Question

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 treatment 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

Answer 1

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