I am an EIT and studying for an upcoming exam and was hoping someone could help me a question that i have been stuck on for more hours/days than I would like to admit. I am having a hard time understanding what is meant by a quasi steady state condition.

the question I am stuck on is as follows:
A rainwater harvesting and reuse system is being designed for a single family house. The system consist of roof gutters that convey the water to small reservoir at the side of the house. The reservoir is cylindrical with a 4m diam and has a free water surface. The reservoir is connected to a type "L" copper pipe with a 76mm interior diameter that conveys the water by gravity to an outside tap (Qout) for outdoor garden irrigation.The total length of copper pipe is 25m and has a "C" factor of 120. the flow through the pipe is controlled by a tap. The flow through the tap is estimated with the valve equation Qv= T*Es*SQRT(H-Ho) where Qv= valve discharge(m^3 sec). Es= valve discharge constant Es=0.95m^5/2/sec), H= hydraulic pressure upstream of valve, Ho= hydraulic pressure downstream of valve (tap discharges to the atmosphere.) and (I assume T stands for amount that valve is opened I.E valve fully closed T=0.0, half open T=0.5, fully open t=1.0.) Both the bottom of the reservoir and the center line of the tap are at the same elevation.
Given the following info on 6 hour hydro graph:
0-1hour Inflow Qin=5L/sec =0.005m^3 sec
1-2 hour Q=10L/s
2-3 hour Q=0
3-4 hour Q=0
4-5 hour Q=10L/sec
5-6 hour Q=7L/sec

a) write the governing equations that describe the quasi-steady state conditions in the reservoir-pipe valve system.

From what I understand a quasi steady condition exists when the system is designed in a way that the it responds to the changes in flow conditions such that there is always a balance in the system or in other words that the you should adjust the valve according to the Qin during that specified period.I may be completely wrong with my above assumption and was hoping someone would have some information on this topic or could point to the correct resources. I have tried to perform some research on this topic but did not have a whole lot of luck. I was able to find an article of some use http://hydratek.com/wp-content/uploa.../12/period.pdf but was beyond the scope of this theoretical simple type of system.

I have went ahead with my above assumption hoping that is correct. For the system in design there are three different flow rates that the valve will need to be opened accordingly to maintain equilibrium or a quasi steady flow condition. Once again I am not sure if this correct or if a quasi steady could means something completely different possibly (and more than likely in hind site) it is referring to finding a flow rate that maintains a constant rate regardless on variable demand but from what I understand I will continue on(sorry to blab on just confused).

Regardless I 1st calculated the head losses due to friction in the outlet pipe using hazen-williams formula re arranged to calc head loss as follows Hf=10.7[L/(D^4.87)]*(Q/C)^1.85
I.E. time 0-1,Q=5L/sec Hf=10.7*(25/0.076^4.87)*0.005/120^1.85= 0.595m
Q=10L/sec Hf=2.144m
Q=7L/sec Hf=1.108m
I believe I can assume that Qin=Qout for a quasi steady state condition and can use the Qin flows given in the valve equation given. I have assumed that Ho hydraulic head downstream of valve =0 since it discharges into the atmosphere. I am confused with what value I should use for H (hydraulic head upstream of valve) but have assumed that since the reservoir and the center line tap elevation are a the same elevation therefore there will be zero static head if the valve was set to match the discharge the flows at the same rate that they are receiving them.I have also been given the equation Total Dynamic Head TDH= Hs+Hf, were Hs = static head and Hf= friction head. At 1st I assumed that head losses would need to be subtracted but another way of looking at it is that in order to compassionate for the loss in pressure due to friction loss more pressure is required at the start of the system. therefor head loss needs to be added but I think that may only apply to pump scenarios. I however am confused as to were this head energy is coming from but will move on once again regardless of my confusion. Rea ranging the the valve equation to solve for the unknown variable T (valve opening) we get the following equation:
T=Qv/(0.95*SQRTof H) where H= Hf calculated above.
the values I calculated for T are as follows:
for Qv=0.005m^3/sec T=0.0068
for Q=0.01 T=0.00718
for Q=0.007 T=0.007.
but these values seem very small which makes me think that I am incorrect. possibly unit issues or just my theory in general. Going back to the original question I am no further off at answering the question for part a) asking for the governing equations that describe the quasi steady state condition.
I will await some help until I tackle part B that asks given the hydro graph info. Assuming that the tap is fully opened (t=1.0) at time 2 hours and closed (t=0)at time 4 hours, determine the design height of the reservoir to prevent overflow. But this is far more than enough writing for one night and seems like a more straight forward to answer considering I actually understand what they are asking for.

I feel like I may be missing some fundamental concepts and that some of my assumptions may be incorrect. If any one could help out a struggling recent grad your help would be truly appreciated as my time is limited as I have other material that I need to prepare before the exam in a fast approaching month [surprise]