Fundamentals Of Thermal-fluid Sciences In Si Units
5th Edition
ISBN: 9789814720953
Author: Yunus Cengel, Robert Turner, John Cimbala
Publisher: McGraw-Hill Education
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Chapter 15, Problem 42P
To determine
The fluid property that is responsible for the development of velocity boundary layer; the effect of velocity on the thickness of the boundary layer.
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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 15 Solutions
Fundamentals Of Thermal-fluid Sciences In Si Units
Ch. 15 - Prob. 1PCh. 15 - Define the frontal area of a body subjected to...Ch. 15 - Prob. 3PCh. 15 - Prob. 4PCh. 15 - Prob. 5PCh. 15 - Prob. 6PCh. 15 - Name some applications in which a large drag is...Ch. 15 - Prob. 8PCh. 15 - Prob. 9PCh. 15 - Prob. 10P
Ch. 15 - Prob. 11PCh. 15 - What is the difference between skin friction drag...Ch. 15 - Prob. 13PCh. 15 - What is the effect of streamlining on (a) friction...Ch. 15 - What is flow separation? What causes it? What is...Ch. 15 - Prob. 16PCh. 15 - Prob. 17PCh. 15 - Prob. 18PCh. 15 - The resultant of the pressure and wall shear...Ch. 15 - Prob. 20PCh. 15 - Prob. 21PCh. 15 - Prob. 22PCh. 15 - Prob. 24PCh. 15 - Prob. 25PCh. 15 - Prob. 26PCh. 15 - Prob. 27PCh. 15 - A submarine can be treated as an ellipsoid with a...Ch. 15 - Prob. 29PCh. 15 - During major windstorms, high vehicles such as RVs...Ch. 15 - Prob. 31PCh. 15 - Prob. 32PCh. 15 - Prob. 33PCh. 15 - Prob. 34PCh. 15 - Prob. 35PCh. 15 - Prob. 36PCh. 15 - Prob. 37PCh. 15 - Prob. 38PCh. 15 - Prob. 39PCh. 15 - Prob. 40PCh. 15 - Prob. 41PCh. 15 - Prob. 42PCh. 15 - Prob. 43PCh. 15 - Prob. 44PCh. 15 - Prob. 45PCh. 15 - Prob. 46PCh. 15 - Prob. 48PCh. 15 - Prob. 49PCh. 15 - Prob. 51PCh. 15 - Prob. 52PCh. 15 - Prob. 53PCh. 15 - Prob. 54PCh. 15 - Prob. 55PCh. 15 - Prob. 56PCh. 15 - Prob. 57PCh. 15 - Prob. 58PCh. 15 - Prob. 59PCh. 15 - Prob. 60PCh. 15 - Prob. 61PCh. 15 - Prob. 62PCh. 15 - A 2-m-long, 0.2-m-diameter cylindrical pine log...Ch. 15 - Prob. 64PCh. 15 - Prob. 65PCh. 15 - Prob. 66PCh. 15 - Prob. 67PCh. 15 - Prob. 68PCh. 15 - Prob. 69PCh. 15 - Prob. 70PCh. 15 - Prob. 71PCh. 15 - Prob. 72PCh. 15 - What is induced drag on wings? Can induced drag be...Ch. 15 - Prob. 74PCh. 15 - Prob. 75PCh. 15 - Prob. 76PCh. 15 - Prob. 77PCh. 15 - Consider an airplane whose takeoff speed is 220...Ch. 15 - Prob. 79PCh. 15 - Prob. 80PCh. 15 - Prob. 82PCh. 15 - Prob. 83PCh. 15 - Prob. 84PCh. 15 - Prob. 85PCh. 15 - Prob. 86RQCh. 15 - A 1.2-m-external-diameter spherical tank is...Ch. 15 - Prob. 88RQCh. 15 - Prob. 89RQCh. 15 - Prob. 91RQCh. 15 - Prob. 92RQCh. 15 - Prob. 93RQCh. 15 - Prob. 94RQCh. 15 - Prob. 95RQCh. 15 - Prob. 96RQCh. 15 - Prob. 97RQCh. 15 - Prob. 98RQCh. 15 - Prob. 99RQCh. 15 - Prob. 100RQCh. 15 - Prob. 102RQ
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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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