Question 10.32: KNOWN: Cylinder of 120 mm diameter at 1000K quenched in satu......
KNOWN: Cylinder of 120 mm diameter at 1000K quenched in saturated water at 1 atm
FIND: Describe the quenching process and estimate the maximum heat removal rate per unit length during cooling.
ASSUMPTIONS: Water exposed to 1 atm pressure, \mathrm T_{\text{sat}} = 100°C.
SCHEMATIC:
ANALYSIS: At the start of the quenching process, the surface temperature is \mathrm T_{\mathrm s}(0) = 1000 K.
Hence, \Delta\mathrm T_{\mathrm e} = \mathrm T_{\mathrm s} – \mathrm T_{\text{sat}} = 1000 K – 373 K = 627 K, and from the typical boiling curve of Figure 10.4, film boiling occurs.
As the cylinder temperature decreases, \Delta\mathrm T_{\mathrm e} decreases, and the cooling process follows the boiling curve sketched above. The cylinder boiling process passes through the Leidenfrost point D, into the transition or unstable boiling regime (D → C).
At point C, the boiling heat flux has reached a maximum, \mathrm q_{\max}^{\prime\prime} = 1.26 MW/m² (see Example 10.1). Hence, the heat rate per unit length of the cylinder is
q _{ s }^{\prime}= q _{\max }^{\prime}= q _{\max }^{\prime \prime}(\pi D )=1.26 MW / m ^2[\pi(0.120 m )]=0.475 MW / m.
As the cylinder cools further, nucleate boiling occurs (C → A) and the heat rate drops rapidly. Finally, at point A, boiling no longer is present and the cylinder is cooled by free convection.
COMMENTS: Why doesn’t the quenching process follow the cooling curve of Figure 10.3?