The decrease in available energy is calculated using the formula:
\( \Delta Q = \dot{Q} \left( 1 - \frac{T_2}{T_1} \right) \)
Where: \( \dot{Q} = 7200 \, \text{kJ/min}, \, T_1 = 1000 \, K, \, \text{and} \, T_2 = 300 \, K \). After plugging in the values:
\( \Delta Q = 7200 \times \left( 1 - \frac{300}{1000} \right) = 7200 \times 0.7 = 5040 \, \text{kJ/min} \)
| LIST I | LIST II |
| A. Reynold’s Number | III. Inertia force to viscous force |
| B. Mach Number | I. Inertia force to elastic force |
| C. Froude’s Number | II. Inertia force to gravity force |
| D. Weber’s Number | IV. Inertia force to surface tension force |
| LIST I | LIST II |
| A. Subcooled water | I. 1 bar and 134°C |
| B. Superheated steam | II. Dryness fraction = 1 and 100°C |
| C. Steam at critical state | III. 20°C and 1.01325 bar |
| D. Saturated steam | IV. 374.15°C and 220.8 bar |
Choose the correct answer from the options given below: