Fundamentals Of Engineering Thermodynamics
9th Edition
ISBN: 9781119391388
Author: MORAN, Michael J., SHAPIRO, Howard N., Boettner, Daisie D., Bailey, Margaret B.
Publisher: Wiley,
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Chapter 4, Problem 4.13E
To determine
If the throttling process model still applies when the expansion valve of a refrigerator becomes ice-encased.
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Heat engine and refrigerator. Consider a heat engine operating between temperatures Th and
Tj. During each cycle with time At, work W is extracted, so Pout = W/At.
(a) Assuming the processes are all reversible, what is the efficiency of this heat engine, n =
Wout/Qn?
(b) Now assume that the low temperature of the heat engine is lowered by a reversible
refrigerator, such that the heat engine operates between Th and T. The refrigerator takes
input power Pin = Win/At and operates between T and T, where Ti < T. Draw an energy-
entropy flow diagram.
(c) Calculate the net efficiency (Wout - Win)/Qh. Is the efficiency of this system higher, lower, or
the same as your answer for (a)?
questão1.
Refrigerator 1 simulates an ideal refrigerator and therefore operates on a Carnot cycle using R-134a as the refrigerant at a flow rate of 1.4 kg/sec. The condensing and evaporating temperatures are 30 °C and -10 °C, respectively. To assess the performance of Refrigerator 1, you must submit the report. of the project with the following information: Enthalpies corresponding to the states indicated in the cycle (1, 2, 3 and 4);
1.a. The cooling rate (QL);
1.b. The work provided to the fluid by the compressor;
1.c. The work generated by the turbine;
1.d. The condenser heat rejection rate (QH);
1e. The coefficient of performance of the cycle;
Steady-state operating data are shown in the figure for an open feedwater heater.Heat transfer from the feedwater heater to its surroundings occurs at an average outer surfacetemperature of 50°C at a rate of 100 kW. Ignore the effects of motion and gravity and let T 0 =25°C, p0 = 1 bar. Determine(a) the ratio of the incoming mass flow rates, ?̇# /?̇ $ .(b) the rate of exergy destruction, in kW.
Chapter 4 Solutions
Fundamentals Of Engineering Thermodynamics
Ch. 4 - Prob. 4.1ECh. 4 - Prob. 4.2ECh. 4 - Prob. 4.3ECh. 4 - Prob. 4.4ECh. 4 - Prob. 4.5ECh. 4 - Prob. 4.6ECh. 4 - Prob. 4.7ECh. 4 - Prob. 4.8ECh. 4 - Prob. 4.9ECh. 4 - Prob. 4.10E
Ch. 4 - Prob. 4.11ECh. 4 - Prob. 4.12ECh. 4 - Prob. 4.13ECh. 4 - Prob. 4.14ECh. 4 - Prob. 4.15ECh. 4 - Prob. 4.1CUCh. 4 - Prob. 4.2CUCh. 4 - Prob. 4.3CUCh. 4 - Prob. 4.4CUCh. 4 - Prob. 4.5CUCh. 4 - Prob. 4.6CUCh. 4 - Prob. 4.7CUCh. 4 - Prob. 4.8CUCh. 4 - Prob. 4.9CUCh. 4 - Prob. 4.10CUCh. 4 - Prob. 4.11CUCh. 4 - Prob. 4.12CUCh. 4 - Prob. 4.13CUCh. 4 - Prob. 4.14CUCh. 4 - Prob. 4.15CUCh. 4 - Prob. 4.16CUCh. 4 - Prob. 4.17CUCh. 4 - Prob. 4.18CUCh. 4 - Prob. 4.19CUCh. 4 - Prob. 4.20CUCh. 4 - Prob. 4.21CUCh. 4 - Prob. 4.22CUCh. 4 - Prob. 4.23CUCh. 4 - Prob. 4.24CUCh. 4 - Prob. 4.25CUCh. 4 - Prob. 4.26CUCh. 4 - Prob. 4.27CUCh. 4 - Prob. 4.28CUCh. 4 - Prob. 4.29CUCh. 4 - Prob. 4.30CUCh. 4 - Prob. 4.31CUCh. 4 - Prob. 4.32CUCh. 4 - Prob. 4.33CUCh. 4 - Prob. 4.34CUCh. 4 - Prob. 4.35CUCh. 4 - Prob. 4.36CUCh. 4 - Prob. 4.37CUCh. 4 - Prob. 4.38CUCh. 4 - Prob. 4.39CUCh. 4 - Prob. 4.40CUCh. 4 - Prob. 4.41CUCh. 4 - Prob. 4.42CUCh. 4 - Prob. 4.43CUCh. 4 - Prob. 4.44CUCh. 4 - Prob. 4.45CUCh. 4 - Prob. 4.46CUCh. 4 - Prob. 4.47CUCh. 4 - Prob. 4.48CUCh. 4 - Prob. 4.49CUCh. 4 - Prob. 4.50CUCh. 4 - Prob. 4.51CUCh. 4 - Prob. 4.1PCh. 4 - Prob. 4.2PCh. 4 - Prob. 4.3PCh. 4 - Prob. 4.4PCh. 4 - Prob. 4.5PCh. 4 - Prob. 4.6PCh. 4 - Prob. 4.7PCh. