The design product will be a description of the most efficient thermodynamic cycle required to supply 750 MW of electric power.
Maximum efficiency is achieved by determining the best pressure at which to operate flash evaporator (2, 3, 7, 8) within the limits imposed by the specified parameters.
You should provide:
1. temperature, pressure, and entropy at the state points;
2. overall flow rate of water through the nuclear reactor;
3. heat transfer in the nuclear reactor;
4. % of the overall flow rate passing through the steam turbine;
5. power required for condensate pump 1;
6. power required for condensate pump 2;
7. cooling water flow rate.
Situation: Attached
Diagram: Attached
Water is used as the working fluid of a pressurized-water nuclear power plant (shown in the figure) designed to produce 750 MW of electric power. Saturated liquid exiting the reactor enters a flash evaporator at a selected pressure. The pressure reduction is achieved by a constant enthalpy throttle valve. Thus, a fraction of the water flashes into saturated steam (state 2) and the remainder flows as liquid into the mixing chamber (state 3, p3=p2). The steam enters the turbine,
PLEASE FIND THE SOLUTION WITH CALCULATIONS for 7 parts in total.
You should provide:
1. temperature, pressure, and entropy at the state points;
2. overall flow rate of water through the nuclear reactor;
3. heat transfer in the nuclear reactor;
4. % of the overall flow rate passing through the steam turbine;
5. power required for condensate pump 1;
6. power required for condensate pump 2;
7. cooling water flow rate.
Situation: Attached
Diagram: Attached
Water is used as the working fluid of a pressurized-water nuclear power plant (shown in the figure) designed to produce 750 MW of electric power. Saturated liquid exiting the reactor enters a flash evaporator at a selected pressure. The pressure reduction is achieved by a constant enthalpy throttle valve. Thus, a fraction of the water flashes into saturated steam (state 2) and the remainder flows as liquid into the mixing chamber (state 3, p3=p2). The steam enters the turbine,
PLEASE FIND THE SOLUTION WITH CALCULATIONS for 7 parts in total.
1 answer:
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Answer:
[012] :
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[721] :
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[110] :
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Answer:
The power cost savings for the 8 inches pipe offsets the increased cost for the pipe therefore to save costs the 8-inch pipe should be used
Explanation:
The given parameters in the question are;
The distance of the river from the the site, d = 2,500 ft.
The planned flow rate = 600 gal/min
The diameter of the pipe, d = 6-in.
The pipe material = Steel
The cost of pumping = 3 cents per kilowatt-hour
The Bernoulli's equation is presented as follows;
Vₐ - 0 m/s (The river is taken as an infinite source)
= 0
The head loss in 6 inches steel pipe of at flow rate of 600 gal/min = 1.19 psi/100 ft
Therefore; = 1.19 × 2500/100 = 29.75 psi
= Q/
= 600 gal/min/(π·(6 in.)²/4) = 6.80829479 ft./s
≈ 6.81 ft./s
The pressure of the pump = P = 62.4 lb/ft³× (6.81 ft./s)²/2 + 29.75 psi ≈ 30.06 psi
The power of the pump = Q·P ≈ 30.06 psi × 600 gal/min = 7,845.50835 W
The power consumed per hour = 7,845.50835 × 60 × 60 W
The cost = 28,243.8301 kW × 3 = $847.31 per hour
Annual cost = $847.31 × 8766 = $7,427,519.46
Pipe cost = $15/ft × 2,500 ft = $37,500
Annual charges = 20% × Installed cost = 0.2 × $37,500 = $7,500
Total cost = $37,500 + $7,500 + $7,427,519.46 = $7,475,519.46
For the 8-in pipe, we have;
= Q/
= 600 gal/min/(π·(8 in.)²/4) = 3.83 ft./s
= 1.17 ft/100 feet
Total head loss = (2,500 ft/100 ft) × 1.17 ft. = 29.25 ft.
∴ = (3.83 ft./s)²/(2 × 32.1740 ft/s²) + 29.25 ft. ≈ 29.5 ft.
The power of the pump = ρ·g·h × Q
Power of pump = 62.4 lb/ft³ × 32.1740 ft/s² × 29.5 ft.× 600 gal/min = 3,363.8047 W
The cost power consumed per annum = 3,363.8047 W × 60 × 60 × 3 × 8766 = $3,184,608.1
The Cost of the pipe = $20/ft × 2,500 ft. = $50,000
The total cost plus charges = $3,184,608.1 + $50,000 + $10,000 = $3,244,608.1
Therefore it is more affordable to use the 8-in pipe which provides substantial savings in power costs