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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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Recursion refers to the act of calling a function itself. With the use of this strategy, complex problems can be reduced to more manageable, simpler ones. Recursion might be a little challenging to comprehend. The best method to figure out how it works is to experiment with it.

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The process of making a function call itself is known as recursion. With the use of this strategy, complex problems can be reduced to more manageable, simpler ones. Recursion might be a little challenging to comprehend. Experimenting with it is the most effective way to learn how it functions.

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A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027
Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = \frac{d}{2} = 3 inches

density,  \rho} = 493 lbm/ ft^{3}

S.G = 1.0027

g = 9.8 m/ m^{2} = 386.22 inch/ s^{2}

Solution:

Using the formula for terminal velocity,

v_{T} = \sqrt{\frac{2V\rho  g}{A \rho C_{d}}}              (1)

(Since, m = V\times \rho)

where,

V = volume of sphere

C_{d} = drag coefficient

Now,

Surface area of sphere, A = 4\pi r^{2}

Volume of sphere, V = \frac{4}{3} \pi r^{3}

Using the above formulae in eqn (1):

v_{T} = \sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho  g}{4\pi r^{2} \rho C_{d}}}

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3 years ago
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