5. A 4500 kW steam heat engine has an efficiency of 025. Determine the hourly fuel con- sumption if its calorific value is Q=33MJ/kg

To determine the hourly fuel consumption of the steam heat engine, we can use the formula:<br /><br />\[ \text{Fuel Consumption} = \frac{\text{Power Output} \times \text{Time}}{\text{Efficiency} \times \text{Calorific Value}} \]<br /><br />Given:<br />- Power Output = 4500 kW<br />- Efficiency = 0.25<br />- Calorific Value = 33 MJ/kg<br />- Time = 1 hour<br /><br />Substituting the given values into the formula:<br /><br />\[ \text{Fuel Consumption} = \frac{4500 \, \text{kW} \times 1 \, \text{hour}}{0.25 \times 33 \, \text{MJ/kg}} \]<br /><br />\[ \text{Fuel Consumption} = \frac{4500 \, \text{kJ}}{8.25 \, \text{MJ/kg}} \]<br /><br />\[ \text{Fuel Consumption} = 545.45 \, \text{kg/hour} \]<br /><br />Therefore, the hourly fuel consumption of the steam heat engine is approximately 545.45 kg.