| D. E. Richards | |
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| Prob | Answers | Hints |
| 8.1 |
(a) 254.4 Btu/h out of system (b) 0.45027 Btu/(h-oR) (c) 0.029735 Btu/(h-oR) (d) 0.48000 Btu/(h-oR) |
In each part of this problem you are applying the entropy
and energy balance to a closed, steady-state system. Again be careful to use
the temperature of the boundary where heat transfer occurs in K or oR
(absolute temperature) to evaluate the entropy transfer rate with heat
transfer.
(a) System is just the gear box. |
| 8.2 |
700 kW electric power output -2.50 kW/K entropy production rate |
Apply entropy accounting principle and conservation of energy to the closed, steady-state system. Whether the system is possible depends on the value of the entropy production rate. What's the significance of a negative entropy production rate? |
| 8.3 | ? ? ? | Apply conservation of energy and the entropy accounting principle to this device. FYI - a transformer is a device that take an input of electrical power at a given voltage and current and transforms it into an output electrical power at a different voltage and current. Step-down transformers that reduce the voltage and increase the current are found everywhere in the electric grid that supplies your home. Unfortunately this transformation process is not perfect and some of the input energy is lost by heat transfer from the system. This also results in entropy being generated. (What would the output power be if the transformer was internally reversible?) Remember to be careful with the signs for the energy and entropy transfers and to use absolute temperature when calculating entropy flows with heat transfer. (This applies to all entropy problems.) |
| 8.4 |
Shaft power output = 2.7 kW Entropy production rate = 0.9434 W/k |
Apply both the entropy balance and the energy balance to a closed, steady-state system. Be careful to use the temperature of the boundary where heat transfer occurs in K or oR (absolute temperature) to evaluate the entropy transfer rate with heat transfer. |
| 8.5 | ||
| 8.6 | ? ? ? | Application of conservation of energy and entropy accounting to two gearboxes connected in series, i.e. the output of the first gearbox is the input of the second gearbox. Each gearbox has a single input power shaft and a single output power shaft. Assume steady-state behavior. |
| 8.7 | Impossible | Any real system must satisfy both the conservation of energy principle and the entropy accounting principle. |
| 8.8 | (a) COP = 4.0 (b) 6.36 Btu/(h-oR), Irreversible & Possible |
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| 8.9 | (a) Power in = 13.1 hp Entropy gen. rate = 51.9 Btu/(h-oR) (b) Best COP = 12.7 |
(a) Apply definition of COP for a heat pump. Then apply
the conservation of energy and entropy accounting principle to the heat
pump system. (b) Now assume ideal performance, i.e. no entropy generation, and calculate ideal COPHP using known boundary temperatures. Then use definition of COPHP to solve for power (c) Substitute the known information, for actual power, ideal power, and entropy production rate into the given relation and check out the equality. |
| 8.10 |
(a) ηth = 25.0 %; (b) ηth = 55.0 %; (c) ηth,ideal = 40.0
%; (d) What do you think? |
(a) Apply definition of
thermal efficiency and simple entropy balance for a cycle.
(b) Apply conservation of energy, simple entropy balance, and definition of thermal efficiency. (c) Apply definition of Ideal Efficiency for a Heat Engine, and simple energy balance and entropy balance for the cycle. |
| 8.11 | (a) 22.0 %; 22 MW (b) 33.5 % (c) 31.5 % |
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| 8.12 | ||
| 8.13 | ||
| 8.14 | ||
| 8.15 | (a) s2
– s1 = 0; 30 kW (b) 50.9 kW; 0.0697 kW/K |
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| 8.16 | (a) T2s = 474oC,
V2s = 686 m/s
(b) Entropy gen. rate = 0.204 kW/K (c) ??? |
(a) This is a hypothetical process. Apply the
conservation of energy and entropy accounting principle to the system.
If the process is adiabatic and internally reversible, what does that
tell us about entropy generation term? Then what does this say about the
entropy change s2-s1 for the
process. Now that you know the value for s2-s1,
use the ideal gas relation for entropy change to relate the known values
T1, P1, and P2 to
the temperature T2s -- they hypothetical outlet
temperature if the process had been adiabatic and internally reversible.
(b) Now use the actual outlet temperature given here to find the actual operating conditions for the process. (c) Now compare the results from the two processes. Which one is better? How could I improve the performance of the actual nozzle. |
| 8.17 |
(a) 17560 kJ (b) Sgen = 8.91 kJ/K (c) Sgen = 56.4 kJ/K |
(a) Finite-time closed system
including water and resistor.
(b) Finite-time closed system including just water. Must also consider resistor separately to see what heat transfer is between resistor and water (c) System is water plus the resistor, i.e. there is no heat transfer for this system. (d) Look at the various systems and how they are related. |
| 8.18 | ||
| 8.19 |
(a) 8508 kJ (b) 0.0013 K |
(a) Apply conservation of
energy and entropy accounting to the closed system containing the gas and
apply the appropriate modeling assumptions for the system and the process.
(b) Actually found as part of (a). Unrealistically low as water vapor in the "air" would undoubtedly condense and cv values could vary drastically over this large a temperature difference. |
| 8.20 | ||
| 8.21 | 987.8 kW | Write conservation of mass and energy and the entropy accounting equation for the open system indicated on the drawing. Now what's known and what's variable? Solve the entropy equation for the heat transfer rate to the heat exchanger and then substitute this into the energy equation. Now you have an equation for max power out where the only variable is the entropy generation rate. Vary the entropy generation rate over its allowed values and determine the maximum value for the net power out of the system. |
| 8.22 | (a) 300 K and 100 kPa (b) 0.0436 kJ/K |
Consider closed system that includes all of the gas within both tanks. (If your boundary is outside the tanks, your closed system will include the mass of the tanks. Unfortunately we know nothing about the mass of the tanks so the problem would be unsolvable. However, if we either neglect the mass of the tanks with the larger system or assume that the adiabatic boundary is inside the tank wall we can solve the problem.) Note that as expected, entropy was generated during this irreversible mixing process. (Is it possible to have a reversible mixing process? What would happen to the entropy generated if the initial tank pressures only differed each other by a very small difference? What if there was no difference in pressure?) |
| 8.23 | (a) 1008oR (b) 184.7 psi (c) 138 ft-lbf |
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| 8.24 | ||
| 8.25 |
(a) 2918.5 kW into the system by electric work; entropy production rate = 6.940 kW/K (b) 2083 kW into the system by heat transfer; 836.5 kW into the system by electric work; Entropy generation rate is zero since system is internally reversible. (c) Mode I -- $1868/day |
(a) Open system has one inlet,
one outlet and electrical power crossing the boundary. Apply conservation of
energy and entropy accounting to this system.
(b) Same system as for Part (a) only this time there is heat transfer at boundary with temperature To. Apply entropy accounting equation with zero entropy generation rate to determine the amount and direction of heat transfer under these conditions. Then use energy balance to calculate electric power required. (c) For Mode I all "heating" is being done by the irreversible conversion of electric energy into the change in state of the air flowing through the heater. For Mode II, the electric power only provides part of the energy for heating with the majority coming from the surroundings through heat transfer at the boundary of the system. If one were to construct these two systems, Mode I would be a simple electric resistance heating element and Mode II would be an electrically driven heat pump. |
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| Homepage URL:
www.rose-hulman.edu/~richards
Revised -- 27 September 2002 -- by D. E. Richards |
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