A steam power cycle with regeneration as shown below on the 7-s diagram employs a single open feedwater heater for efficiency improvement. The fluids mix with each other in an open feedwater heater. The turbine is isentropic and the input (bleed) to the feedwater heater from the turbine is at state 2 as shown in the figure. Process 3-4 occurs in the condenser. The pump work is negligible. The input to the boiler is at state 5. The following information is available from the steam tables:

State	1	2	3	4	5	6
Enthalpy (ki/kg) |	3350	2800	2300	175	700	1000

                                                                  

The mass flow rate of steam bled from the turbine as a percentage of the total mass flow rate at the inlet to the turbine at state 1 is ________.

The correct answer is 20.

To find the mass flow rate of steam bled from the turbine as a percentage of the total mass flow rate at the inlet to the turbine, we can perform an energy balance on the open feedwater heater (OFWH).

Let us analyze the T-s diagram provided:
- State 1 is the inlet to the turbine. Let the mass flow rate at this state be $m_1$ (or 100% of the flow).
- State 2 is the state at which steam is bled from the turbine and sent to the open feedwater heater. Let the fraction of steam bled be $y$.
- State 3 is the turbine exit, where the remaining fraction $(1-y)$ of steam enters the condenser.
- State 4 is the saturated liquid exiting the condenser.
- State 5 is the liquid entering and exiting the open feedwater heater. Since pump work is negligible, the liquid from the condenser is pumped to the intermediate pressure of the open feedwater heater with a negligible enthalpy change, so the enthalpy of the cold water entering the heater is $h_4$. The state of the saturated liquid leaving the open feedwater heater is state 5, with enthalpy $h_5$.

From the given steam table data, the enthalpies at the respective states are:
- $h_2=2800\text{kJ/kg}$
- $h_4=175\text{kJ/kg}$
- $h_5=700\text{kJ/kg}$

Applying the conservation of mass and energy to the open feedwater heater:
$y·h_2+(1-y)·h_4=h_5$

Substitute the given values into the equation:
$y·2800+(1-y)·175=700$
$2800y+175-175y=700$
$2625y=700-175$
$2625y=525$
$y=\frac{525}{2625}=0.2$

Expressing this fraction as a percentage:
$\text{Percentage of bled steam}=0.2·100=20\%$