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10PCh. 4 - Prob. 4.11PCh. 4 - Prob. 4.12PCh. 4 - Prob. 4.13PCh. 4 - Prob. 4.14PCh. 4 - Prob. 4.15PCh. 4 - Prob. 4.16PCh. 4 - Prob. 4.17PCh. 4 - Prob. 4.18PCh. 4 - Prob. 4.19PCh. 4 - Prob. 4.20PCh. 4 - Prob. 4.21PCh. 4 - Prob. 4.22PCh. 4 - Prob. 4.23PCh. 4 - Prob. 4.24PCh. 4 - Prob. 4.25PCh. 4 - Prob. 4.26PCh. 4 - Prob. 4.27PCh. 4 - Prob. 4.28PCh. 4 - Prob. 4.29PCh. 4 - Prob. 4.30PCh. 4 - Prob. 4.31PCh. 4 - Prob. 4.32PCh. 4 - Prob. 4.33PCh. 4 - Prob. 4.34PCh. 4 - Prob. 4.35PCh. 4 - Prob. 4.36PCh. 4 - Prob. 4.37PCh. 4 - Prob. 4.38PCh. 4 - Prob. 4.39PCh. 4 - Prob. 4.40PCh. 4 - Prob. 4.41PCh. 4 - Prob. 4.42PCh. 4 - Prob. 4.43PCh. 4 - Prob. 4.44PCh. 4 - Prob. 4.45PCh. 4 - Prob. 4.46PCh. 4 - Prob. 4.47PCh. 4 - Prob. 4.48PCh. 4 - Prob. 4.49PCh. 4 - Prob. 4.50PCh. 4 - Prob. 4.51PCh. 4 - Prob. 4.52PCh. 4 - Prob. 4.53PCh. 4 - Prob. 4.54PCh. 4 - Prob. 4.55PCh. 4 - Prob. 4.56PCh. 4 - Prob. 4.57PCh. 4 - Prob. 4.58PCh. 4 - Prob. 4.59PCh. 4 - Prob. 4.60PCh. 4 - Prob. 4.61PCh. 4 - Prob. 4.62PCh. 4 - Prob. 4.63PCh. 4 - Prob. 4.64PCh. 4 - Prob. 4.65PCh. 4 - Prob. 4.66PCh. 4 - Prob. 4.67PCh. 4 - Prob. 4.68PCh. 4 - Prob. 4.69PCh. 4 - Prob. 4.70PCh. 4 - Prob. 4.71PCh. 4 - Prob. 4.72PCh. 4 - Prob. 4.73PCh. 4 - Prob. 4.74PCh. 4 - Prob. 4.75PCh. 4 - Prob. 4.76PCh. 4 - Prob. 4.77PCh. 4 - Prob. 4.78PCh. 4 - Prob. 4.79PCh. 4 - Prob. 4.80PCh. 4 - Prob. 4.81PCh. 4 - Prob. 4.82PCh. 4 - Prob. 4.83PCh. 4 - Prob. 4.84PCh. 4 - Prob. 4.85PCh. 4 - Prob. 4.86PCh. 4 - Prob. 4.87PCh. 4 - Prob. 4.88P
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- 1. Two reversible engines operate in series between a high-temperature (T) and a low-temperature (T) reservoir. Engine A rejects heat to engine B, which in tum rejects heat to the low-temperature reservoir. The high-temperature reservoir supplies heat to engine A. Let T,,- 1000°K and T, - 400°K and the engine thermal efficiencies are equal. The heat received by engine A is 500 kJ. Determine: a) the temperature of heat rejection by engine A in K, b) the work of engine A and B in kJ, and c) the cycle efficiency.arrow_forwardExplain into details why thermodynamics depends on the entropy?arrow_forwardShown below is P-V diagram for a reversible cycle enclosed by 4 reversible process curves. The curve 1-2 and the curve 3-4 are reversible isothermal processes, and the curve 2-3 and the curve 1-4 are reversible adiabatic processes. If the cycle direction is counter clockwise, answer the question below. Select curve(s) which represent process(es) having heat interactions?____ A. curve 1-2 B. curve 2-3 C. curve 1-4 D. curve 3-4arrow_forward
- (2) Describe briefly why the following statements are wrong. (a) "Hot cup of coffee becoming cold spontaneously is an entropy-decreasing process. So the law of entropy increase can be violated." (b) “Air conditioners require bulky external unit, exhausting heat to outside. With the technology constantly advancing, external unit will be eliminated in the future."arrow_forwardThe. relations are valid for both reversible and irreversible processes. h. ds T ds P. ds P. dvarrow_forwardTwo reversible cycles operate between hot and cold reservoirs at temperatures TH and TC, respectively. (a)If one is a power cycle and the other is a refrigeration cycle, what is the relation between the coefficient of performance of the refrigeration cycle and the thermal efficiency of the power cycle? (b)If one is a refrigeration cycle and the other is a heat pump cycle, what is the relation between their coefficients of performance?arrow_forward
