Consider 1.0 kg of austenite containing 1.15 wt% C, cooled to below 727\(^{\circ}\)C.
(a) What is the proeutectoid phase?
(b) How many kilograms each of total ferrite and cementite form?
(c) How many kilograms each of pearlite and proeutectoid phase form?
\( m_{alloy} \) =1.0 kg
\(C_{0}\) = 1.15 wt% C
(a) proeutectoid phase = ?
(b) \(m_{\alpha}\) = ? \(m_{Fe_{3}C}\) = ?
(c) \(m_{pearlite}\) = ? \(m_{proeutectoid}\) = ?
Hypereutectoid alloy
\(m_{\gamma}\) = 1.0 kg : before cooling
Below eutectoid temp: \(W_\alpha \) = \(C_{Fe_{3}C} - C_0 \over C_{Fe_{3}C} - C_\alpha \)
Below eutectoid temp: \(W_{Fe_{3}C} \) = \(C_{0} - C_\alpha \over C_{Fe_{3}C} - C_\alpha \)
Above eutectoid temp: \(W_{pearlite} \) = \(W_{\gamma} \) = \(C_{Fe_{3}C} - C_0 \over C_{Fe_{3}C} - C_\gamma \)
Above eutectoid temp: \(W_{proeutectoid} \) = \(C_{0} - C_\gamma \over C_{Fe_{3}C} - C_\gamma \)
(a) proeutectiod phase is Cementite
(b) \(m_\alpha \) = 0.83 kg   \(m_{Fe_{3}C} \) = 0.17 kg
(c) \(m_{pearlite}\) = 0.93 kg   \(m_{proeutectoid}\) = 0.07 kg