rcise 3.5 A steam plant produces steam at 8 MPa and \( 450^{\circ} \mathrm{C} \) at a rate of \( 5040 \mathrm{~kg} / \mathrm{h} \) by burning fuel at a rate of \( 540 \mathrm{~kg} / \mathrm{h} \). The feedwater enters the economiser at \( 40,3^{\circ} \mathrm{C} \). The fuel has a calorific value of \( 34 \mathrm{MJ} / \mathrm{kg} \). Calculate the: 1.1 rate of heat absorbed by the steam 1.2 rate of heat supplied by the fuel

To solve the problem, we will break it down into two parts: calculating the rate of heat absorbed by the steam and the rate of heat supplied by the fuel. ### 1.1 Rate of Heat Absorbed by the Steam The rate of heat absorbed by the steam can be calculated using the formula: \[ Q_{\text{steam}} = \dot{m}_{\text{steam}} \cdot h_{\text{steam}} \] Where: - \( Q_{\text{steam}} \) is the rate of heat absorbed by the steam (in kW). - \( \dot{m}_{\text{steam}} \) is the mass flow rate of the steam (in kg/s). - \( h_{\text{steam}} \) is the specific enthalpy of the steam at the given pressure and temperature. #### Step 1: Convert mass flow rate of steam Given: - Mass flow rate of steam = \( 5040 \, \text{kg/h} \) Convert to kg/s: \[ \dot{m}_{\text{steam}} = \frac{5040 \, \text{kg/h}}{3600 \, \text{s/h}} = 1.4 \, \text{kg/s} \] #### Step 2: Find the specific enthalpy of the steam To find \( h_{\text{steam}} \), we can use steam tables or thermodynamic properties for steam at \( 8 \, \text{MPa} \) and \( 450^{\circ} \text{C} \). For \( 8 \, \text{MPa} \) and \( 450^{\circ} \text{C} \), the specific enthalpy \( h_{\text{steam}} \) is approximately \( 2800 \, \text{kJ/kg} \). #### Step 3: Calculate the rate of heat absorbed by the steam Now we can calculate \( Q_{\text{steam}} \): \[ Q_{\text{steam}} = 1.4 \, \text{kg/s} \cdot 2800 \, \text{kJ/kg} = 3920 \, \text{kJ/s} = 3920 \, \text{kW} \] ### 1.2 Rate of Heat Supplied by the Fuel The rate of heat supplied by the fuel can be calculated using the formula: \[ Q_{\text{fuel}} = \dot{m}_{\text{fuel}} \cdot CV \] Where: - \( Q_{\text{fuel}} \) is the rate of heat supplied by the fuel (in kW). - \( \dot{m}_{\text{fuel}} \) is the mass flow rate of the fuel (in kg/s). - \( CV \) is the calorific value of the fuel (in kJ/kg). #### Step 1: Convert mass flow rate of fuel Given: - Mass flow rate of fuel = \( 540 \, \text{kg/h} \) Convert to kg/s: \[ \dot{m}_{\text{fuel}} = \frac{540 \, \text{kg/h}}{3600 \, \text{s/h}} = 0.15 \, \text{kg/s} \] #### Step 2: Use the calorific value Given: - Calorific value \( CV = 34 \, \text{MJ/kg} = 34000 \, \text{kJ/kg} \) #### Step 3: Calculate the rate of heat supplied by the fuel Now we can calculate \( Q_{\text{fuel}} \): \[ Q_{\text{fuel}} = 0.15 \, \text{kg/s} \cdot 34000 \, \text{kJ/kg} = 5100 \, \text{kJ/s} = 5100 \, \text{kW} \] ### Summary of Results 1. **Rate of heat absorbed by the steam**: \( 3920 \, \text{kW} \) 2. **Rate of heat supplied by the fuel**: \( 5100 \, \text{kW} \)