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The determination of k is not a strait forward exercise. Model it from first principles using Bernoulli's equation would produce a square root relationship and would probably have little relationship to the real system. The rate of discharge could be measured at different heads, but would require diverting the flow into a separate receptacle for a given time period. It is possible to log the level with time, as the tank drains freely through a fully open drain valve and while the pump is off. Knowing the area at each level ( see drawing of insert Figure 5) it is possible to determine the instantaneous volume against time. This has been done using the areas of the top and bottom prismatic sections of the tank and is shown in Figure.
Figure Plot showing example of instantaneous volume as V tank drains (using top and bottom areas of prismatic sections)
By determining the flow rate at the top and bottom of the tank, we can plot the graph shown in Figure. We have made the assumption that QL is directly proportional to the head. This is not true in actual fact, but will simplify the analysis and controller design. Figure 3 should be interpreted with care. Between about 10 and 20 seconds, the area is changing as we are in the V section and neither curve represents the instantaneous volume. After about 25 seconds, we are within the range of the outlet hole diameter, so effectively the orifice changes to channel like flow, but is complicated by the valve and pipe work arrangement.
To calculate the e.m.f of the cell when does the nelsons equation is use? Ans) The Nernst equation is useful to measure the e.m.f of the cell when the concentrations of solution
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Electron occupies the available orbital singly before pairing in any one orbital occurs, it is : (1) Pauli's exclusion principle (2) Hund's Rule (3) Heisenberg's principle
Illustrate with examples: (a) Lyophilic and Lyophobic sols (b) Multimolecular and Macromolecular colloids (c) Homogeneous and Heterogeneous catalysis.
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The number of unpaired electrons in an O 2 molecule is: (1) 0 (2) 1 (3) 2 (4) 3 Ans: 2
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