Use of the Simplified Equation to Calculate the Energy Released in a Two-Phase Explosion (BLEVE)
Rework the previous illustration using only the fact that the density of water at 212.4°C is about 0.85 g/cm ^{3}, and that over the temperature range
C_{ P }^{ L }=4.184 \frac{ J }{ g K } \text { and } \Delta_{ vap } \hat{H}=2250 \frac{ J }{ g }=2250 \frac{ kJ }{ kg }Given the density of water above, we have
M=0.197 m ^{3} \times 0.85 \frac{ g }{ cc } \times 10^{6} \frac{ cc }{ m ^{3}} \times 10^{-3} \frac{ kg }{ g }=167.5 kgNext, we use
\begin{aligned}\omega_{f} &=1-\exp \left[\frac{C_{ P }^{ L }}{\Delta_{\text {vap }} \hat{H}}\left(T_{ b }-T_{i}\right)\right] \\&=1-\exp \left[\frac{4.184}{2250}(100-212.42)\right]=0.189\end{aligned}which is equal to the value found in the previous illustration using accurate thermodynamic tables.
Next we have
compared with 15 772 kJ from the more accurate calculation of the previous example. Since the total mass and the fraction vaporized have been correctly calculated, all the error is a result of the approximate calculation of the internal energies. Nonetheless, such approximate calculations are useful when detailed thermodynamic data are not available.