Cooling Calculations: 5 Amp vs. 30 Amp Units

[Tylercc]

Homework Statement

If a 5 amp unit can cool 4 cups of water from 75 to 50 degrees fahrenheit in one hour, what temperature could a 30 amp unit cool 128 cups of water in the same amount of time? Is there enough information here to get an idea about what the 30 amp unit would do? If so where would one start?

Homework Equations

The Attempt at a Solution

I am not too sure where to start.

 

[lychette]

cups are not recognised units...this could be a problem but not necessarily. also degrees fahrenheit is not good but not disasterous

 

[Tylercc]

so what units could I convert them to so the problem can make more sense?

 

[gleem]

For this problem the unit of volume is immaterial. How much more energy is necessary to cool 128 cups from 75 deg to 50 deg in one hour? How much more power does 30 amps provide compared to 5 amps?

 

[Tylercc]

For this problem the unit of volume is immaterial. How much more energy is necessary to cool 128 cups from 75 deg to 50 deg in one hour? How much more power does 30 amps provide compared to 5 amps?

So how would I find out how much more energy is necessary to cool 128 cups? An extra 100 watts (The 30 amp unit has an output cooling power of 253 watts compared to the 5 amps which has 154 watts.) So sense the difference is 100 watts, what equation would I use to find out what kind of a difference that would make?

 

[gleem]

From where did you get your power figures? The currents given do not reflect your power difference.

 

[Tylercc]

In the section where it says product description.
https://www.amazon.com/dp/B00EQ1X5EC/?tag=pfamazon01-20
https://www.amazon.com/dp/B0105Y8TK8/?tag=pfamazon01-20

 

[gleem]

Not to drag this out but where did the currents 5 and 30 amps come from?

 

[Tylercc]

same box.

 

[haruspex]

Is there enough information here

Strictly speaking, no. The work needed to cool a body over a given temperature range depends on the ambient temperature.
If we overlook that, this becomes a very simple question of ratios. With 32 times as much water to cool, how much cooling do you think the 5 amp unit could achieve in the hour?

 

[gleem]

Ok I think I get it. These are two different types of cooler and must be running at different voltages.

The current spec is immaterial. It is the power spec that is important. To cool anything one must provide energy to do the work of removing heat. If the 5 amps is associated with the 154 watt cooler and the 30 amps with the 253 watt cooler then assuming that they are equally efficient the higher power cooler will remove heat how much faster than the lower power unit? How much more water are you cooling?

 

[Tylercc]

Strictly speaking, no. The work needed to cool a body over a given temperature range depends on the ambient temperature.
If we overlook that, this becomes a very simple question of ratios. With 32 times as much water to cool, how much cooling do you think the 5 amp unit could achieve in the hour?

I would estimate maybe .5 to 1 degrees per hour.

 

[Tylercc]

Ok I think I get it. These are two different types of cooler and must be running at different voltages.

The current spec is immaterial. It is the power spec that is important. To cool anything one must provide energy to do the work of removing heat. If the 5 amps is associated with the 154 watt cooler and the 30 amps with the 253 watt cooler then assuming that they are equally efficient the higher power cooler will remove heat how much faster than the lower power unit? How much more water are you cooling?

I am trying to cool 8 gallons of water, and the speed at which that happens is not too important to me, I was just using 1 hour to try and get some ratio of what I might be able to expect

 

[gleem]

Keep in mind that the currents given in this problem are for two different types of cooling units and not related to actual cooling ability. Use must compare the actual power ratings of the units not the currents. So you use 32 times more energy but do it at a rate that is only about 1.64 ? times the lower power unit.

 

[Tylercc]

Keep in mind that the currents given in this problem are for two different types of cooling units and not related to actual cooling ability. Use must compare the actual power ratings of the units not the currents. So you use 32 times more energy but do it at a rate that is only about 1.64 ? times the lower power unit.

well this is turning out to be a lot more complex than I thought, you lost me on your last sentence.

 

[gleem]

sorry, 8 gals is 32 time greater than 4 cups. so you need to remove 32 time more heat(energy). What takes 1 hr to cool 4 cups will take 32 time longer with the same unit. But the bigger unit has more power and can cool things 253/154 =1.64 times faster.or take only 1/1.64 = 0.61 times as long to cool the 8 gals.

 

[Tylercc]

so if we make some big assumptions we could guess that if I gave the 8 gals 2 hours to cool with the bigger unit we might be in the 50 degree range?

 

[gleem]

not quite, as an analogy if you took 1 hour to go 4 miles at 4 mph how long would it take for you to go 128 miles at 6.56 (=1.64x4) mph?

 

[Tylercc]

19.5 hours to travel 128 miles. So it could take all day?

 

[gleem]

That's what I'm thinking. And since cooling device is a plate taking heat only from the bottom you mus take into account that the heat at the top must be transported to the bottom further delaying its removal.

 

[Tylercc]

So I was actually thinking of building one of these

which I am sure adds some complications to the whole concept, but at least I now know what I might be getting in to. thanks for the advice.

 

[haruspex]

Use must compare the actual power ratings of the units not the currents.

Are they not operating off the same voltage?
@Tylercc, if this is a practical question, not a homework problem, then you should consider the first line in my post #10.
But surely you would want the same cooling range and want to know how long it will take, not how much cooling would be achieved in a given time?

 

[Tylercc]

Are they not operating off the same voltage?
@Tylercc, if this is a practical question, not a homework problem, then you should consider the first line in my post #10.
But surely you would want the same cooling range and want to know how long it will take, not how much cooling would be achieved in a given time?

So according to amazon they are operating off the same voltage, and I have consider the first line in your post which is why if I build the cooling device it will most likely turn out to be just a fun project, instead of being useful for my cooling needs. And sure it would be nice to have the same cooling range, but because of the project I was going to put the peltier device in at the end I was not really expecting that to happen. Also the end project is more about how cold the water can be not how long it takes.

 

[haruspex]

So according to amazon they are operating off the same voltage, and I have consider the first line in your post which is why if I build the cooling device it will most likely turn out to be just a fun project, instead of being useful for my cooling needs. And sure it would be nice to have the same cooling range, but because of the project I was going to put the peltier device in at the end I was not really expecting that to happen. Also the end project is more about how cold the water can be not how long it takes.

Ok.
As it gets colder, the work required for each further degree of cooling will increase. The standard thermodynamic equation requires you to know the temperature of the available heat sink. But that equation does not allow for any sense of urgency. E.g. if the heat sink is cooler than your target temperature it will tell you you do not need to do any work at all, just let it cool down. Indeed, in that situation, you would probably do better just to pump the water through a heat exchanger rather than do any active cooling.