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One very simplistic calculation is the following, with the following assumptions: There is no heat insulation of the tank (no double wall etc) the heat tank has infinite strength (will not blow up) The heat tank temperature and its contents is uniform (e.g. the water and the walls of the tank do not have more that a few tenths of degrees K) then you can ...

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Does the copper melt? Possibly, we can't be sure. Assuming 0.25m² of surface for the vessel, this formula for radiative heat transfer and an emissivity of 0.03, at 1068°C (melting point of copper) the vessel radiates about 1.4 kW, less then the 2.5kW input. So ignoring convective heat transfer, it would reach a high enough temperature. We don't know enough ...

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Regarding the question title, How to compute heat produced by 2.5 kW heating element. That's easy. The heat generated will be 2.5 kW. For temperature generated see below. After water is boiled away and bit of steam released. Steam has a volume of 1760 times the water volume. If that's a 1 L tank then that's 1.7 m3 of steam. What would happen to the ...

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The formula for $h_{rad}$ (shown below) is one I found online but I'm not sure if that is correct. You can try turning off radiation entirely to see the effect. You can also try fixing the surface temperature and estimating the relative heat flows by convection compared to radiation, since the two equations are well-defined. Use that estimate to substitute ...

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you take the difference because the force due to pressure acting on the top surface has direction upwards, while the force on the bottom surface has the opposite direction (same direction as gravity). So the two forces cancel each other. Becauce the $F_{top}= P_{Ext}\cdot A_{top}$ and $F_{bot}= P_{Ext}\cdot A_{bot}$ the net force is: F_{net}=F_{bot}- F_{...

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There is a simple reason and a useful added benefit. As you compress air of a fixed mass it occupies less volume. If you give it more space again, it will immediatly decompress and the multiple stages become useless. If you give it just enough space to occupy at its current pressure, the air will compress further. You could reduce space by reducing the ...

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Heat is the transfer of energy from a hotter region to a cooler region. The surface in contact with the earth enjoys the benefit of conductive heating. Being cooler, the energy of the warmer earth will be transferred to the earth-bridge, ostensibly until the energy imbalance is removed. The air-bridge is going to become colder due to the colder air and will ...

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Thanks for a nicely presented first question. First you need a model for the equation of state for the tank. Given the initial conditions, develop formulas that yield static pressure, temperature, work, etc as a function of the mass remaining in the tank. (You need at least one fully defined state point for the tank, I'm assuming it's the initial condition). ...

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To answer your last question, I would classify this process as an unsteady adiabatic expansion through a nozzle. The $\dot W$ term in the energy balance usually represents forms of work other than flow work. Since you have a rigid tank ($dV=0$), there is no other work being done on/by the system, so you can eliminate $\dot W$ from the energy balance. It then ...

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The polytropic equation is derived using the ideal gas EOS, so it would probably not accurately describe a real gas in a thermodynamic process. One way to analyze a non-polytropic process is by solving the energy balance between two states to get the temperature change of the working fluid, then using that temperature change in the equation of state to ...

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