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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?

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Answer:

V = 1.23 m/s

Step-by-step 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* 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* 10^3 kg/m^3


A = 2* 10^(-3) * 3* 10^(-2) m^2

therefore VELOCITY V will be


120* 10^(3) = 7.9* 10^(3) * 6* 10^(-5) * V* 502* 410

solving for V we get

V = 1.23 m/s

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