liquid cold plate design

Your description is difficult to visualize. If you are trying to keep the cold side cold, why is it in contact with a 250°C environment on the hot side? A diagram of the arrangement would help.

If I understand correctly, he's Using Peltier Modules, presumably to generate electric current. He's keeping the hot side @ 250°C, and the cold side @ 50°C,m for a ∆T of 200C°.

The answer is Yes: provided there is good intimate (large surface) contact between the cold side of the module and the water-cooled heat sink, and the heat sink is made of a material which has good thermal conductivity, then that surface of the module should have a temperature very close to that of the heat sink and its internal fluid.

I suspect you're right about the Peltier modules, though I thought also of something like a heat pipe. I don't know how high a temperature difference (or gradient) can be maintained with such modules, but it seems like 200°C would be quite a stretch. If the OP can clarify....

Without knowing the wattage to be transferred it is not possible to calculate the surface temperature.

If water is the liquid and it is well stirred, you can expect a 1ºC temperature difference between the water and the surface for 0.1 w/cm2 wattage density at the metal surface.

Based on your system characteristics you can estimate the surface temperature using this.

yes I am using peltier module, and also will try to draw schematic and will post it. Now if it is possible then how would i know the amount of heat energy i will have to remove from cold side? I mean, to maintain cold side temp(50 C) I need to remove some amount of heat from it. And my question is how to measure heat transfer from high to cold side, because we are generating electricity so some amount of heat will be used in that and also Thermoelectric device uses array of semiconductor material. So, simple heat conduction principle will not work. I guess I am not explaining my problem properly I would try to submit drawing so can explain in better. thank you very much for all replies

What you are describing does not make sense, except for testing purposes.

Peltier devices have very low efficiency. It will take significantly more electric energy to heat the hot side of the Peltier device than you will get out of it. Unless you can utilize a source of solar, waste or otherwise nearly free heat, and also have a relatively large sink (the ground, a tank of water, etc.) to absorb the heat from the cold side, you will gain nothing.

You have to continuously remove ALL the heat you introduce into the hot side, from the cold side. This is to say you have a flow of heat from the hot side to the cold side, and the Peltier device is able to convert a small part of that heat energy into electrical energy.

Have you investigated the ability of Peltier devices to withstand 250°C for extended periods? Most semiconductors don't last long at that kind of temperature, if they can survive it at all.

Good luck on your project!

Since you are using the device as an electric source, the amount of heat depends on the power drawn from the system. The cold side film coefficient is fixed by the agitation and cleanliness of that surface and not on what is happening elsewhere. If you know the wattage and the area available then it is easy to calculate after assuming a permissible temperature difference between the surface and the bulk water.