- Ts diagrams for two reversible thermodynamic power cycles are shown in the following figure. Both cycles operate between a high temperature reservoir at 500 K and a low temperature reservoir at 300 K. The process on the left is the Carnot cycle described in Section 2.9. The process on the right is a Stirling cycle, which is similar to a Carnot cycle, except that the two steps (state 4 to state 1) and (state 2 to state 3) are at constant volume. Which cycle, if either, has a greater efficiency? Explain.arrow_forwardb) A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Take the initial pressure and initial temperature of the air before it is put in the tire to be P = 1.00 bar and T; = 298 K. The final volume of the air in the tire is V = 1.00 L and the final pressure is Pf = 5.00 bar. Hints: • Cy.m = 5/2 R. Heat Capacity ratio, y = Cp.m Cy,m = 1.4 %3D Cy,m For a reversible adiabatic change, 4 = (4 )°, where c = R For perfect gas: CP.m - Cv,m = R CP.m and Cv.m denote molar heat capacity at constant pressure and volume, respectively. Tand V are temperature and volume at initiate (i) state ad final (f) state. i) Derive a mathematical expression to show the relationship of temperature, volume and heat capacity ratio (y) for the transformation of this process from initial to final states.arrow_forwardProperties of Carbon Dioxide: h@ 220 KPa and -5°C = 350 KJ/kg bi@ 25°C = 223.65 KJ/kg b:@ 22°C = 220.75 KJ/kg ha @ 220 KPa = 347.13 KJ/kg %3D %3D Saturated Vapor Carbon Dioxide refrigerant at 220 KPa leaves the evaporator and enters the compressor at -5 °C. The refrigerant leaves the condenser as saturate liquid at 25 °C and enters the expansion valve at 22 °C. Heat rejected from the condenser amount 90 KW. The work to the compressor is 60 KJ/kg while the heat lost from the compressor is 5 KJ/kg. If 2.5 KJ/kg of heat are lost in the piping between the compressor and condenser. Solve for the mass flowrate of the refrigerant in kg/hr.arrow_forward
- 1.Refrigerator 1 simulates an ideal refrigerator and therefore operates on a Carnot cycle using R-134a as the refrigerant at a flow rate of 1.4 kg/sec. The condensing and evaporating temperatures are 30 °C and -10 °C, respectively. To assess the performance of Refrigerator 1, you must submit the report. of the project with the following information: a) Enthalpies corresponding to the states indicated in the cycle (1, 2, 3 and 4); b) The cooling rate (Q L ); c) The work supplied to the fluid by the compressor; d) The work generated by the turbine; e) The condenser heat rejection rate (Q H ); f) The coefficient of performance of the cycle. Important detail: For the energy balance, make the following considerations: ✓ Permanent regimen; ✓ Kinetic and potential energy variations are negligible; ✓ Compressor and turbine operate adiabatically; ✓ Evaporation and condensation steps are labor-free.arrow_forwardTwo reversible engines operate in series between a high-temperature (T) and a low-temperature (T) reservoir. Engine A rejects heat to engine B, which in turn rejects heat to the low-temperature reservoir. The high-temperature reservoir supplies heat to engine A. Let T= 1000°K and T, = 400°K and the engine thermal efficiencies are equal. The heat received by engine A is 500 kJ. Determine: a) the temperature of heat rejection by engine A in K, b) the work of engine A and B in kJ, and c) the cycle efficiency.arrow_forwardThermodynamics: - Clear writing please!!! Thank you - Sketch and label all numbers on the turbine. - Sketch and label all numbers on the process on a T-s diagram. 7.133E A large condenser in a steam power plant dumps 15 000 Btu/s by condensing saturated water vapor at 115 F with an ambient temperature of 77 F. What is the entropy generation rate?arrow_forward
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What is entropy? - Jeff Phillips; Author: TED-Ed;https://www.youtube.com/watch?v=YM-uykVfq_E;License: Standard youtube license