New to the forum - Stored energy to melt ice

[chiggs24]

Hi All,

New to the forum, but requiring a certain amount of assistance if possible.

In relation to ice defrost time, 334 Joules per gram is required to change the state of ice to water and 2.03 joules per gram is required to raise the temperature of ice by one degree?

With this in mind, is it safe to calculate, that hypothetically, if you have around 2kg of ice and the ambient temperature is minus 5degreesC, in order to melt the ice, you require 20,300J to raise the temperature to 0, then 668,000J to complete the phase change to water, which equates to a total of 0.191194kWh?

If a 50litre vessel of water held at 50 degrees C indoors (20 degree room temperature) contains 1.75kWh, can you then calculate that the vessel will give up 0.20416kW in 7 minutes, enough to melt the 2 kilos of ice?

Im not a physicist, so if any of the above calculations have been wildly misunderstood, please do not hesitate to correct me :)

Thank you all in advance

 

[Andrew Mason]

Hi All,

New to the forum, but requiring a certain amount of assistance if possible.

In relation to ice defrost time, 334 Joules per gram is required to change the state of ice to water and 2.03 joules per gram is required to raise the temperature of ice by one degree?

With this in mind, is it safe to calculate, that hypothetically, if you have around 2kg of ice and the ambient temperature is minus 5degreesC, in order to melt the ice, you require 20,300J to raise the temperature to 0, then 668,000J to complete the phase change to water, which equates to a total of 0.191194kWh?

Welcome to PF. This part is correct.

If a 50litre vessel of water held at 50 degrees C indoors (20 degree room temperature) contains 1.75kWh, can you then calculate that the vessel will give up 0.20416kW in 7 minutes, enough to melt the 2 kilos of ice?

Where do you get the 7 minutes from? The rate of heat flow will depend upon the way the ice is put in thermal contact with the water. A very thin sheet of ice spread over a 5 m^2 surface of the water 1 cm deep will melt faster than a 2kg ice cube.

Also, 50 L of water at 50 degrees contains more than 1.75 kWh of internal energy. I assume you are referring to the heat energy required to raise the temperature of 50L of water from 20C to 50C.

AM

 

[chiggs24]

Hello Andrew,

Thank you for the quick reply.

The scenario is based upon a plate heat exchanger which has a coating of ice over it. Heated water from a buffer vessel will pass through the heat exchanger in a bid to defrost the formed ice.

The 7 minutes was calculated by; (1.75kWh/60mins)*7mins = 0.20416kW? > 0.191194kW?

In relation to the energy in 50L, I was unintentionally referring to the energy required to heat 20 to 50 degrees C. If this is the case, how do I calculate the energy contained within amount of water n at temperature n?

I would eventually like to get to a point where I can utilize average local relative humidity and prolonged low ambient temperature to calculate how much energy is required to melt ice which will have collected on a Air Source Heat Pump Evaporator.

 

[Andrew Mason]

Hello Andrew,

Thank you for the quick reply.

The scenario is based upon a plate heat exchanger which has a coating of ice over it. Heated water from a buffer vessel will pass through the heat exchanger in a bid to defrost the formed ice.

The 7 minutes was calculated by; (1.75kWh/60mins)*7mins = 0.20416kW? > 0.191194kW?

Why not use 7 seconds, or any time at all? (You are dealing with 6300 KJ. of energy. You only need 688.3 KJ to melt the ice at -5C. You could express 6300 KJ as 6.3 MWs. Since you only need .688 MWs to melt the ice, would you then only need .688/6.3 = .1 second to melt the ice?)

There is enough internal energy in 50 L of water at 50C to melt 2 Kg of -5 degree C ice. So the question is how fast can you get that heat energy to flow into the ice? To get maximum rate of heat flow you need maximum thermal contact between the water and the ice.

In relation to the energy in 50L, I was unintentionally referring to the energy required to heat 20 to 50 degrees C. If this is the case, how do I calculate the energy contained within amount of water n at temperature n?

If you want to measure the amount of energy required to heat 50L of water at absolute 0 to 50C (323K) it is very complicated. But that would be the amount in internal energy it contains. You don't need to know that.

I would eventually like to get to a point where I can utilize average local relative humidity and prolonged low ambient temperature to calculate how much energy is required to melt ice which will have collected on a Air Source Heat Pump Evaporator.

You have to have a lot of information about your system to calculate that. You might be able to ballpark it with some simple assumptions.

AM

 

[chiggs24]

Hi Andrew,

Why not use 7 seconds, or any time at all? (You are dealing with 6300 KJ. of energy. You only need 688.3 KJ to melt the ice at -5C. You could express 6300 KJ as 6.3 MWs. Since you only need .688 MWs to melt the ice, would you then only need .688/6.3 = .1 second to melt the ice?)

How do you get from 6300KJ to mWs? If 1 kW is equal to 3,600KJ, how do you then convert that to mW?
Also, do you not require 20.3KJ to raise the temperature of -5°C ice to 0, then the 688.3KJ to change state to water? Or is it solely the 688.3KJ to go from ice at -5 to water at 0 degrees C?

There is enough internal energy in 50 L of water at 50C to melt 2 Kg of -5 degree C ice. So the question is how fast can you get that heat energy to flow into the ice? To get maximum rate of heat flow you need maximum thermal contact between the water and the ice.

I would be flowing at around 50litres per minute back through the heat exchanger.
I take on board what you are saying about thermal contact. If the layer of ice on face of the heat exchanger is 3mm, potentially, only a small amount of that ice needs to be melted in order for the whole layer to shift and fall off. Likewise, if the ice has worked through the heat exchanger, i.e. between the plates, there may be more mass to melt, but also potentially more thermal contact?

If you want to measure the amount of energy required to heat 50L of water at absolute 0 to 50C (323K) it is very complicated. But that would be the amount in internal energy it contains. You don't need to know that.

You have to have a lot of information about your system to calculate that. You might be able to ballpark it with some simple assumptions.

AM

Would calculating from absolute zero require taking into account the differences in specific heat capacities of the various states (2.03j/g°C for ice and 4.18j/g°C for water)?
If i knew how much energy was available in the vessel and if I calculated how much energy was required to defrost, could I not then store that energy in a vessel?

Thank you again Andrew.

 

[Andrew Mason]

Hi Andrew,

How do you get from 6300KJ to mWs? If 1 kW is equal to 3,600KJ, how do you then convert that to mW?

A watt-second is simply a joule. 6300KJ = 6.3MJ = 6.3MWs.

Also, do you not require 20.3KJ to raise the temperature of -5°C ice to 0, then the 688.3KJ to change state to water? Or is it solely the 688.3KJ to go from ice at -5 to water at 0 degrees C?

668KJ + 20.3 KJ = 688.3KJ

Would calculating from absolute zero require taking into account the differences in specific heat capacities of the various states (2.03j/g°C for ice and 4.18j/g°C for water)?

No.

If i knew how much energy was available in the vessel and if I calculated how much energy was required to defrost, could I not then store that energy in a vessel?

Yes.

AM