Grain elevator. You need to convince your boss that this is a very inefficient system.

Transcription

1 Grain elevator You are working for a grain storage comany over the summer, and they have a roblem with the grain elevator, that kees breaking down. This morning, your boss woke u feeling like a genius and decides that he is going to design one on his own, so he gathers the engineering grou in his office to exlain his idea. The rojected elevator consists of a latform mounted on a iston that can move along a cylinder. The cylinder is oen to the room at the to. The lower art of the cylinder is also full of air, but the tight fitting iston seals this chamber, so that no air can escae. The iston has a cross-sectional area of 9.8 m. The elevator begins unloaded. The air inside the air chamber starts at room temerature (5C) and a ressure of 0 kpa. Then, 0 tons of grain are loaded on the latform. During the loading rocess, the thermometer remains at 5C and the iston sinks. The air in the chamber is then heated u to 75C, and the elevator rises slowly and steadily. This high temerature is ket constant while the elevator is unloaded. Finally, the air is cooled back to room temerature, and the iston returns to its starting level. You need to convince your boss that this is a very inefficient system. Hint: In your calculation for the absorbed heat, consider only the heat absorbed by the air in the chamber while the temerature is being increased. In reality, heat is also absorbed while the grain is unloaded a constant temerature, but this turns out of be much smaller, and makes the calculations significantly more comlicated. nd, in any case, taking this heat into account would only make the efficiency even worse! Loading latform Cylinder Piston ir chamber

2 Solution The first thing we need to do is identify the thermodynamic cycle of the system. There are 4 rocesses (, B, C and D) in a cycle that takes the system through 4 states (,, and 4) that we can identify from the text: Process ( to ): The grain is loaded while the temerature remain the same. The iston sinks, i.e., volume decreases. Pressure increases. This is an isothermal comression. Process B ( to ): The iston is lifted because the air is warmed u. Because the cylinder is oen to the room at the to and the latform raises steadily (constant velocity), we conclude that the ressure must remain constant the entire time. This is an isobaric exansion. Process C ( to 4): The grain is unloaded at constant temerature. The iston goes u and the ressure decreases. This is an isothermal exansion. Process D (4 to ): The air cools down and the iston goes down. gain, the ressure must remain constant. This is an isobaric comression. B D C 4 W Our goal is to comute the efficiency of the cycle: e, so we need find the work done Q during the cycle, and the heat absorbed from the energy source (whatever warms u the air during rocess B). in

3 . Work The total work is the sum of the work in all four rocesses: W W WB WC WD For an isothermal rocess, f f nrt f W d d nrt ln i i. For an isobaric rocess, W f i nrtf Ti Using these exressions: W nrt ln. W nr T T nr T T ( is an isothermal, so T T ) B 4 WC nrtln W nr T T nr T T (C is an isothermal, so T T ) D 4 4 i Note that WD WB, so the total work is: W W W nrt ln nrt ln 4 C Since both and C are isothermals, [] 4 and because B and D are isobarics. Therefore,. But 4 With all this, we can rewrite the total work as: W nr T T ln 4

4 The increase in ressure in rocess is due to the weight of the grain added onto the latform: [] where M is the mass of the grain and is the area of the iston. Taking this into account, our final exression for the work is: W nr T T ln []. Heat Heat is absorbed during rocesses B and C. s suggested in the hint, we are mostly concerned with rocess B. Process B is an isobaric exansion of a diatomic ideal gas (air is 99% N and O, both diatomic): where we used that C 7 QB nc T T nrt T, [4] 7 R for a diatomic ideal gas. We are now ready to comute the efficiency: Using the given numerical values: M 0 kpa 0 0 kg 9.8 m, nrt T ln W W e ln 7 QB QC QB nr 7 T T we obtain a very disaointing efficiency (even without the heat absorbed in rocess C!) e (4.4%) Less than 4.4% of the energy that we ay for is actually transformed into work.

5 few more numerical values can hel us better understand why the efficiency of this engine is so low. Note that we are given the initial temerature and ressure. We do not know the initial volume of the air chamber,. How big a chamber do we need? ll the remaining volumes in the rocess are related to the initial one. s shown above in [], since rocess is isothermal, we have: The required change in volume from states to is what takes the grain u. If the iston (of area ) goes u a height h, then nd thus, h h [5] On the other hand, simly using the ideal gas equation for states and : But =, so T T T [6] T Combining [5] and [6], we obtain: T h T T h T T h T T [7]

6 For the rojected elevator, we had: M 0 kp 0 0 kg 9.8 m T 5C 98K T 75C 48K Equation [] gives us a ressure for states and of: 40 kpa If we wanted to take the grain u a height of, say, h = 0 m, equation [7] tells us that the required initial volume for the air chamber is: 40 kpa 98K 9.8 m 0 m 68 m 0 kpa 50K That is comarable to the room you are sitting in!!! In order to increase the temerature of all that air by 50K, we are going to need a LOT of heat, and that s a big rice to ay for so little work. You can try to lay around with the numbers a bit. If we decrease the initial volume, we obtain a ridiculous elevation h, or we need to resort to such high temeratures and ressures that the system is likely to exlode and/or melt. Engineering ain t that easy, tell your boss. In case you want the actual values: the number of moles is now known, n, which you RT can substitute into [] and [4], and you ll obtain that you need 48 MJ of heat to get a meager MJ of work.