Mindyard | Consider the open feed water heater (FWH) shown in the figure given below: Specific enthalpy of steam at location 2 is 2624 kJ/kg, specific enthalpy of water at location 5 is 226.7 kJ/kg and specific enthalpy of saturated water at location 6 is 708.6 kJ/kg. If the mass flow rate of water entering the open feed water heater (at location 5) is 100 kg/s then the mass flow rate of steam at location 2 will be __________

Consider the open feed water heater (FWH) shown in the figure given below: Specific enthalpy of steam at location 2 is 2624 kJ/kg, specific enthalpy of water at location 5 is 226.7 kJ/kg and specific enthalpy of saturated water at location 6 is 708.6 kJ/kg. If the mass flow rate of water entering the open feed water heater (at location 5) is 100 kg/s then the mass flow rate of steam at location 2 will be ___________ kg/s (round off to one decimal place).

Correct Answer :

Correct answer is 25.16

h₂ = 2624 kJ/kg, h₅ = 226.7 kJ/kg, h₆ = 708.6 kJ/kg, m₅= 100 kg/s

m₂ h₂ + m₅ h₅ = (m₂ + m₅) h₆

m₂ × 2624 + 100 × 226.7 = (m₂ + 100) × 708.6

m₂ = 25.16 kg/s

Solution :

The correct answer is 25.16.

Analysis of the Diagram:
As shown in the cycle diagram, the system consists of a steam generator, a turbine, a condenser, a Low-Pressure (LP) pump, a High-Pressure (HP) pump, and an Open Feed Water Heater (Open FWH).
Looking at the flow paths entering and leaving the Open FWH:
1. Extraction steam from the turbine enters the Open FWH at location 2 with mass flow rate $m_{2}$ and specific enthalpy $h_{2}$ .
2. Feed water from the LP pump enters the Open FWH at location 5 with mass flow rate $m_{5}$ and specific enthalpy $h_{5}$ .
3. The combined mixture leaves the Open FWH at location 6 and flows toward the HP pump with mass flow rate $m_{6}$ and specific enthalpy $h_{6}$ .

1. Mass Balance Equation:
Under steady-state conditions, the mass entering the open feed water heater must equal the mass leaving it:
$m_{6} = m_{2} + m_{5}$

2. Energy Balance Equation:
Assuming steady-flow conditions and no heat exchange with the surroundings (adiabatic), the energy balance is:
$m_{2} h_{2} + m_{5} h_{5} = m_{6} h_{6}$

Substituting the mass balance relation into the energy equation yields:
$m_{2} h_{2} + m_{5} h_{5} = (m_{2} + m_{5}) h_{6}$

3. Substituting the Given Numerical Values:
The given values from the problem description are:
- Specific enthalpy of steam at location 2: $h_{2} = 2624 kJ/kg$
- Specific enthalpy of water at location 5: $h_{5} = 226.7 kJ/kg$
- Specific enthalpy of saturated water at location 6: $h_{6} = 708.6 kJ/kg$
- Mass flow rate of water at location 5: $m_{5} = 100 kg/s$

Substituting these values gives:
$m_{2} \times 2624 + 100 \times 226.7 = (m_{2} + 100) \times 708.6$

4. Solving for the Mass Flow Rate of Steam ( $m_{2}$ ):
First, expand the terms on both sides:
$2624 m_{2} + 22670 = 708.6 m_{2} + 70860$
Next, group all the terms with $m_{2}$ on the left side of the equation:
$2624 m_{2} - 708.6 m_{2} = 70860 - 22670$
Simplify the coefficients:
$1915.4 m_{2} = 48190$
Divide both sides by $1915.4$ to find $m_{2}$ :
$m_{2} = \frac{48190}{1915.4} \approx 25.16 kg/s$

Therefore, the mass flow rate of steam at location 2 is 25.16 kg/s.

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