The power developed by a turbine in a certain steam power plant is 1206 kW. The heat supplied to boiler is 3500 kJ/kg. The heat rejected by steam to cooling water is 2900 kJ/kg. The feed pump work required to condense condensate back into the boiler is 6 kW. What will be mass flow rate of cycle ? (1) 2 kg/s (2) 0.002 kg/s (3) 6.22 kg/s (4) 0.00622 kg/s (5) Question not attempted

The net work output of a thermodynamic cycle is the difference between the total heat supplied and the total heat rejected. Net work done by the cycle, $W_{net} = Q_{supplied} - Q_{rejected}$. Given: Heat supplied to boiler, $Q_{in} = 3500 \text{ kJ/kg}$ (per unit mass) Heat rejected to cooling water, $Q_{out} = 2900 \text{ kJ/kg}$ (per unit mass) The net work per unit mass is: $w_{net} = Q_{in} - Q_{out} = 3500 \text{ kJ/kg} - 2900 \text{ kJ/kg} = 600 \text{ kJ/kg}$.The power developed by the turbine is given as $P_{turbine} = 1206 \text{ kW}$. The feed pump work is given as $W_{pump} = 6 \text{ kW}$.In a steam power plant, the turbine produces power, and the pump consumes power. The net power output of the plant is the turbine power minus the pump power. $W_{net\_power} = P_{turbine} - W_{pump}$. $W_{net\_power} = 1206 \text{ kW} - 6 \text{ kW} = 1200 \text{ kW} $.The net work per unit mass ($w_{net}$) is related to the net power output ($W_{net\_power} $) by the mass flow rate ($\dot{m}$) of the working fluid: $W_{net\_power} = \dot{m} \times w_{net}$.We need to find the mass flow rate $ \dot{m}$. $\dot{m} = \frac{W_{net\_power}} {w_{net}}$.Substitute the values: $\dot{m} = \frac{1200 \text{ kW}}{600 \text{ kJ/kg}}$.To ensure units are consistent, note that $1 \text{ kW} = 1 \text{ kJ/s}$. So, $\dot{m} = \frac{1200 \text{ kJ/s}}{600 \text{ kJ/kg}}$. $ \dot{m} = 2 \text{ kg/s}$.