $=73\times 10^{-3} m$

Part length: $l=250 mm=250\times 10^{-3} m$

Feed: $f=0.30 mm/rev= 0.3\times 10^{-3} m/rev$

Depth of cut: $d=3.5 mm$

For the HSS tool:

Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

So, $C_e=$ $\$20/15+2=\$3.33/edge$

Tool changing time, $T_t=3$ min.

$C= 80 m/min$

$n=0.130$

(a) From equation (i), cutting speed for the minimum cost:

$V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}$

$\Rightarrow 47.7$ m/min

(b) From equation (ii), the tool life,

$T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}$

$\Rightarrow T=53.4$ min

(c) Cycle time: $T_c=T_h+T_m+\frac{T_t}{n_p}$

where,

$T_m=$ Machining time for one part

$n_p=$ Number of pieces cut in one tool life

$T_m= \frac{l}{fN}$ min, where $N=\frac{V_{opt}}{\pi D}$ is the rpm of the spindle.

$\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc$

So, the number of parts produced in one tool life

$n_p=\frac {T}{T_m}$

$\Rightarrow n_p=\frac {53.4}{4.01}=13.3$

Round it to the lower integer

$\Rightarrow n_p=13$

So, the cycle time

$T_c=2.5+4.01+\frac{3}{13}=6.74$ min/pc

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times6.74+\frac{3.33}{13}=\$4.75/pc$

(e) Total time to complete the batch= Sum of setup time and production time for one batch

$=2\times60+ {50\times 6.74}{50}=457 min=7.62 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times4.01}{457}=0.4387=43.87\%$

Now, for the cemented carbide tool:

Cost per edge,

$C_e=$ $\$8/6=\$1.33/edge$

Tool changing time, $T_t=1min$

$C= 650 m/min$

$n=0.30$

(a) Cutting speed for the minimum cost:

$V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min$ [from(i)]

(b) Tool life,

$T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min$ [from(ii)]

(c) Cycle time:

$T_c=T_h+T_m+\frac{T_t}{n_p}$

$T_m= \frac{\pi D l}{fV_{opt}}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 363}=0.53min/pc$

$n_p=\frac {7}{0.53}=13.2$

$\Rightarrow n_p=13$ [ nearest lower integer]

So, the cycle time

$T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc$

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times3.11+\frac{1.33}{13}=\$2.18/pc$

(e) Total time to complete the batch $=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times0.53}{275.5}=0.0962=9.62\%$

Similarly, for the ceramic tool:

$C_e=$ $\$10/6=\$1.67/edge$

$T_t-1min$

$C= 3500 m/min$

$n=0.6$

(a) Cutting speed:

$V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}$

$\Rightarrow V_{opt}=2105 m/min$

(b) Tool life,

$T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min$

(c) Cycle time:

$T_c=T_h+T_m+\frac{T_t}{n_p}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc$

$n_p=\frac {2.33}{0.091}=25.6$

$\Rightarrow n_p=25 pc/tool\; life$

So,

$T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc$

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times2.63+\frac{1.67}{25}=$1.82/pc$

(e) Total time to complete the batch

$=2\times60+ {50\times 2.63}=251.5 min=4.19 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times0.091}{251.5}=0.0181=1.81\%$

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3 years ago