Degarmo's Materials And Processes In Manufacturing
13th Edition
ISBN: 9781119492825
Author: Black, J. Temple, Kohser, Ronald A., Author.
Publisher: Wiley,
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Chapter 4, Problem 15RQ
To determine
Three information that can be obtained at any point in phase diagram.
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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)
Chapter 4 Solutions
Degarmo's Materials And Processes In Manufacturing
Ch. 4 - What kind of questions can be answered by...Ch. 4 - Prob. 2RQCh. 4 - Supplement the examples provided in the text with...Ch. 4 - Prob. 4RQCh. 4 - What three primary variables are generally...Ch. 4 - Use the pressure–temperature diagram for water...Ch. 4 - Prob. 7RQCh. 4 - What form of equilibrium phase diagram is most...Ch. 4 - What is a cooling curve?Ch. 4 - Prob. 10RQ
Ch. 4 - Prob. 11RQCh. 4 - Prob. 12RQCh. 4 - Prob. 13RQCh. 4 - What types of changes occur upon cooling through a...Ch. 4 - Prob. 15RQCh. 4 - What is a tie�line? For what types of phase...Ch. 4 - What points on a tie�line are used to determine...Ch. 4 - Prob. 18RQCh. 4 - What is a cored structure? Under what conditions...Ch. 4 - What is the difference between a cored structure...Ch. 4 - Prob. 21RQCh. 4 - Prob. 22RQCh. 4 - Prob. 23RQCh. 4 - Prob. 24RQCh. 4 - For the various three�phase reactions, what does...Ch. 4 - Prob. 26RQCh. 4 - Prob. 27RQCh. 4 - Prob. 28RQCh. 4 - Prob. 29RQCh. 4 - Prob. 30RQCh. 4 - Prob. 31RQCh. 4 - Prob. 32RQCh. 4 - Prob. 33RQCh. 4 - Prob. 34RQCh. 4 - Prob. 35RQCh. 4 - Prob. 36RQCh. 4 - Prob. 37RQCh. 4 - What is carbon equivalent, and how is it computed?Ch. 4 - Prob. 39RQCh. 4 - Prob. 40RQCh. 4 - Prob. 1PCh. 4 - Prob. 2PCh. 4 - Prob. 3PCh. 4 - Consider the manufacture of a fishhook beginning...Ch. 4 - If a stainless steel were to be used, what type of...Ch. 4 - A wide spectrum of coatings and surface treatments...Ch. 4 - Prob. 1.4CSCh. 4 - Prob. 2.1CSCh. 4 - Prob. 2.2CSCh. 4 - Prob. 2.3CSCh. 4 - Prob. 2.4CS
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- Problem (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_forwardAn experimental setup is being built to study the flow in a large water main (i.e., a large pipe). The water main is expected to convey a discharge (Qp). The experimental tube will be built at a length scale of 1/20 of the actual water main. After building the experimental setup, the pressure drop per unit length in the model tube (APm/Lm) is measured. Problem (20): Given the value of APm/Lm [kPa/m], and assuming pressure coefficient similitude, calculate the drop in the pressure per unit length of the water main (APP/Lp) in [Pa/m]. Givens: AP M/L m = 590.637 kPa/m meen Answers: ( 1 ) 59.369 Pa/m ( 2 ) 73.83 Pa/m (3) 95.443 Pa/m ( 4 ) 44.444 Pa/m *******arrow_forward
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