# Find the irrigation pumping rate (mass flow rate of a pump) versus time

Here is what we are given. Time ranges from 0 to 10 hours, with t=5 being solar noon. $$\dot Q = 90Asin(\pi t/10)$$ Watts, where A is the solar collector area in square meters and t is the time after sunrise in hours. The solar collector efficiency is 40% of the incident solar flux. The flow rate of the water ($$\dot m(t)$$) through the collector varies so the collector outlet condition is saturated vapor at 150 Celsius. The isentropic (only entropy is constant) efficiency of the turbine is 70% and the outlet pressure is 8 kPa. The reservoir water pump lifts water 8 meters with negligible friction. The isentropic efficiency of the pumps is 80%. I found the solar collector area required to provide $$\dot m_{Pmax}$$ = 500 kg/min of irrigation water at solar noon (t=5). That is the max irrigation pumping rate. Then, I need to plot the irrigation pumping rate $$\dot mP(t)$$ versus time over the 10 hours. We are given the hint that the irrigation pumping rate is proportional to the maximum rate (500 kg/min) times the relative solar flux at any time compared to the solar noon flux.

Would that just be (500 kg/min)*$$\dot Q(t)/\dot Q(t=5)$$? Where does the height come into this equation? I have all of the enthalpy values for the system from a basic Rankine cycle analysis, but I am confused about the second part.

• I assume you used the height to determine the flow rate at noon, so now it is disregarded. All you need is the change in heat form the sun due to the time of day. If your flow rate at noon is a given then you don't need to know anything about the thermals in the system. – Tiger Guy Apr 19 at 19:47