from reservoir A (surface elevation 80 ft) to reservoir D (surface elevation 90 ft). In week 14, we considered a specific example with a cast iron pipe (A = 0.00085 ft), flow rate of Q = 40 ft/s, and average velocity of V = 5.66 ft/s. For these conditions, the friction factor was 2= 0.015 and the pump added 85 ft of head. Assume the water temperature is always 60°F. Part A Assuming the pump is 75% efficient and electricity costs 10¢/kW hr, how much will it cost to run this pumpfor a year? Hint, 1 kW hr = (1000 watts)(1 hr) =(1000 J/s)(3600 s) = 3.6x10° J. Part BNow let us assume we need to double the discharge to Q=80 ft’/s, but unlike the example in week 13, we are not going to make the false assumption that the friction factor 1 is constant. You may, however, assume that the roughness height A is constant. We will consider three options: Option #1: Build a parallel system with the same pipe and pump, with a flow rate of Q= 40 ft’/sin each system. What is the annual electricity cost? Option #2: Use the original pipe but double the velocity. What is the annual electricity cost? Option #3: Build a new pipe with diameter of 50.9 in. The cross-sectional area of this pipe is twice that of a 36 in pipe, so the velocity will be V= 5.66 ft/s for a discharge of Q= 80 ft’/s.What is the annual electricity cost? Part C Briefly comment on the pros and cons of options 1-3 (<30 words).
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