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the outside air is colder, but the flame temperature is still much much hotter than that so the difference cold inlet air makes on the outlet temperature of the furnace will be small. The same commentary applies to the density argument: yes, but the effect is small. The fact that the outlet air is at 43 C represents the 8% efficiency loss of your furnace. ...

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In the steady-state case you describe, the thermal conductivity determines how much heat (kW) a material will trasnmit per unit surface for a given temperature difference (K). In heat exchanger applications, the thermal conductivity is one of three heat transfers in series: primary medium to pipe, transfer through pipe, pipe to secondary medium. To my best ...

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There is quite a bit to unpack here, so I may need to deconstruct your question a bit. Your question is surrounding the fouling rate of a Shell & Tube (S&T) heat exchanger which I will get to, but there are a few things with respect to the RO that must be addressed. If you have sand in the feedwater to your RO, you have much bigger issues than heat ...

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I assume you are referring to a exchanger like the following: If : the external dimensions of the heat exchanger remain the same. the tubing diameter remains the same. the number of tubes changes (halves) $n_2 = \frac{n_1}{2}$ In that case like you mentioned the exchange surface will halve, and that will reduce the heat exchange. As a result because the ...

2

you should look into district heating solutions. For your application the lengths are quite small so you probably can get away with using an air duct and a fan to push the air through. However, a better way is (probably optimal) to actually use a air/liquid heat pump which extracts the heat (cools down the air in one building), and heats up water which is ...

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Your specification is that you know the two inlet temperatures, $T_{h,i}$ (hot in) and $T_{c,i}$ (cold in). Interestingly, you also say that you want to cool the hot stream to a defined temperature $T_{h,o}$. So, you have at least three defined parameters for your "real world" case. Let's take also as a given that you are specifying the fluids for ...

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Most of the calculations that I find in books are about calculating the heat transfer rate, which I'm not sure why I would care about. Let's see if that's true. What I imagine is the case in a real world situation is that I have a fluid that I want to cool down from temperature A to temperature B, and I want to find an exchanger that can do the job. What I ...

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It depends on the length of the heat exchanger. generally, in the steady state there will be an steady temperature difference across the length of the heat exchanger. The difference will be smaller for higher lengths. If the length is long enough, then eventually: the out temperature of the hot $t_{h,o}$ will be equal to the inlet of the cool $t_{c,i}$ the ...

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Nothing will change beyond L3. As you increase the length of the heat exchanger, the hot will reach the cool temperature, and beyond that point there will be no change. So beyond L3, the temperature of the cool flow and the hot flow will be equal to the cool flow inlet temperature $t_{c,i}$.

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Example Since the rest of the answer is too long I will start with an example. Imagine you could build a heat exchanger from wooden pipes. If you run through the liquid you would observe very little efficiency (i.e. the hot liquid would remain hot and the cold cold). There would be too little heat exchange, because the wood would not allow heat to pass ...

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I didn't realize how confusing the terms could be for a heat exchanger until I started typing it out: I will refer to the hot side as: the side where fluid leaves cooler than it enters the side that ends up being cooled the side that heat flows out of I will refer to the cold side as: the side where fluid leaves hotter than it enters the side that is ...

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My answer will focus on your statement: Most of the calculations that I find in books are about calculating the heat transfer rate, which I'm not sure why I would care about. Assume you have a hot(ter) fluid with temperature $t_{h1}$, that exits at temperature $t_{h2}$, and a cool(er) fluid that comes in temperature $t_{c1}$, that exits at temperature $t_{... 1 Photons either pass through a given material or they don't. If they pass through, it's as if the material isn't there as far as they're concerned (ignoring refraction). If they get absorbed, then they dump all their energy into the material. This might excite the incident atom such that it then re-emits the photon with the same or different wavelength, such ... 1 Outside air is colder. Does this decrease the maximum temperature of the flames? Your furnace will always need air for combustion. This combustion air must ultimately come from the outside. Combustion air can either come directly from outside (say it's 30F outside) or it can leak into your house, go through the heater and be warmed up from 30F to 70F, and ... 1 You need to consult a phase diagram for nitrogen to answer this. Yes, it could be done, but nitrogen is a terrible refrigerant, and you'll have trouble finding somewhere to discharge heat to at the extreme low temps. https://www.engineeringtoolbox.com/nitrogen-d_1421.html shows the properties. In the graph, you want the area where pressure and temperature ... 1 Background Total energy lost from fluid The heat transfer rate$\dot{Q} $if you know for a pipe with fluid$f$(where f: Red, Yellow, Blue) the temperature at input and output, the mass rate, and the heat capacity of the material is given by: $$\dot{Q}_f = \dot{m}_f\cdot C_{p,f}(T_{f,o}- T_{f,i})$$ So for the Red-yellow pipe you have:$$\dot{Q}_{ry} = -\... 1 Edit to add: The OP states the applicaiton is a heat pipe, so this answer assumed the yellow fluid is a liquid and moves by free convection. It turns out the yellow fluid will undergo a phase change, which changes a lot. This is an outline for a solution that should get you going: Step one - find an equation describing the convection of the yellow medium.$\...

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In order to calculate geothermal pumps, the best place to start is from ASHRAE Technical Committee 6.8. Having gone through the ropes myself, and since you already know the heating loads, I'll provide you with what is -in IMHO- a very useful shortcut. (Keep in mind that this should not be a substitute for a proper geothermal study following the guidelines ...

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