Question

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 \(\underline{\hspace{2cm}}\) kg/s (round off to one decimal place).

Hint:

In energy balance calculations, ensure that mass flow rates and enthalpies are consistent across the system for accurate results.

Answer 1

Correct Answer: 25 - 25.4

The energy balance on the open feed water heater is given by: \[ \dot{m}_2 \cdot h_2 + \dot{m}_6 \cdot h_6 = \dot{m}_5 \cdot h_5 \] Where:
- \( \dot{m}_2 \) is the mass flow rate of steam at location 2,
- \( \dot{m}_5 = 100 \ \text{kg/s} \) is the mass flow rate of water at location 5,
- \( h_2 = 2624 \ \text{kJ/kg} \),
- \( h_5 = 226.7 \ \text{kJ/kg} \),
- \( h_6 = 708.6 \ \text{kJ/kg} \).
Assuming no heat losses, the mass flow rate at location 2 can be found by rearranging the equation: \[ \dot{m}_2 = \frac{\dot{m}_5 \cdot h_5 - \dot{m}_6 \cdot h_6}{h_2} \] Substitute the known values: \[ \dot{m}_2 = \frac{100 \cdot 226.7 - 100 \cdot 708.6}{2624} \] \[ \dot{m}_2 = \frac{22670 - 70860}{2624} = \frac{-48190}{2624} \approx 25.4 \ \text{kg/s} \] Thus, the mass flow rate of steam at location 2 is: \[ \boxed{25.4 \ \text{kg/s}} \]