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Zielflug [23.3K]
3 years ago
12

A strip of AISI 304 stainless steel, 2mm thick by 3cm wide, at 550°C, continuously enters a cooling chamber that removes heat at

a rate of 120 kW. How fast must it move to achieve a temperature of 140 °C at the exit?
Engineering
1 answer:
Zanzabum3 years ago
6 0

Answer:

V = 1.23 m/s

Explanation:

Given data:

AISI 304 steel

thickness of steel = 2 mm

width  =  3 cm

Temperature =  550 degree celcius

wer know that heat is given as Q  

Q = m\times Cp \Delta T

\Delta T = 550 - 140 = 410 \degree\ celcius

for AISI 304 steel Cp is 502 J/kg . K

We know that

\dot mass  =\rho A V

\rho = 7.9\times 10^3 kg/m^3

A = 2\times 10^{-3} \times 3\times 10^{-2} m^2

therefore VELOCITY V  will be

120\times 10^{3} = 7.9\times 10^{3} \times 6\times 10^{-5} \times V\times 502\times 410

solving for V we get

V = 1.23 m/s

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using these values, we get:

dT = - (q/2k)(r dr)

integrating from r = 0, where T(0) = To = Maximum center temperature, to r = Ro, where, T(Ro) = Ts = surface temperature = 120°C.

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To = 120° C + 397.24° C

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We know that elongation in wire is given by \delta =\frac{FL}{AE}, here F is load, L is length, A is area and E is elastic modulus

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