Thermodynamics Lab Report 1 Results & Conclusion PDF

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Results & Discussion Provided Data Unit Load Cell N Motor Speed RPM Water flow Rate l/hr Air Flow Rate l/min Motor Power Watts P1 kPa P2 kPa T1 °C T2 °C T3 °C T4 °C T5 °C T6 °C T7 °C System Definitions & Rate of Energy Losses

Average 22.17 1253.67 29.03 115.4 652.23 102.33 246.77 22.97 102.83 68.67 23.13 21 23.87 22.33

Note: All equations and full working out procedures can be found in the appendices.

Compressor System (Boundary A) This first system contains the Compressor and the Belt Drive. The compressor functions by exerting work done externally in order to increase pressure internally. This produces heat, on account of the increasing pressure. This process is explicitly steady state steady flow, and since it also has a single flow stream passing through the system, the continuity equation can apply. This process can be assumed to be Steady-State Steady Flow, so therefore

( dEdt )

=0 . Whilst the work boundary

CV

of this system is moving, the work done is still negative, and therefore the heat transferred is also negative. We are allowed to assume single flow of air, due to the boundary conditions, resultant known flow rates, and the SSSF condition. As per the continuity equation, the mass flowing in is equal to the mass flowing out, proven by the known single flow rates and SSSF condition.

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1 ´ ´ A =m + Q ( V 2−V 12) + 1 ×9.81 ×( Z 2−Z1 ) + W´ air ( h2 − h1 ) 2000 2 1000 Q´ A =−1.46777 ×10−2 kW It’s important to note the difference of heat transferred with potential and kinetic energy is negligibly different to the heat transferred without incorporating potential and kinetic energy. This is proven in the Significance of Energy Changes in the Compressor. The rate of energy loss within this system is the smallest out of all of the systems by a fair margin.

Heat Exchanger System (Boundary B) This system only contains the Heat Exchanger. This system’s function is to exchange the heat energy from air to water, or any other form of coolant. The one style of heat exchangers (closed style) has an input and output, used to heat or cool the fluids flowing through. Since they are kept separate, the process is steady state and steady flow. Open style heat exchangers do not follow this rule.

This process can be known to be Steady-State Steady Flow, so therefore

dE dt

( )

=0 . The work boundary of this

CV

system is non-moving, therefore the assumption of the work transfer across the boundary equal to 0 is proved. Finally, similar to the compressor system, the kinetic and potential energy changes are proven negligible, and are omitted from further equations.

´ B=m ´ water ( h6 −h5) +m´ air (h 4−h3 ) Q ´ B = (8.04391 ×10−3 ×12.01382 ) +( 2.0645 ×10−3 ×− 45.72 ) Q ´ =−0.02249 kW Q B This result makes sense with the operational goals of the heat exchanger. A negative sign means while the water temperature may be going up (+12.01 enthalpy change), the air temperature is cooled significantly more (-45.72 enthalpy change), therefore the assumption of steady-state steady-flow is reasonable, and so heat energy is lost.

Overall System (Boundary C) This is the overall system, containing systems A, B and D, as well as various other components, notably the receiver, inputs, and outputs. Each internal system is considered to be independent if there is no work acting upon it. For this reason, this larger system has the 3 smaller systems within, which are all SSSF on their own. This system is defined as open, as mass is able to cross this boundary and enter or exit. The assumption of steady state steady flow still applies here, as the system does not accrue mass or energy over time, due to the fact both Q and W are negative. The continuity equation also applies, as the mass flow rates are conserved between system transmissions. Finally, according to the magnitude found for the kinetic and potential energy changes, they are negligible. Work done in the system is non-zero and negative, so still considered.

´ =m ´ water ( h6−h5 ) − m ´ air ( h7−h 1 ) +W´ Q C ´ C =−0.556689 kW Q This result is the largest of all systems, and this makes sense. The rate of energy change between all systems is smaller than each system combined into one, on account of the fact that no system’s energy transfer is ideal, and energy losses are going to occur.

Electric Motor System (Boundary D) This system is the motor. Energy is produced and output through a motor shaft, whilst also outputting heat. It’s a simple system, and because there’s no flow of fluids, the mechanical transfer of energy is the main source of energy losses.

´ D =−|Pinput|+|Poutput| Q ´ D =−0.1545 kW Q This result is negative, and so proves that energy is being lost. The work is negative, and so is the energy transfer rate.

Significance of Energy Changes in the Compressor The change in kinetic energy in compressor is calculated below. Now this value is relatively insignificant, on account of the fact of its position in the heat transfer equation. This value is added to the much larger specific enthalpy value, which has a value of 80.1784. That’s a difference in magnitude of 10 4.

1 ( V 2−V 12 )=−0.02462405 2000 2 The height difference is found as Z 2 – Z1. Since we don’t have these values, we assume the height as 1m. The significance of potential energy difference can be estimated, expressed below. Similarly to the kinetic energy, the potential energy change is even more distant, with a magnitude of 105. When added to the specific enthalpy value, the effect is negligible.

