Applied Fluid Mechanics: Global Edition
7th Edition
ISBN: 9781292019611
Author: Robert Mott
Publisher: Pearson Higher Education
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Chapter 4, Problem 4.26PP
Refer to Fig.4.38
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using the theorem of three moments, find all the reactions and supports, I need concise calculations only. the answers are at the bottom, I need concise steps and minimal explanations
In an industrial facility, a counter-flow double-pipe heat exchanger uses superheated steam at a temperature of 155°C to
heat feed water at 30°C. The superheated steam experiences a temperature drop of 70°C as it exits the heat exchanger.
The water to be heated flows through the heat exchanger tube of negligible thickness at a constant rate of 3.47 kg/s. The
convective heat transfer coefficient on the superheated steam and water side is 850 W/m²K and 1250 W/m²K,
respectively. To account for the fouling due to chemical impurities that might be present in the feed water, assume
a fouling factor of 0.00015 m² K/W for the water side.
The specific heat of water is determined at an average temperature of (30+70)°C/2 = 50°C and is taken to be Cp
J/kg-K.
Water
Steam
Determine the heat exchanger area required to maintain the exit temperature of the water to a minimum of 70°C.
The heat exchanger area required is
Stress, ksi
160
72
150-
140
80
70
༄ ྃ ༈ ཎྜ རྦ ༅ ཎྜ ྣཧྨ ➢
130
120
110
100
90
2.0
2.8
3.6
4.4
5
Wire diameter, mm
6.0
6.8
2
7.6
8.4
Compression and extension springs.
ASTM A227 Class II
Light service
Average service
0.020
0.060
0.100
0.140
0.180
0.220
0.260
0.300
0.340
0.380
0.420
0.460
0.500
Wire diameter, in
Torsional stress due to initial tension, ksi
10
४
20
Preferred
range
100
Stress, MPa
9.2
10.0
10.8
11.6
12.4
1100
1035
965
895
825
760
Severe service
690
620
550
50
150
3456789 10 11 12 13 14 15 16
Spring index, C = DJD
FIGURE 18-21 Recommended torsional shear stress in an extension spring due to initial tension (Data from Associated
Spring, Barnes Group, Inc.)
50
200
485
Stress, MPa
Chapter 4 Solutions
Applied Fluid Mechanics: Global Edition
Ch. 4 - figure 4.2 shows a vacuum tank with a flat...Ch. 4 - The flat left end of the tank shown in Fig. 4.21...Ch. 4 - An exhaust system for a room creates a partial...Ch. 4 - A piece of 14 -in Schedule 40 pipe is used as a...Ch. 4 - A pressure relief valve is designed so that the...Ch. 4 - A gas-powered cannon shoots projectiles by...Ch. 4 - The egress hatch of a manned spacecraft is...Ch. 4 - A tank containing liquid ammonia at 77F has a flat...Ch. 4 - The bottom of a laboratory vat has a hole in it to...Ch. 4 - A simple shower for remote locations is designed...
Ch. 4 - Calculate the total force on the bottom of the...Ch. 4 - If the length of the tank in Fig. 4.24 is 1.2m,...Ch. 4 - An observation port in a small submarine is...Ch. 4 - A rectangular gate is installed in a vertical wall...Ch. 4 - '4.15 A vat has a sloped side, as shown in Fig....Ch. 4 - The wall shown in Fig. 4.28 is 20 ft long, (a)...Ch. 4 - If the wall in Fig. 4.29 is 4m long, calculate the...Ch. 4 - Refer to Fig. 4.30Ch. 4 - Refer to Fig. 4.31Ch. 4 - Refer to Fig.4.32Ch. 4 - Refer to Fig 4.33Ch. 4 - Refer to Fig. 4.34Ch. 4 - Refer to Fig. 4.35 (?Ch. 4 - Swimming poo!WilierGlasswindow2 ft diameterFigure...Ch. 4 - 4.25 Refer to Fig 4.37Ch. 4 - Refer to Fig.4.38Ch. 4 - Refer to Fig.4.39Ch. 4 - Refer to Fig.4.40Ch. 4 - Refer to Fig 4.41Ch. 4 - figure 4.42i5 shows a gasoline tank filled into...Ch. 4 - If the tank in Fig. 4.42 is filled just to the...Ch. 4 - If the tank in Fig. 4.42 is only half full of...Ch. 4 - For the water tank shown in Fig. 4.43, compute the...Ch. 4 - For the water tank shown in Fig. 4.43, compute the...Ch. 4 - For the water tank shown in Fig. 4.43, compute the...Ch. 4 - For the orange-drink tank shown in Fig. 4.32,...Ch. 4 - For the orange-drink tank shown in Fig. 4.32,...Ch. 4 - For the oil tank shown in Fig. 4.35, compute the...Ch. 4 - For the oil tank shown in Fig. 4.35; compute the...Ch. 4 - figure 4.44 shows a rectangular gate holding water...Ch. 4 - figure 4.45 shows a gate hinged at its bottom and...Ch. 4 - figure 4.46 shows a tank of water with a circular...Ch. 4 - Repeat Problem 4.19(Fig. 4.31), except that the...Ch. 4 - Repeat Problem 4.22 (Fig. 4.32), except that the...Ch. 4 - Repeat Problem 4.26 (Fig. 4.38 ). except that the...Ch. 4 - Repeat Problem 4.28 (Fig. 4.40 ), except that the...Ch. 4 - Use Fig 4.47. The surface is 2.00m long.Ch. 4 - Use Fig.4.48. The surface is 2.50m long.Ch. 4 - Use Fig.4.49. The surface is 5.00 ft longCh. 