A fan that consumes 50 W of electric power when operating discharges air from a ventilated room at a rate of 0.50 kg/s at a discharge velocity of 12 m/s. Determine the mechanical efficiency of the fan.

The useful power output is the rate at which the kinetic energy of the air increases.
Given:
Mass flow rate ($\dot{m}$) = 0.50 kg/s
Inlet velocity ($V_1$) = 0 m/s
Discharge velocity ($V_2$) = 12 m/s

$$\dot{W}_{out} = \dot{m} \left( \frac{V_2^2}{2} - \frac{V_1^2}{2} \right)$$ $$\dot{W}_{out} = (0.50 \text{ kg/s}) \left( \frac{(12 \text{ m/s})^2}{2} - \frac{(0 \text{ m/s})^2}{2} \right)$$ $$\dot{W}_{out} = (0.50 \text{ kg/s}) \left( \frac{144 \text{ m}^2/\text{s}^2}{2} \right)$$ $$\dot{W}_{out} = (0.50 \text{ kg/s}) (72 \text{ m}^2/\text{s}^2)$$ $$\dot{W}_{out} = 36 \text{ W}$$

The mechanical efficiency is the ratio of the useful mechanical power output to the electric power input.
Given:
Electric power input ($P_{in}$) = 50 W
Useful mechanical power output ($\dot{W}_{out}$) = 36 W

$$\eta_{mech} = \frac{\dot{W}_{out}}{P_{in}}$$ $$\eta_{mech} = \frac{36 \text{ W}}{50 \text{ W}}$$ $$\eta_{mech} = 0.72$$

To express as a percentage:

$$\eta_{mech} = 0.72 \times 100\% = 72\%$$

The mechanical efficiency of the fan is $\boxed{72\%}$.

b) A system undergoes a power cycle while receiving 1000 kJ by heat transfer from a high-temperature reservoir and rejecting 800 kJ by heat transfer to a low-temperature reservoir. Determine the net work done by the system.

For a power cycle, the net work output ($W_{net}$) is the difference between the heat received ($Q_{in}$) and the heat rejected ($Q_{out}$).
Given:
Heat received ($Q_{in}$) = 1000 kJ
Heat rejected ($Q_{out}$) = 800 kJ

$$W_{net} = Q_{in} - Q_{out}$$ $$W_{net} = 1000 \text{ kJ} - 800 \text{ kJ}$$ $$W_{net} = 200 \text{ kJ}$$

The net work done by the system is $\boxed{200 \text{ kJ}}$.