Question

Find the solidification time of the two iron castings when both are poured




(with no superheats) into the sand molds at initial temperature 28°C.




(i) A slab shaped casting of 10 cm thickness, (ii) a sphere of 10 cm in dia. Iron: freezing temp: 1540°C; L= 272 kJ/kg; density = 7850 kg/m3 Sand: c = 1.17 kJ/kg-K; k = 0.865 W/mk; density = 1600 kg/m3

Answer 1

The solidification time for a slab shaped casting is 161.1min and for a sphere with 10cm diameter is 4.475min

Data;

• ρ(m) = Density of metal (kg/m^3) = 7850kg/m^3
• T(o)= Initial temperature of mold = 28 + 273 = 301K
• Tm = melting or freezing temperature of liquid = 1540 degree
• ΔT(s) = T(p) - T(m) = 0 (No superheat)
• L = latent heat of fusion (JKg^-1) = 273*10^3 Jkg^-1
• K = Thermal conductivity of mold = 0.8655 w/mK
• ρ = density of mold = 1600 kg/m^3
• c = specific heat of mold = 1.17 kJ/kg K
• Cm = specific heat of metal = 0.45 kJ/kg K

Solidification Time

Using the formula

$\gamma = [\frac{\rho m L}{(T_m - T_o)}]^R[\frac{\pi}{4k\rho c} ][1+(\frac{C_m\delta T_s}{L})]$

Substituting the values into the equation;

$\gamma = [\frac{7850*272*10^3}{(1540-28)}]^2[\frac{\pi}{4*0.8655*1600*1170}][1+0]\\ \gamma= 9.66697*10^5 s/m^2\\ \gamma = 1.611 min/cm^2$

The formula for solidification time is given as

$T_s = \gamma (\frac{v}{A})^2$

i) The casting of 10cm thickness

For slab shape casting,

$(v/A)= \frac{A*t}{A} = t\\ (\frac{v}{A})^2 = A^2 = (0.1)^2 = 0.01\\ t = 16111.6*0.01 = 161.1min$

ii) For a sphere of 10cm diameter

$(\frac{v}{A}) =(\frac{R}{3})\\ (\frac{v}{A})^2 = (\frac{0.05}{3})^2\\ t_s = 16111.6*(\frac{0.05}{3})^2\\ t_s= 4.475min$

From the calculations above, the solidification time for a slab shaped casting is 161.1min and for a sphere with 10cm diameter is 4.475min

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