
Fundamentals of Engineering Thermodynamics
8th Edition
ISBN: 9781118412930
Author: Michael J. Moran, Howard N. Shapiro, Daisie D. Boettner, Margaret B. Bailey
Publisher: WILEY
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Chapter 5.11, Problem 86P
To determine
Whether the operation of the refrigerator is reversible, irreversible or impossible.
Whether the operation of the refrigerator is reversible, irreversible or impossible when there is no power input in the cycle.
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The evaporator of a vapor compression refrigeration cycle utilizing R-123 as the refrigerant isbeing used to chill water. The evaporator is a shell and tube heat exchanger with the water flowingthrough the tubes. The water enters the heat exchanger at a temperature of 54°F. The approachtemperature difference of the evaporator is 3°R. The evaporating pressure of the refrigeration cycleis 4.8 psia and the condensing pressure is 75 psia. The refrigerant is flowing through the cycle witha flow rate of 18,000 lbm/hr. The R-123 leaves the evaporator as a saturated vapor and leaves thecondenser as a saturated liquid. Determine the following:a. The outlet temperature of the chilled waterb. The volumetric flow rate of the chilled water (gpm)c. The UA product of the evaporator (Btu/h-°F)d. The heat transfer rate between the refrigerant and the water (tons)
(Read image) (Answer given)
Problem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a
mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and
(y2), respectively.
Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s].
Givens:
y1 = 4.112 m
y2 =
0.387 m
b = 0.942 m
Answers:
( 1 ) 1880.186 lit/s
( 2 ) 4042.945 lit/s
( 3 ) 2553.11 lit/s
( 4 ) 3130.448 lit/s
Chapter 5 Solutions
Fundamentals of Engineering Thermodynamics
Ch. 5.11 - Prob. 1ECh. 5.11 - 2. Are health risks associated with consuming...Ch. 5.11 - Prob. 3ECh. 5.11 - Prob. 4ECh. 5.11 - Prob. 5ECh. 5.11 - 6. Does the second law impose performance limits...Ch. 5.11 - Prob. 7ECh. 5.11 - 8. What is delaying the appearance in new car...Ch. 5.11 - Prob. 9ECh. 5.11 - 10. How significant is the roughness at a pipe’s...
Ch. 5.11 - Prob. 11ECh. 5.11 - 12. What factors influence the actual coefficient...Ch. 5.11 - Prob. 13ECh. 5.11 - 14. How does the thermal glider (Sec. 5.4) sustain...Ch. 5.11 - 1. A reversible heat pump cycle operates between...Ch. 5.11 - Prob. 2CUCh. 5.11 - 3. Referring to the list of Sec. 5.3.1,...Ch. 5.11 - 4. Uses of the second law of thermodynamics...Ch. 5.11 - Prob. 5CUCh. 5.11 - Prob. 6CUCh. 5.11 - Prob. 7CUCh. 5.11 - Prob. 8CUCh. 5.11 - Prob. 9CUCh. 5.11 - Prob. 10CUCh. 5.11 - Prob. 11CUCh. 5.11 - Prob. 12CUCh. 5.11 - Prob. 13CUCh. 5.11 - Prob. 14CUCh. 5.11 - Prob. 15CUCh. 5.11 - Prob. 16CUCh. 5.11 - Prob. 17CUCh. 5.11 - 18. Referring to Fig. 5.15, if the boiler and...Ch. 5.11 - Prob. 19CUCh. 5.11 - Prob. 20CUCh. 5.11 - Prob. 21CUCh. 5.11 - 22. A cell phone initially has a fully charged...Ch. 5.11 - Prob. 23CUCh. 5.11 - Prob. 24CUCh. 5.11 - Prob. 25CUCh. 5.11 - Prob. 26CUCh. 5.11 - Prob. 27CUCh. 5.11 - 28. As shown in Fig. P5.28C, energy transfer...Ch. 5.11 - 29. As shown in Fig. P5.29C, a rigid, insulated...Ch. 5.11 - 30. As shown in Fig. P5.30C, when the steam in the...Ch. 5.11 - Prob. 31CUCh. 5.11 - Prob. 32CUCh. 5.11 - Prob. 33CUCh. 5.11 - Prob. 34CUCh. 5.11 - Prob. 35CUCh. 5.11 - Prob. 36CUCh. 5.11 - Prob. 37CUCh. 5.11 - Prob. 38CUCh. 5.11 - Prob. 39CUCh. 5.11 - Prob. 40CUCh. 5.11 - Prob. 41CUCh. 5.11 - Prob. 42CUCh. 5.11 - 43. The maximum coefficient of performance of any...Ch. 5.11 - Prob. 44CUCh. 5.11 - Prob. 45CUCh. 5.11 - Prob. 46CUCh. 5.11 - 47. When an isolated system undergoes a process,...Ch. 5.11 - Prob. 48CUCh. 5.11 - Prob. 49CUCh. 5.11 - Prob. 50CUCh. 5.11 - 5.1 Complete the demonstration of the equivalence...Ch. 5.11 - 5.2 Shown in Fig. P5.2 is a proposed system that...Ch. 5.11 - 5.3 Classify the following processes of a closed...Ch. 5.11 - Prob. 4PCh. 5.11 - Prob. 5PCh. 5.11 - Prob. 6PCh. 5.11 - 5.7 Provide the details left to the reader in the...Ch. 5.11 - 5.8 Using the Kelvin–Planck statement of the...Ch. 5.11 - Prob. 9PCh. 5.11 - Prob. 10PCh. 5.11 - Prob. 11PCh. 5.11 - Prob. 12PCh. 5.11 - Prob. 13PCh. 5.11 - Prob. 14PCh. 5.11 - 5.15 To increase the thermal efficiency of a...Ch. 5.11 - Prob. 16PCh. 5.11 - Prob. 17PCh. 5.11 - Prob. 18PCh. 5.11 - 5.19 A power cycle operating at steady state...Ch. 5.11 - 5.20 As shown in Fig. P5.20, a reversible power...Ch. 5.11 - Prob. 21PCh. 5.11 - Prob. 22PCh. 5.11 - Prob. 23PCh. 5.11 - Prob. 24PCh. 5.11 - Prob. 25PCh. 5.11 - Prob. 26PCh. 5.11 - Prob. 27PCh. 5.11 - Prob. 28PCh. 5.11 - Prob. 29PCh. 5.11 - Prob. 30PCh. 5.11 - Prob. 31PCh. 5.11 - Prob. 32PCh. 5.11 - Prob. 33PCh. 5.11 - 5.34 A power cycle operates between hot and cold...Ch. 5.11 - Prob. 35PCh. 5.11 - 5.36 An inventor claims to have developed a power...Ch. 5.11 - Prob. 37PCh. 5.11 - Prob. 38PCh. 5.11 - 5.39 As shown in Fig. P5.39, a system undergoing a...Ch. 5.11 - Prob. 40PCh. 5.11 - Prob. 41PCh. 5.11 - Prob. 42PCh. 5.11 - Prob. 43PCh. 5.11 - 5.44 A reversible refrigeration cycle operates...Ch. 5.11 - Prob. 45PCh. 5.11 - 5.46 A heating system must maintain the interior...Ch. 5.11 - Prob. 47PCh. 5.11 - 5.48 The thermal efficiency of a reversible power...Ch. 5.11 - 5.49 Shown in Fig. P5.49 is a system consisting of...Ch. 5.11 - 5.50 An inventor has developed a refrigerator...Ch. 5.11 - 5.51 An inventor claims to have developed a food...Ch. 5.11 - 5.52 An inventor claims to have developed a...Ch. 5.11 - 5.53 An inventor claims to have devised a...Ch. 5.11 - 5.54 Data are provided for two reversible...Ch. 5.11 - 5.55 By removing energy by heat transfer from its...Ch. 5.11 - 5.56 At steady state, a refrigeration cycle...Ch. 5.11 - Prob. 57PCh. 5.11 - 5.58 At steady state, a refrigeration cycle...Ch. 5.11 - Prob. 59PCh. 5.11 - Prob. 60PCh. 5.11 - Prob. 61PCh. 5.11 - Prob. 62PCh. 5.11 - Prob. 63PCh. 5.11 - 5.64 As shown in Fig P5.64, an air conditioner...Ch. 5.11 - Prob. 65PCh. 5.11 - Prob. 66PCh. 5.11 - 5.68 The refrigerator shown in Fig. P5.68 operates...Ch. 5.11 - Prob. 69PCh. 5.11 - 5.70 By supplying energy at an average rate of...Ch. 5.11 - 5.71 A heat pump with a coefficient of performance...Ch. 5.11 - 5.72 As shown in Fig. P5.72, a heat pump provides...Ch. 5.11 - 5.73 As shown in Fig. P 5.73, a heat pump receives...Ch. 5.11 - Prob. 74PCh. 5.11 - Prob. 75PCh. 5.11 - Prob. 76PCh. 5.11 - Prob. 77PCh. 5.11 - Prob. 78PCh. 5.11 - Prob. 79PCh. 5.11 - Prob. 80PCh. 5.11 - 5.81 A quantity of water within a piston–cylinder...Ch. 5.11 - Prob. 82PCh. 5.11 - 5.83 Two kilograms of air within a piston–cylinder...Ch. 5.11 - Prob. 84PCh. 5.11 - Prob. 85PCh. 5.11 - Prob. 86PCh. 5.11 - Prob. 87PCh. 5.11 - Prob. 88PCh. 5.11 - Prob. 89PCh. 5.11 - 5.90 Figure P5.90 gives the schematic of a vapor...Ch. 5.11 - Prob. 91PCh. 5.11 - Prob. 92PCh. 5.11 - 5.93 As shown in Fig. P5.93, a system executes a...Ch. 5.11 - Prob. 94P
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- Problem (14): A pump is being used to lift water from an underground tank through a pipe of diameter (d) at discharge (Q). The total head loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h where (V) is the flow velocity in the pipe. The elevation difference between the pump and tank surface is (h). Given the values of h [cm], d [cm], and K [-], calculate the maximum discharge Q [Lit/s] beyond which cavitation would take place at the pump entrance. Assume Turbulent flow conditions. Givens: h = 120.31 cm d = 14.455 cm K = 8.976 Q Answers: (1) 94.917 lit/s (2) 49.048 lit/s ( 3 ) 80.722 lit/s 68.588 lit/s 4arrow_forwardProblem (13): A pump is being used to lift water from the bottom tank to the top tank in a galvanized iron pipe at a discharge (Q). The length and diameter of the pipe section from the bottom tank to the pump are (L₁) and (d₁), respectively. The length and diameter of the pipe section from the pump to the top tank are (L2) and (d2), respectively. Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m], calculate total head loss due to friction (i.e., major loss) in the pipe (hmajor-loss) in [cm]. Givens: L₁,d₁ Pump L₂,d2 오 0.533 lit/s L1 = 6920.729 m d1 = 1.065 m L2 = 70.946 m d2 0.072 m Answers: (1) 3.069 cm (2) 3.914 cm ( 3 ) 2.519 cm ( 4 ) 1.855 cm TABLE 8.1 Equivalent Roughness for New Pipes Pipe Riveted steel Concrete Wood stave Cast iron Galvanized iron Equivalent Roughness, & Feet Millimeters 0.003-0.03 0.9-9.0 0.001-0.01 0.3-3.0 0.0006-0.003 0.18-0.9 0.00085 0.26 0.0005 0.15 0.045 0.000005 0.0015 0.0 (smooth) 0.0 (smooth) Commercial steel or wrought iron 0.00015 Drawn…arrow_forwardThe flow rate is 12.275 Liters/s and the diameter is 6.266 cm.arrow_forward
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