Calculating Energy Requirements for Solar-Powered Equipment Lift

[goli12]

Homework Statement

A scientist wants to life 1000kg equipment tot the top of the hill he is working on, a rise of 50 metres. He sets up a solar energy system with parabolic mirrors that collects sunlight and warms his working fluid to 120°C (the working fluid does not freeze and does not biol in the temperature range over which it is being used). The exhaust temperature is -20°C. What is the minimum amount of energy required to heat his working fluid and run his engine to lift the mass?

Homework Equations

E=mgh

The Attempt at a Solution

Energy required to lift mass=1000*50*9.81=490500J
W=Q(hot)-Q(cold)=E(required to lift mass)+E(required to warm working fluid to 120°C)

 

[Simon Bridge]

So far so good - can you describe where you get stuck and how you are thinking about the problem?
The usual approach would be to sketch a heat-flow diagram and relate the heats to the question.

 

[goli12]

Hi Simon

I'm not sure how to calculate the required energy to warm the working fluid as neither mass or the specific heat capacity. I'm assuming that heat from the working fluid @ 120°C would be the source of energy [Q(hot)] for the engine to use as work and the rest is expelled to the cold reservoir @ -20°C. But then this raises another question, how do i go about calculating the energy flowing to the engine, and how much is going to the cold reservoir? Thinking about it now, Efficiency=1-T(cold)-T(hot)=1-Q(cold)/Q(hot) seems like an important equation that i need to implement. Am I going in the right direction?

 

[Simon Bridge]

You don't need that - you just need to sketch the diagram, label the heat reservoirs, and work out how much heat has to go in the flow arrows at each stage.

So the hot reservoir provides $Q_{in}$. $Q_{out}$ goes to the cold reservoir, while W goes to work.
You need the minimum $Q_{in}$.

Hint:
The amount of energy required to do a particular amount of work depends on the efficiency.
What is the most efficient engine you can possibly get?

 

[goli12]

Since efficiency is 1-T(cold)/T(hot), the most efficient engine i could get in this system would be:

Eff= 1-(253.25K/393.15K) = 0.3558.

So only 35.58% of Q(hot) would be converted to work, while the rest goes to Q(cold)

 

[Simon Bridge]

Since efficiency is 1-T(cold)/T(hot)

For a Carnot Engine - yes.
All other engines are less efficient - so they'd need more input energy to get the same work done.
You add heat to the working fluid - some of that heat gets used to do work and some leaves via the exhaust.

 

[Simon Bridge]

Since efficiency is 1-T(cold)/T(hot)

For a Carnot Engine - yes.
All other engines are less efficient - so they'd need more input energy to get the same work done.
You add heat to the working fluid - some of that heat gets used to do work and some leaves via the exhaust.

 

[goli12]

Ok i think I've got this...

W/Q(hot)=1-(T(cold)/T(hot))
mgh/Q(hot)=1-(T(cold)/T(hot))
Q(hot)=mgh/(1-(T(cold)/T(hot)))
Q(hot)=1.378*10^6J

Is this correct Simon?

 

[Simon Bridge]

That what I'd have done with the same information.
If this is a long answer, you actually want to use the word "Carnot" somewhere in relation to minimizing Q(hot) needed... i.e. link it to the question.

 

[goli12]

Sweet thanks for the help Simon!