Water at 20°C is pumped at a constant rate of 9 m3/h from a large reservoir resting on the floor to the open top of an experimental absorption tower. The point of discharge is 5 m above the floor, and friction losses in the 50-mm pipe from the reservoir to the tower amount to 2.5 J/kg. At what height in the reservoir must the water level be kept if the pump can deliver only 0.1 kW.Determine the mass flow rate of the flowing water in kg/s

A vehicle travelling at 100 km/h has air passing through its radiator at the rate of 100 kg mol per minute. The hot water from cooling the engine is circulated through the radiator at the rate of 10 L per minute, and as it passes through the radiator it is cooled from 78 °C to 32 °C. Assume the car is in good condition so the radiator has no leaks and the air and water do not contact. Given the heat capacity of the air is 1.0035 J·g-1·K-1, what is the temperature of the air exiting the radiator (assume air has a molecular weight of 29 g·mol-1).

Air (MW=29 g/mol) at 115.00 kPa and 285.00  is compressed steadily to 600.0 kPa. The mass flow rate of the air is 2.00 kg/s and a heat loss of 32.1 kW occurs during the process. You may assume that changes in kinetic and potential energy are negligible, the temperature of the surroundings is 25 ∘C, and that the CP of air is 3.5 R. Given the compressor operates with a second law (reversible) efficiency of 0.60,calculate the following. What is the actual work interaction term in kW? What is the actual exit temperature of the air in Celcius?

A site is being considered for wind power generation. At this site, the wind blows steadily at 9 m/s for 2,579 hours per year. Assuming the wind velocity is negligible at other times for simplicity, determine the kWh/year that can be produced at the site for a turbine with a mechanical efficiency of 100% and an effective flow area of 5 m2. Use a value of 1.25 kg/m3 for the density of the air. Give your answer in killowatt-hours per year (kWh/year). Hint: I reccomend starting by finding the kinetic energy (kJ/kg) of the wind. Hint 2: Multiply the power generation (kW) by the number of hours of wind per year (h/year) to get kWh/year.