- #1
Sorbik
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Homework Statement
My simplified version of the equation:
Power plant proposes output power increase from 188MW to 535 MW
Excess heat is dumped into river with smelt. Smelt can withstand a 3.44C temp increase
River = 222ft wide at dump site. Do not know depth or cross section here
Further down the river, there is a concrete area that is 88.7ft wide and 36.2ft deep. The flow rate here is 10.4mph (4.649216 m/s)
The planned efficiency of the plant is 37.30%.
What is the max power output w/o endangering the smelt?
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The Actual question:
A battle has erupted over a local power plant owner\'s decision to raise the power delivered by the plant from 188 MW to 535 MW (The delivered power accounts for energy losses). The local population is increasing, so the area needs more power. However, the plant expends its heated waste water* into the Frustration River; local citizens are concerned that the heated water will raise water temperatures to the point where the endangered river smelt will die from oxygen depletion. Local biologists called into the fray report that the smelt can only withstand an average increase of 3.44°C. The river is 222 ft wide at the point where the waste water is dumped, but the depth and cross section is not well known. Farther downstream, however, the water flows through a narrow man-made concrete channel of rectangular cross-section, which has a width of 88.7 ft and depth of 36.2 ft. The water flows through this channel at 10.4 mph. The power plant claims a planned efficiency of 37.30% for its new power plant design. Based on this information, what is the maximum power that the power plant can output without endangering the river smelt?
Homework Equations
Related equations:
Q= mc delta T
C = 4186 J/ kg* K
Work = F*d -- Power = work/time -- power = (f*d)/time -- Force = m*g -- P=m*g*Velocity -- Density=m/Volume - - mass = density/volume - - Power = (density/volume) * gravity * velocity - -Density of water = 1000 kg/m^3 - - Watt = Joule/second - - Joule = (kg m^2)/ s^2
Efficiency of any engine = 1 – (Qc/Qh ) … Efficiency of carnot = 1 – (Tc/Th) - -celsius to Kelvin = C + 273.15
The Attempt at a Solution
my attempt at this problem was first finding all those related equations. I don't know how to piece the equations together to get to where I need to be.
The problem gives you a width at the dump site and then the area of a rectangular cross section at a place farther down the river? Why do I care about what goes on at a different spot when the width isn't the same?
Random guess:
First use the 3 measurements to find volume. 222 * 88.7*36.2 = 7.1282868 x10^5 ft^3. converting this to SI units (m^3) we multiply this by 0.02831684659 to get 20185.06 m^3
mass = density / volume
Using Q = mcΔT I would guess to do Q = [itex]\frac{density}{volume}[/itex]*4186 * 3.44 +273.15 =>> Q= (1000/20185.06) * 4186 * 276.59 = 57359.538 J
or: if using P = (density/volume) * gravity * velocity: P=.04954 * 9.8 * 0.44704 = 0.217 watts (really low...)
For a cyclic process heat engine: efficiency = work/heat input. Solving for heat input in joules we would do Qh = work/efficiency or Qh = 0.217/0.373 = 0.58188 joules