The value of L will be equal to the square root of the diffusion coefficient of oxygen in water times the reaction rate constant. The steady-state oxygen consumption rate in the pond is given by: Q = S*rA = −S*kCAo*2πL2.
a. Steady-state oxygen concentration distribution in the pond: Air oxygen (A) dissolves in a shallow stagnant pond and is consumed by microorganisms. The rate of the consumption can be approximated by a first order reaction, i.e. rA = −kCA, where k is the reaction rate constant in 1/time and CA is the oxygen concentration in mol/volume. The pond can be considered dilute in oxygen content due to the low solubility of oxygen in water (B). The diffusion coefficient of oxygen in water is DAB. Oxygen concentration at the pond surface, CAo, is known. The depth and surface area of the pond are L and S, respectively.
The equation for steady-state oxygen concentration distribution in the pond is expressed as:r''(r) + (1/r)(r'(r)) = 0where r is the distance from the centre of the pond and r'(r) is the concentration gradient. The equation can be integrated as:ln(r'(r)) = ln(A) − ln(r),where A is a constant of integration which can be determined using boundary conditions.At the surface of the pond, oxygen concentration is CAo and at the bottom of the pond, oxygen concentration is zero, therefore:r'(R) = 0 and r'(0) = CAo.The above equation becomes:ln(r'(r)) = ln(CAo) − (ln(R)/L)*r.Substituting for A and integrating we have:CA(r) = CAo*exp(-r/L),where L is the characteristic length of oxygen concentration decay in the pond. The value of L will be equal to the square root of the diffusion coefficient of oxygen in water times the reaction rate constant, i.e. L = √DAB/k.
b. Steady-state oxygen consumption rate in the pond: Oxygen consumption rate in the pond can be calculated by integrating the rate of oxygen consumption across the pond surface and taking into account the steady-state oxygen concentration distribution obtained above. The rate of oxygen consumption at any point in the pond is given by:rA = −kCA.
The rate of oxygen consumption at the pond surface is given by: rA = −kCAo.
Integrating the rate of oxygen consumption across the pond surface we have: rA = −k∫∫CA(r)dS = −k∫∫CAo*exp(-r/L)dS.
Integrating over the surface area of the pond and substituting for the steady-state oxygen concentration distribution obtained above we have: rA = −kCAo*∫∫exp(-r/L)dS.
The integral over the surface area of the pond is equal to S and the integral of exp(-r/L) over the radial direction is equal to 2πL2.Therefore,rA = −kCAo*S*2πL2. The steady-state oxygen consumption rate in the pond is given by:Q = S*rA = −S*kCAo*2πL2.
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A geothermal power plant uses dry steam at a temperature of 308 °C and cooling water at a temperature of 23 °C. What is the maximum % efficiency the plant can achieve converting the geothermal heat to electricity?
The maximum efficiency the geothermal power plant can achieve in converting geothermal heat to electricity is approximately 49.09%
The maximum efficiency of a heat engine is determined by the Carnot efficiency, which depends on the temperatures of the hot and cold reservoirs. In this case, the hot reservoir is the geothermal steam at 308 °C (581 K), and the cold reservoir is the cooling water at 23 °C (296 K).
The Carnot efficiency (η_Carnot) is given by the formula:
η_Carnot = 1 - (T_cold / T_hot)
where T_cold is the temperature of the cold reservoir and T_hot is the temperature of the hot reservoir.
Substituting the given temperatures:
η_Carnot = 1 - (296 K / 581 K)
η_Carnot ≈ 0.4909 or 49.09%
Therefore, the maximum efficiency the geothermal power plant can achieve in converting geothermal heat to electricity is approximately 49.09%
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Why did the flame of a candle go out when a jar was put on top of it
These byproducts can accumulate within the closed jar, further contributing to the depletion of oxygen and ultimately causing the flame to go out.
When a jar is placed on top of a candle, it creates a closed environment within the jar. This closed environment leads to a depletion of oxygen, which is necessary for combustion to occur. As the candle burns, it consumes oxygen from the surrounding air to sustain the flame.
When the jar is placed over the candle, it limits the availability of fresh air and restricts the flow of oxygen into the jar. As the candle burns and consumes the available oxygen, it eventually uses up the oxygen trapped inside the jar. Without sufficient oxygen, the combustion process cannot continue, and the flame extinguishes.
Additionally, the combustion process produces carbon dioxide and water vapor as byproducts. These byproducts can accumulate within the closed jar, further contributing to the depletion of oxygen and ultimately causing the flame to go out.
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The number of moles of CO² which contain 8. 00g of oxygen is