Applied Fluid Mechanics (7th Edition)
7th Edition
ISBN: 9780132558921
Author: Robert L. Mott, Joseph A. Untener
Publisher: PEARSON
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Chapter 9, Problem 9.5PP
A small velocity probe is to be inserted through a pipe wall. If we measure from the outside of the DN
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Chapter 9 Solutions
Applied Fluid Mechanics (7th Edition)
Ch. 9 - Compute points on the velocity profile from the...Ch. 9 - s9.2 Compute points on the velocity profile from...Ch. 9 - Compute points on the velocity profile from the...Ch. 9 - Compute points on the velocity profile from the...Ch. 9 - A small velocity probe is to be inserted through a...Ch. 9 - If the accuracy of positioning the probe described...Ch. 9 - An alternative scheme for using the velocity probe...Ch. 9 - Prob. 9.8PPCh. 9 - For the flow of 12.9L/min of water at 75C in a...Ch. 9 - A large pipeline with a 1,200m inside diameter...
Ch. 9 - Prob. 9.11PPCh. 9 - Prob. 9.12PPCh. 9 - Prob. 9.13PPCh. 9 - Prob. 9.14PPCh. 9 - Using Eq. (9-4), compute the ratio of the average...Ch. 9 - Prob. 9.16PPCh. 9 - Repeat Problem 9.16 for the same conditions,...Ch. 9 - Prob. 9.18PPCh. 9 - A shell-and-tube heat exchanger is made of two...Ch. 9 - Figure 9.14 shows a heat exchanger in which each...Ch. 9 - Figure 9.15 shows the cross section of a...Ch. 9 - Air with a specific weight of 12.5N/m3 and a...Ch. 9 - Carbon dioxide with a specific weight of...Ch. 9 - Water at 90F flows in the space between 6 in...Ch. 9 - Refer to the shell-and-tube heat exchanger shown...Ch. 9 - Refer to Fig. 9.14, which shows two DN 150...Ch. 9 - Refer to Fig. 9.15, which shows three pipes inside...Ch. 9 - Water at 10C is flowing in the shell shown in Fig....Ch. 9 - Figure 9.19 shows the cross section of a heat...Ch. 9 - Figure 9.20 shows a liquid-to-air heat exchanger...Ch. 9 - Glycerin ( sg=1.26 ) at 40C flows in the portion...Ch. 9 - Each of the square tubes shown in Fig. 9.21...Ch. 9 - A heat sink for an electronic circuit is made by...Ch. 9 - Figure 9.23 shows the cross section of a cooling...Ch. 9 - Prob. 9.35PPCh. 9 - The blade of a gas turbine engine contains...Ch. 9 - For the system described in Problem 9.24. compute...Ch. 9 - For the shell-and-tube heat exchanger described in...Ch. 9 - For the system described in Problem 9.26 compute...Ch. 9 - For the system described in Problem 9.27 compute...Ch. 9 - For the shell-and-tube heat exchanger described in...Ch. 9 - For the heat exchanger described in Problem 9.29...Ch. 9 - For the glycerin described in Problem 9.31 compute...Ch. 9 - For the flow of water in the square tubes...Ch. 9 - If the heat sink described in Problem 9.33 is 105...Ch. 9 - Compute the energy loss for the flow of water in...Ch. 9 - In Fig. 9.26 ethylene glycol ( sg=1.10 ) at 77F...Ch. 9 - Figure 9.27 shows a duct in which methyl alcohol...Ch. 9 - Prob. 9.49PPCh. 9 - Figure 9.29 shows a system in which methyl alcohol...Ch. 9 - A simple heat exchanger is made by welding...Ch. 9 - Three surfaces of an instrument package are cooled...Ch. 9 - Figure 9.32 shows a heat exchanger with internal...
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- The volume of 1 kg of helium in a piston–cylinder device is initially 5 m3. Now helium is compressed to 3 m3 while its pressure is maintained constant at 130 kPa. Determine the initial and final temperatures of helium as well as the work required to compress it, in kJ. The gas constant of helium is R = 2.0769 kJ/kg·K. The initial temperature of helium is K. The final temperature of helium is K. The work required to compress helium is kJ.arrow_forwardA piston-cylinder device initially contains 0.4 kg of nitrogen gas at 160 kPa and 140°C. Nitrogen is now expanded isothermally to a pressure of 80 kPa. Determine the boundary work done during this process. The properties of nitrogen are R= 0.2968 kJ/kg-K and k= 1.4. N₂ 160 kPa 140°C The boundary work done during this process is KJ.arrow_forward! Required information An abrasive cutoff wheel has a diameter of 5 in, is 1/16 in thick, and has a 3/4-in bore. The wheel weighs 4.80 oz and runs at 11,700 rev/min. The wheel material is isotropic, with a Poisson's ratio of 0.20, and has an ultimate strength of 12 kpsi. Choose the correct equation from the following options: Multiple Choice о σmax= (314) (4r2 — r²) - о σmax = p² (3+) (4r² + r²) 16 σmax = (314) (4r² + r²) σmax = (314) (4² - r²)arrow_forward
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