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2 years ago

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A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are w

ell insulated; the front surface is exposed to the flow of a coolant at T[infinity] = 15°C. From reliability considerations, the chip temperature must not exceed T = 85°C.f the coolant is air and the corresponding convection 200 W/m2 K, what is the maximum coefficient is h allowable chip power? If the coolant is a dielectric liquid for which h 3000 W/m2 K, what is the maxi- mum allowable power?

1 answer:

nirvana33 [79]2 years ago

5 0

Answer:

Q(h=200)=0.35W

Q(h=3000)=5.25W

Explanation:

first part h=200W/Km^2

we must use the convection heat transfer equation for the chip

Q=hA(Ts-T∞)

h= convective coefficient=200W/m2 K

A=Base*Leght=5mmx5mm=25mm^2

Ts=temperature of the chip=85C

T∞=temperature of coolant=15C

Q=200x2.5x10^-5(85-15)=0.35W

Second part h=3000W/Km^2

Q=3000x2.5x10^-5(85-15)=5.25W

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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: $d=3.5 mm$

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$n=0.130$

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

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(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

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where,

$T_m=$ Machining time for one part

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

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

Round it to the lower integer

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

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

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$T_t-1min$

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