FUND OF ENG THERMODYN(LLF)+WILEYPLUS
9th Edition
ISBN: 9781119391777
Author: MORAN
Publisher: WILEY
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Figure P6.22 provides the T-s diagram of a Carnot refrigeration cycle for which the substance is Refrigerant 134a. Determine the coefficient of performance.
At steady state, a refrigeration cycle removes 800 kJ/min of energy by heat transfer from acold space maintained at -15o C and discharges energy by heat transfer to its surroundings at20o C. If the coefficient of performance of the actual cycle is 80 percent of that of areversible refrigeration cycle (ideal) operating between thermal reservoirs at these twotemperatures (β = 0.80 βmax), determine:a) Coefficient of performance;b) Power input to the cycle, in kW;c) Heat rejected to the surroundings.Draw sketch and show all calculations.
An air conditioner is a device used to cool the inside of a home. It is, in
essence, a refrigerator in which mechanical work is done and heat removed
from the (cooler) inside and rejected to the (warmer) outside.
A home air conditioner operating on a reversible Carnot cycle between
the inside, absolute temperature T2, and the outside, absolute tempera-
ture T1 > T2, consumes P joules/sec from the power lines when operating
continuously.
(a) In one second, the air conditioner absorbs Q2 joules from the house
and rejects Q1 joules outdoors. Develop a formula for the efficiency ratio
Q2/P in terms of T1 and T2.
(b) Heat leakage into the house follows Newton's law Q = A(T, – T2).
Develop a formula for T, in terms of T1, P, and A for continuous operation
of the air conditioner under constant outside temperature T and uniform
(in space) inside temperature T2.
(c) The air conditioner is controlled by the usual on-off thermostat
and it is observed that when the thermostat set at 20°C and an…
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- A power cycle operating between two reservoirs receives energy by heat transfer from a hot reservoir, Qh = 1600 kJ at Th = 1575 K and rejects energy by heat transfer Qc= 800kJ to a cold reservoir at Tc = 495 K. Determine whether the cycle operates reversibly, irreversibly, or does not verify the second law of thermodynamics. O a. irreversibly O b. reversibly does not verify the second law of thermodynamics TOSHIBAarrow_forwardAs shown in the figure, an air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature is 90°F. The rate of heat transfer into the dwelling through the walls and roof is 30,000 Btu/h and the net power input to the air conditioner compressor is 3 hp. Determine a. the coefficient of performance for the air conditioner b. power input required in hp c. coefficient of performance for a reversible air conditioner providing the same cooling effect while operating between the same cold and hot temperatures.arrow_forwardC5 6.arrow_forward
- Which of the following statements best describes the 2nd law of thermodynamics? The total entropy of a a reservoir must stay the same or increase. The change in internal energy for an ideal gas is given as A U=mc _▲T. V Heat energy will always be transferred from a hotter object to a colder object. Around a complete cycle, the net heat and net work additions must sum to zero. O A reversible cyclic engine can convert all the heat input it receives into useful work output.arrow_forwardThe following processes occur in a reversible thermodynamic cycle: 1-2: 0.2 kg heating at constant pressure 1.05 bar at specific volume 0.1 m3/kg and work done -515 J. 2-3: Isothermal compression to 4.2 bar. 3-4: Expansion according to law pv1./= constant. 4-1: heating at constant volume back to the initial conditions. Calculate the work done for the isothermal process in J.arrow_forward8Q Prove that, whenever a system undergoes a cycle, p<0 (b) T.arrow_forward
- An open feedwater heater is a direct-contact heat exchanger used in vapor power plants. Shown in the figure below are feedwater heater with H20 operating data for an open 31 °C as the working fluid operating at steady state, where T1 P2=3 bar = 0.92 +2 P3 3 bar Saturated liquid m380 kg/s T, PT=3 bar Open feedwater 3 heater Ignoring stray heat transfer from the outside of the heat exchanger to its surroundings and kinetic and potential energy effects, determine the rate of entropy production, in kW/K.arrow_forwardGive solution to the thermodynamics problem.arrow_forwardSelect the Kelvin-Plank statement of the second law of Thermodynamics___________ A. It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. B. It is impossible for any device that operates on a cycle to receive heat form a single reservoir and produce a net amount of work.arrow_forward
- An ideal gas within a piston–cylinder assembly undergoes a Carnot refrigeration cycle, as shown in the figure below. The isothermal compression occurs at 325 K from 2 bar to 6 bar. The isothermal expansion occurs at 275 K. Determine: (a) the coefficient of performance, (b) the heat transfer to the gas during the isothermal expansion, in kJ per kmol of gas, (c) the magnitude of the net work input, in kJ per kmol of gas.arrow_forwardAs shown in the figure, an air conditioner operating at steady state maintains a dwelling at 70°F on a day when the outside temperature is 103.5°F. The rate of heat transfer into the dwelling through the walls and roof is 30,000 Btu/h and the net power input to the air conditioner compressor is 3.45 hp. Step 1 B Inside, 70° F Determine the coefficient of performance for the air conditioner. = Evaporator i 30,000 Btu/h Determine the coefficient of performance of the air conditioner. Determine the power input required, in hp, and the coefficient of performance for a reversible air conditioner providing the same cooling effect while operating between the same cold and hot temperatures. Refrigerant loop Compressor Outside Q + Condenserarrow_forwardA power cycle operating between two reservoirs receives energy Qu by heat transfer from a hot reservoir at TH = 2000 K and rejects energy Qc by heat transfer to a cold reservoir at Tc = 400 K. For the cases below you will be asked to determine the cycle n and whether the cycle operates Reversibly, Irreversibility, or is Impossible. Сycle Сycle Сycle Assume: п — пСarnot n пСarnot 1. The maximum thermal efficiency nCarnot for the cycle is equal to а. 0.2 b. 0.8 с. 1.0 d. none of the above. 2. If QH = 1100 kJ and the Weycle = 900 kJ then the cycle is Reversible b. Irreversible c. Impossible а. d. none of the above. 3. If QH = 1000 kJ and Qc = 200 kJ then the cycle is a. Reversible b. Irreversible c. Impossible d. none of the above. 4. If Wq a. Reversible b. Irreversible c. Impossible d. none of the above. суcle 1400 kJ and Qc= 600 kJ then the cycle is 5. If n = 50% then the cycle is a. Reversible b. Irreversible c. Impossible d. none of the above.arrow_forward
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