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Zepler [3.9K]
2 years ago
14

A round bar of chromium steel, (ρ= 7833 kg/m, k =48.9 W/m-K, c =0.115 KJ/kg-K, α=3.91 ×10^-6 m^2/s) emerges from a heat treatmen

t process at 285°C and is quenched by an air stream at 15 m/s and at 35°C. The bar has a diameter of 40 cm and is 200 cm long. Determine the initial rate of heat
Engineering
1 answer:
Lerok [7]2 years ago
7 0

Answer:

Q = 424523.22 kw

Explanation:

\rho =7833 kg/m

k = 48.9 W/m - K

c = 0.115 KJ/kg- K

\alpha = 3.91*10^{-6} m^2/s

T_s = 285 degree celcius

T_∞ = 35 degree celcius

velocity of air stream = 15 m/s

D = 40 cm

L = 200 cm

mass flow rate\dot m = \rho AV = 7833 *\frac{\pi}{4} 0.4^2*15

\dot m = 14764.85 kg/s

A_s = \pi DL = \pi 0.4*2 = 2.513 m^2

Q = \dot m C \Delta T = h A_s \Delta T

\dot m C \Delta T = h A_s \Delta T

solving for h

h = \frac{14764.85*0.115*(285-35)}{2.513*(285-35)}

h = 675.6 kw/m^2K

Q = h A_s\Delta T

Q = 675.6*2.513*(285-35)

Q = 424523.22 kw

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Now as the water gets filtered out of the membrane,the water's volume decreases and concentration C of NacL increases, thus osmotic pressure also increases.Thus, at 50% water been already filtered out, the osmotic pressure at the membrane will be maximum

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<u>Answer:</u>

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