Open System Carnot Efficiency

[ShamelessGit]

Homework Statement

The original is in German, but I will try my best to translate.

A power generator uses the temperature difference between surface ocean temperature and sea bottom temperature (27C and 6C) to generate electricity. The power outage is 50kW at 10% of the Carnot efficiency. The sea water that is pumped up is warmed from 6 C to 7 C. How much mass per second must be pumped from the ocean floor for this process to occur?

The teacher solved this a few weeks ago and I remember that it turns out that the mass necessary actually goes down if you decrease its temperature increase as it goes through the process (for example if it went from 6C to 6.5 instead of 7). I thought this was counter-intuitive, so I tried to solve it again after Christmas break and I don't get the same result.

Homework Equations

kg/s(Δh) = Q/s + Power
Δh/s = C(kg/s)ΔT
Carnot efficiency = 1 - Hot/Cold

The Attempt at a Solution

I'm pretty sure this solution is wrong but I don't know why. The Carnot efficiency is 1- 279/300 = .11, and ten percent of that is .011. I figure the energy being taken from the surface water is responsible for heating the deep sea water and for the 50kW, so I stuck a minus on the P. Also, the heat taken in (Q/s) times the efficiency is supposed to be the work you get out, so P/ε = Q/s = 50kW/.011 = 4545kW of heat from the Ocean. Q - P kg/s(Δh) = 4495kW. Then I figure you just divide both sides by the change in enthalpy to get the mass stream, which means that kg/s is inversely proportional to the temperature increase, which is the opposite of expected.

Plz help. An explanation for why the mass necessary decreases if the temperature increase does would be nice too.

 

[KataKoniK]

Thank you for sharing your question and thought process. It is always great to see students actively thinking and trying to understand a concept. Let me try to help you with this problem.

Firstly, your attempt at a solution seems to be on the right track. However, there are a few things that I think may have caused some confusion. Let me break it down for you step by step.

1. Finding the Carnot efficiency: As you have correctly calculated, the Carnot efficiency in this case is 11%. This means that only 11% of the heat taken from the hot reservoir (surface water) can be converted into work, while the remaining 89% is rejected to the cold reservoir (deep sea water).

2. Finding the heat taken from the surface water: This is where I think you may have made a mistake. The power generated by the power generator is 50kW, but this is not the heat taken from the surface water. Remember that the Carnot efficiency only applies to the heat taken from the hot reservoir, not the total power output. So, to find the heat taken from the surface water, we need to divide the power output by the Carnot efficiency. This gives us 50kW/0.11 = 454.55kW.

3. Finding the mass flow rate: Now, let's look at the equation you have used: kg/s(Δh) = Q/s + Power. This is the correct equation to use, but let's rearrange it to make it easier to understand. We can write it as kg/s(Δh) = Q/s - Power, where Q/s is the heat taken from the surface water and Power is the power output. So, substituting the values we have, we get kg/s(1°C) = 454.55kW - 50kW = 404.55kW. Now, we know that Δh/s = C(kg/s)ΔT, and we know the Carnot efficiency (Carnot efficiency = 1 - Hot/Cold = 1 - 27/6 = 0.77). So, substituting this into our equation, we get kg/s(0.77) = 404.55kW. This means that kg/s = 404.55kW/0.77 = 525.39kg/s. This is the mass flow