To solve the problem, let's go step by step for each part of the calculation.
Given Data:
- Supply water temperature (\( T_w \)) = 32 °C
- Boiler pressure (\( P_b \)) = 2200 kPa
- Superheated steam temperature (\( T_s \)) = 268 °C
- Specific heat capacity of the steam (\( C_s \)) = 2.05 kJ/kg·°C
- Boiler efficiency = 92.2% or 0.922
- Calorific value of the fuel = 30 MJ/kg = 30,000 kJ/kg
- Specific heat capacity of water (\( C_w \)) = 4.2 kJ/kg·°C
3.1.1: The mass of the steam per kilogram of fuel
To find the mass of steam produced per kilogram of fuel, we need to calculate the total heat required to produce steam and the heat generated from the fuel.
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Calculate the heat input from the fuel (\( q_{in} \)):
\[ q_{in} = \text{Calorific value} \times \text{mass of fuel} \]
For 1 kg of fuel: \[ q_{in} = 30,000 \text{ kJ/kg} \]
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Calculate the heat output (\( q_{out} \)):
The heat output must equal the heat absorbed by the water to raise its temperature to boiling, the heat of vaporization, and finally the heat added to superheat the steam.
The specific heat of water, approximately, at 100 °C can be taken as:
- \( C_w \approx 4.2 , \text{kJ/kg}·°C \)
- The enthalpy of vaporization at 100 °C is approximately \( h_{fg} \approx 2260 , \text{kJ/kg} \)
Heat required to heat water from 32 °C to 100 °C:
\[ q_1 = m_{steam} \cdot C_w \cdot (100 - 32) \]
Heat required to vaporize the water:
\[ q_2 = m_{steam} \cdot h_{fg} \]
Heat required to superheat the steam to 268 °C:
\[ q_3 = m_{steam} \cdot C_s \cdot (268 - 100) \]
The total heat required \( q_{out} \):
\[ q_{out} = m_{steam} \cdot (C_w \cdot (100 - 32) + h_{fg} + C_s \cdot (268 - 100)) \]
Substituting values:
- \( q_{1} = m_{steam} \cdot 4.2 \cdot (68) = 285.6 , m_{steam} \)
- \( q_{2} = m_{steam} \cdot 2260 = 2260 , m_{steam} \)
- \( q_{3} = m_{steam} \cdot 2.05 \cdot (168) = 344.4 , m_{steam} \)
Therefore:
\[ q_{out} = m_{steam} \cdot (285.6 + 2260 + 344.4) = m_{steam} \cdot 2890 , \text{kJ/kg} \]
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Set up the equation with boiler efficiency:
The efficiency of the boiler relates the useful output to the input heat:
\[ \text{Efficiency} = \frac{q_{out}}{q_{in}} \]
Therefore: \[ 0.922 = \frac{m_{steam} \cdot 2890}{30000} \]
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Solve for \( m_{steam} \):
Rearranging gives: \[ m_{steam} = \frac{0.922 \cdot 30000}{2890} \approx \frac{27660}{2890} \approx 9.57 , \text{kg of steam/kg of fuel} \]
3.1.2: The heat transferred to the boiler
The heat transferred to the boiler is \( q_{out} \):
\[ q_{out} = m_{steam} \cdot 2890 \]
Replacing \( m_{steam} \):
\[ q_{out} = 9.57 \cdot 2890 \approx 27660 , \text{kJ} \]
3.1.3: The equivalent evaporation at 100 °C
Equivalent evaporation is defined as the mass of steam generated per kg of fuel at 100 °C, taking into account the heat needed to convert water at 100 °C to steam.
\[ \text{Equivalent Evap.} = \frac{q_{out}}{h_{fg}} \]
Using the heat of vaporization at 100 °C:
\[ \text{Equivalent Evap.} = \frac{q_{out}}{2260} = \frac{27660}{2260} \approx 12.24 , \text{kg of steam/kg of fuel} \]
Summary of Results:
- 3.1.1: The mass of steam per kilogram of fuel is approximately \( 9.57 \) kg.
- 3.1.2: The heat transferred to the boiler is approximately \( 27,660 \) kJ.
- 3.1.3: The equivalent evaporation at 100 °C is approximately \( 12.24 \) kg of steam/kg of fuel.