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UPDATED TL;DR: Cast iron has the advantage, because it has a greater capacity to hold heat, and because of its lower conductivity it releases it more slowly, thus its temperature does not fluctuate as much (thus having a more even and steady heat flow) UPDATE: Answering in terms of the advantage (scroll way down) for the following sections Answer to ...


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We cannot obtain negative temperatures in thermodynamics calculations. Only the absolute scales prevent this. The two absolute scales are Kelvin and degrees Rankin. Here are the two calculations to show that both give the same result. I take that $r^{k-1} = 3.45$ based on using your Rankin values. $$T_f = (50 + 459.67)(3.45) = 1763\ ^o\mathrm{R}$$ $$T_i = ((...


4

Use a water phase diagram to work it out. Figure 1. Water phase diagram. Source: SSC Chemistry. If we choose the 100°C and 1.0 atm intersection that represents where you're at when boiling a kettle. Water is undergoing a phase change from liquid to gas as you specified. If you now drop the pressure (going vertically down the dashed line) it should be clear ...


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To my knowledge when you are doing thermodynamic calculations you should always use K (Kelvin), unless stated otherwise in the textbook you are using. Otherwise the results from those relationships (with powers) will always be all over the place. Another way to put it (probably more correct) is, "Always use the units in the textbook. When in doubt use ...


3

Could you get it to work, sure. There are dozens of ways to turn heat into energy. Could you get it to work at scale and cost effectively? Almost certainly not. There's just not enough energy in heated air to do much, and the capital costs of the system break the effort. This is another entry into the "Why can't Stirling engines give us almost free ...


2

The energy equation can be simplified to the two equations you've referenced, given that in the first case there is negligible work, in the second case there is negligible heat transfer, and in both cases there is negligible viscous dissipation. For example, consider a heat exchanger with negligible inlet and outlet velocities and no elevation difference ...


2

Start with the first law. $$dU = \delta q + \delta w$$ Isentropic sets $dS = \delta q /T = 0$, giving $dU = \delta w$. For an ideal gas with a constant specific heat, $dU = C_v dT$. This gives the equation being used. $$\Delta U = C_v \Delta T = w $$ If temperature decreases, work is negative, meaning it is done by the gas. Alternatively, if temperature ...


2

You are adding heat at a constant rate $\dot{q}_o$ (W) to vaporize an amount $n$ (moles) water at a constant temperature. When you drop the external pressure, you also drop the partial pressure of the water vapor in the surrounding gas. This increases the thermodynamic driving force for the water to evaporate (Le Chatelier's principle should come to mind ...


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Your calculations are probably wrong. Peltier effect coolers are known to be extremely inefficient in converting electrical power into useful refrigeration. This is why they are not commonly used in things like home refrigerators and air conditioning units, where heat pumps are far more effective. The figure of merit for refrigeration efficiency in things ...


1

A useful measure for this comparison is the COP (Coefficient of performance) which measures ratio between the heat energy transferred over the energy used. The following is a theoretical prediction of the COP a Peltier element, with respect to temperature difference. For a split unit heat pump, you can find different values but in most cases, the COP of the ...


1

The fundamental relation for any closed system can be written as $$dU=\left(\frac{PC_V}{T\alpha K}\right)dT+\left(T-\frac{P}{\alpha K}\right)dS,$$ where $U$ is energy, $P$ is pressure, $C_V$ is the constant-volume heat capacity, $T$ is temperature, $\alpha$ is the thermal expansion coefficient, $K$ is the bulk modulus, and $S$ is entropy. If heating is ...


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You can't do it with conservation of energy alone: you need another physical principle, namely conservation of momentum in the form of the Navier-Stokes momentum equation (or, if the shear stresses are negligible, the Euler momentum equation). That conservation of momentum equation will give you an expression for the surface traction density at any given ...


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