Answer:
474.59 mg/L
Explanation:
Given that
BOD = 30 mg/L
Original BOD = 30 mg/L × dilution factor
Original BOD = 30 mg/L × 10 = 300 mg/L
here is the ultimate BOD ; BOD is the biochemical oxygen demand ; t = 0.20 /day
provides steady-state operating data for a solar power plant that operates on a Rankine cycle with Refrigerant 134a as its worki
ng fluid. The turbine and pump operate adiabatically. The rate of energy input to the collectors from solar radiation is 0.3 kW per m2 of collector surface area, with 60% of the solar input to the collectors absorbed by the refrigerant as it passes through the collectors. Determine the solar collector surface area, in m2 per kW of power developed by the plant. Discuss possible operational improvements that could reduce the required collector surface area
1 answer:
Answer:
hello some parts of your question is missing attached below is the missing part ( the required fig and table )
answer : The solar collector surface area = 7133 m^2
Explanation:
Given data :
Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2
percentage of solar power absorbed by refrigerant = 60%
Determine the solar collector surface area
The solar collector surface area = 7133 m^2
attached below is a detailed solution of the problem
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Answer:
Heat flux of CO₂ in cgs
= 170.86 x 10⁻⁹ mol / cm²s
SI units
170.86 x 10⁻⁸ kmol/m²s
Explanation:
