Energy Released from a Steam-Water Explosion Joe Udel decides to install his own 52-gallon (0.197 m^3) hot-water heater. However, being cheap, he ignores safety codes and neglects to add safety devices such as a thermostat and a rupture disk or pressure relief valve. In operation, the hot-water hea

Question

Energy Released from a Steam-Water Explosion

Joe Udel decides to install his own 52-gallon (0.197 [latex]m ^{3}[/latex]) hot-water heater. However, being cheap, he ignores safety codes and neglects to add safety devices such as a thermostat and a rupture disk or pressure relief valve. In operation, the hot-water heater is almost completely filled with liquid water and the pressure in the heater tank is the waterline pressure of 1.8 bar or the water saturation pressure at the heater temperature, whichever is greater. The water heater tank will rupture at a pressure of 20 bar. Several hours after Joe completes the installation of the water heater, it explodes.

a. What was the temperature of the water when the tank exploded?

b. Estimate the energy released in the blast.

Accepted Answer

a. From Appendix A.III the thermodynamic properties of saturated liquid water at 20 bar (2 MPa) are

[latex]T_{i}=212.42^{\circ} C \quad \hat{V}_{i}^{ L }=0.001177 \frac{ m ^{3}}{ kg } \quad \hat{U}_{i}^{ L }=906.44 \frac{ kJ }{ kg }[/latex]

and

[latex]\hat{S}_{i}^{ L }=2.4474 \frac{ kJ }{ kg K }[/latex]

Therefore, the water temperature when the tank explodes is 212.42°C. Also, as indicated in the problem statement, the tank contains only liquid, so [latex]\omega_{i}[/latex] = 0 and

[latex]M=\frac{V_{ T }}{\hat{V}_{i}^{ L }}=\frac{0.197 m ^{3}}{0.001177 \frac{ m ^{3}}{ kg }}=167.24 kg[/latex]

b. After the explosion we expect to have a vapor-liquid mixture. Since the pressure is 1 atm (actually, we use 1 bar), the temperature is 100°C and the other thermodynamic properties are

[latex]\begin{gathered}\hat{U}^{ L }=417.36 \frac{ kJ }{ kg } \quad \hat{U}^{ V }=2506.1 \frac{ kJ }{ kg } \\hat{S}^{ L }=1.3026 \frac{ kJ }{ kg K } \quad \hat{S}^{ V }=7.3594 \frac{ kJ }{ kg K }\end{gathered}[/latex]

We first use the entropy balance to determine the fractions of vapor and liquid present

[latex]\hat{S}_{i}^{ L }=\omega_{f} \hat{S}_{f}^{ V }+\left(1-\omega_{f} ight) \hat{S}_{f}^{ L }[/latex]

or

[latex]2.4474=\omega_{f} 7.3594+\left(1-\omega_{f} ight) 1.3026[/latex]

which gives

[latex]\omega_{f}=0.1890[/latex]

Next, to calculate the blast energy we use the energy balance, Eq. 5.4-11:

[latex]\begin{aligned}&M\left[\omega_{f} \hat{U}^{ V }\left(T_{f}, P= ext { ambient } ight)+\left(1-\omega_{f} ight) \hat{U}^{ L }\left(T_{f}, P= ext { ambient } ight) ight] \&-M\left[\omega_{i} \hat{U}^{ V }\left(T_{i}, P_{i} ight)+\left(1-\omega_{i} ight) \hat{U}^{ L }\left(T_{i}, P_{i} ight) ight]=W\end{aligned}[/latex] (5.4-11)

[latex]\begin{aligned}-W &=M\left[\hat{U}_{i}^{ L }-\hat{U}_{f} ight]=M\left[\hat{U}_{i}^{ L }-\left\{\omega_{f} \hat{U}_{f}^{ V }+\left(1-\omega_{f} ight) \hat{U}_{f}^{ L } ight\} ight] \&=167.24 kg [906.44-\{0.1890 imes 2506.1+0.8110 imes 417.36\}] kJ / kg \&=15772 kJ\end{aligned}[/latex]

This is equivalent to 3.43 kg of TNT.

Comment
In January 1982 a large hot-water tank exploded in an Oklahoma school, killing 7 people and injuring 33 others. The tank was found 40 meters from its original location, and part of the school cafeteria was destroyed. It was estimated that the tank failed at a pressure of only 7 bar.

Question 5.4.3: Energy Released from a Steam-Water Explosion Joe Udel decide...

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