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Hello, After Favre averaging the momentum equation for an inviscid flow, the following can be obtained: $$\frac{\partial}{\partial t} \left(\overline{\rho}\tilde{u}_i \right) + \frac{\partial}{\partial x_j}\left( \overline{\rho}\tilde{u}_i \tilde{u}_j \right) + \frac{\partial...

Starting to become clearer. How did you obtain ##T_{LC} = \frac{(T_H + T_C)}{2}## though?

Thanks Chestermiller! My lecturer actually drew a picture like this when he attempted to explain it. Where THC and TLC are the high and low temperatures of the system respectively. "To maximise the heat transfer rate in, let THC = TLC. But then Qin = Qout." I don't really understand this...

No the paper says isothermal processes. I'm still trying to decipher what this means.

Hey guys, I ran into this paper talking about the Maximum power you can obtain from a Carnot cycle: http://aapt.scitation.org/doi/abs/10.1119/1.10023 From what I understood, there are two extremes. To achieve maximum efficiency you have to make sure that the temperature of the system is never...

Thank you for that! I understand it much better now! In my thermo course, I think we just used steam tables for pure substances and the ideal gas equation for air. So I'd never dealt with a situation where the temperature was a function of time. I'm doing a second course later this year which...

Ah! Sorry Here goes, $$\dot{Q} + \dot{m}(h_1 - h_2) = \dot{m}(u_2 - u_1)$$ Heat is from convection so, $$hA_c(T_h-T_o) = \dot{m}(c_p + c_v)(T_o-T_i)$$ Is that correct? Why would ##T_i## be a constant but not ##T_o##?

$$\dot{Q} - \dot{W} = \sum_{out} \dot{m} (h+ke+pe) -\sum_{in} \dot{m} (h+ke+pe) $$ Hence, $$\dot{Q} = \sum_{out} \dot{m} h -\sum_{in} \dot{m} h $$ So for convection, $$hA_c(T_h - T_o) = \dot{m}(h_2-h_1) = \dot{m}c_p(T_o-T_i)$$? Why would ##T_i## be a constant but not ##T_o##??

Thanks Chestermiller. The way I derived your expression is, $$\dot{Q_{in}} - \dot{Q_{out}} = \frac{d(mc(T_o - T_i))}{dt}$$ but there is no ##\dot{Q_{out}}## if you consider the inner chamber as the system and there is no heat loss to the surroundings. So, $$\dot{Q_{in}} = \frac{d(mc(T_o -...

Hey Chestermiller, could you provide more insight into how you derived that equation? Reading online I've only seen the expression $$Ah(T-T_{\infty}) = -\rho Vc \frac{dT}{dt}$$ after googling transient heat convection transfers.

Ah! Thanks so much!

I'm still not quite convinced. Why would ##T_h - T_0## be the overall temperature difference?? Would the "overall" temperature difference be the average temperature difference? Which is $$T_h - \frac{T_0 - T_i}{2}$$ ?

Hello Chestermiller; really apologise I didn't fill out the homework template. What does the over heat transfer coefficient ##h_c## have anything to do with the temperature difference being ##T_h - T_o##? I would argue that the temperature difference used in ##q = h_c A \Delta T## should be...

Hello guys, in the question attached, my understanding is that there is a heat transfer that heats the fluid from an initial at the input, to at the output. This heat transfer is via convection from walls of temperature . Firstly, . Because the walls are and the fluid is of a lower...