1 ×9.81 × ( Z 2−Z1 ) =0.00981 1000 Since the difference of heat transferred without potential and kinetic energy in the equation is roughly equal to heat transferred with PE and KE in the equation. This difference was calculated as around 3 x 10 -5. Therefore, the PE and KE can be ignored in this case. The difference can be seen below (without PE/KE and with, respectively). −2 −2 Q´ A =−1.47103 × 10 kW ≈−1.46777 ×10 kW

Significance of Energy Losses in each system Due to the nature of physics, all conversion of energy is guaranteed to not be completely efficient. This holds true for not only the heat transfer, but for other energies such as light, sound, vibrations, friction within the system, resistance of electrical components, and other methods. Since there is a transfer of water and air throughout the entire system, energy is also lost that way. The compressor, the electric motor and the heat exchanger are all losing energy, as their values are negative. Additionally, the entire system is losing energy. This is not strictly only heat energy, but other forms as well. Since all the losses of systems A, B and D are smaller than the losses of C, this means the system as a whole is losing energy externally as well as internally.

´ A =−1.46777 ×10−2 kW Q ´ B =−0.02249 kW Q ´ C =−0.556689 kW Q ´ =−0.1545 kW Q D ´ C≠Q ´ A +Q´ B + Q ´D ∴Q It is reasonable to ignore some of these losses, as their values are much smaller in respect to others. System A has a remarkably small loss of heat energy in comparison to System D. However, the efficiency of the motor is still significantly high, at 76.3%. This means the losses are negligible for the application in most circumstances.

η=

Poutput =76.3 % P input

Possible Causes of Energy Losses The various forms of energy losses are accountable to a range of causes, including convention heat loss, heat radiation loss, pipe friction, and more. Fundamentally, heat convection occurs by a given fluid, either water, air, or other substance, moving within itself. Usually, the heated fluid circulates away from the heat source, due to pressure changes within the substance. This effectively moves the heat away, allowing for the substance to cool down. It then cycles back to a hotter part of the area, as the hotter fluid pushes the cooler fluid back towards the heat source. This is unlikely to occur in the air

compressor system, as the fluids are moving, and are effectively unable to convect. This isn’t accounting for air surrounding the system/s, as this air does circulate, and remove heat energy radiating from the given system. On the other hand, heat radiation is more significant, propagating energy from the heat source in the form of electromagnetic waves. This occurs in any medium, not specifically a fluid. Therefore, it is harder to prevent, with the easiest and main management tactic consisting of using thermally absorbent materials. Regardless, heat energy will be lost over time, given to the cooler surroundings. Another source of lost heat is pipe friction, from the air, water, and mechanically connected components. This is a more mechanical source of energy loss, whereby moving forces are transformed into heat, sound and vibrations, then lost to the surrounding components. Pipe friction is especially hard to prevent, as the fluids moving within the pipe are contacting with minor imperfections, and transferring energy that way. Generally, higher pressure and velocities of fluids within the pipes cause greater loss. In order to prevent this type of energy loss, the pipes need to be especially smooth, and supported by a sufficiently strong structure, to minimise any movement.

Conclusion This experiment was a great insight into understanding the first law of thermodynamics, and how it applies to steadystate steady-flow processes with respect to analysing the energy conservation in an air compressor. Understanding the meaning behind the equations and thermodynamic models of fluid and energy flow was key to analysing the air compressor. The significance of the boundaries across all the systems helped with comprehending what the work and energy transfer would look like. An interesting finding is how the energy losses for the component systems (A, B and D) were smaller than the energy loss for the system as a whole (C) by a significant margin (almost 3x as much). This helps explain the thermodynamics process, especially how all processes aren’t thermally conserving, and will lose energy to processes such as convection and thermal radiation, and mechanical losses such as friction and vibrations. Acquiring experiences with the assumptions of certain characteristics of the boundaries was also achieved throughout this experiment. Several common assumptions made were that the systems underwent a SSSF process, the mass flow rates were conservative between system inputs and outputs, and that the work transfer across the various heat exchanges equalled zero. Assuming the processes were SSSF is critical, as the amount of unnecessary working that would be done would extend and complicate the mathematical model by a significant amount. Another example is assuming the kinetic and potential energy differences in each system is negligible. This calculation is somewhat simple enough, but rather unnecessary. The difference between the results in the compressor system (boundary A) were very minor, and effectively saved time and effort in showing how the effect had on this system were negligible. This is why it’s important to make assumptions, as they simplify the calculations, reducing the chance for mistakes at the cost of accuracy of the thermodynamic model. Developing the requires skills in measurement, computation and analysis wasn’t entirely appropriate due to the nature of the laboratory experience, as there was no physical experience with the air compressor, and as such, the goals relevant to instrumentation and measurement of the air compressor were not realisable. The computation and analysis aspects were achieved in this project, with regards to thermodynamic laws applying to real systems....