4 - Use Fig.4.50. The surface is 4.50 ft long.Ch. 4 - Use Fig.4.51.The surface is 4.00 m long.Ch. 4 - Use Fig .4.52. The surface is 1.50m longCh. 4 - Use Fig. 4.53. The surface is 1.50m long.Ch. 4 - Use Fig. 4.54. The surface is 60 in longCh. 4 - Repeat Problem 4.47 using Fig. 4.47, except that...Ch. 4 - Repeat Problem 4.48 using Fig. 4.48, except that...Ch. 4 - The tank in Fig. 4.55 has a view port in the...Ch. 4 - Insulated concrete forms (ICFs) are becoming more...Ch. 4 - Lacks are installed in rivers to allow boats to...Ch. 4 - When a dam is installed in a river that has...Ch. 4 - A wealthy eccentric is interested in having an...Ch. 4 - A pneumatic cylinder like the one shown in Fig....Ch. 4 - Determine the magnitude and the location of the...Ch. 4 - For the hinged gate shown in Fig. 4.61, determine...Ch. 4 - Prob. 4.65PPCh. 4 - Write a program to solve Problem 4.41 with any...Ch. 4 - Write a program to solve Problem 4.42 (Fig. 4.46)...Ch. 4 - Write a program to solve curved surface problems...Ch. 4 - For Program 1, cause the depth h to vary over some...
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- Bolted Joint Design Bolted Frames Total Force due to door weight: P = 240 lb Number of Bolts: N = Distance to Bolt C/L: a = 4 N/A Bolt Material - Allowable shear stress of bolt material: T₂ = x Distance from Bolt centroid to bolt: x = y Distance from Bolt centroid to bolt: y = Degrees per Radian- Results y-Load on each bolt: F, = Moment resisted by bolt pattern: M = Radial distance from Bolt centroid to bolt: r = Sum squares of all radial distances: Σr² Force on each bolt to resist moment: F, - Angle for force composition: e= X-Force on each bolt to resist moment: F- y-Force on each bolt to resist moment: Fly Total y-Force on each bolt: Fy = Resultant force on bolt 1: R₁ = Required shear stress area for a bolt: A₂ = ASTM Grade A307 Steel 10,000 0 psi from Table 20-1 3.0 57.296 in degrees lb per bolt lb-in Formula FS-P/N M-Px XB r = (x² + y²)0.5 in² Σ 4r² Mr F₁ = Στ lb degrees lb lb lb Minimum Bolt Diameter: Din = Rounded up Bolt Diameter: D = 55 P. 1.5 in 2 in (3x) 1 in This bracket…arrow_forwardUniversity of Babylon Collage of Engineering/ Al-Musayab Department of Automobiles Final Examination/ Stage: 3rd Notes: Answer 4 questions only 2023-2202 Subject: Theory of vehicles Date: 2023\06\10-Saturday Time: Three Hours Course 2nd Attempt 1st Q1: A Hooke's coupling connects two shafts whose axes are inclined at 30°. The of the driven shaft? Find the maximum value of retardation or acceleration and driving shaft rotates uniformly at 600 rpm. What are the extreme angular velocities state the angle where both will occur. (12.5 Marks) Q2: Four masses, A, B, C, and D), revolve at equal radii and are equally spaced along a shaft. The mass B is 7 kg, and the radius of C and D make angles of 90° and 240°, respectively, with the radius of B. Find the magnitude of the masses A, C, and D and the angular position of A so that the system may be completely balanced. (12.5 Marks) Q3: A cam has straight worked faces that are tangential to a base circle of diameter 90 mm. The follower is a roller…arrow_forwardProblem 18-26 Added Extension Springs Spring Material ASTM A227 Modulus of Elasticity of the Material in Shear: G 1.150E+07 psi Average Service Max Operating Load: F₁ = 100 lb Max Length between attachment points: L₁ = 60.00 in 20.00 lb 26.00 1.400 Min Operating Load: F₁ = Min Length between attachment points: L₁ = Maximum Outside Diameter = in in Results Note: you select a wire diameter from the "US steel wire gage" column in table 18-2 Formula k = AF/AL k = (F0-F1)/(Lo - L₁) Spring Rate: k = lb/in Assumed Trial Outside Diameter: OD = Assumed Trial Mean: D ma Assumed Design Stress in Spring: Tda in 1.070 in 102,000 psi Assumed Wahl Factor: K = 1.2 Calculated Wire Diameter: Dwa Actual Wire Diameter: Dw Actual outer diameter: OD = Actual inner diameter: ID= Spring Index: C = See Figure 18-8 Dw= [8KF Dm πTd 1/3 in 5' 5' 5' 5' This corresponds to US Steel 9 wire gage ID = Dm - Dw C = Dm/Dw 4C - 1 0.615 K = + 4C - с Wahl Factor: K = 8KFDm 8KFC T = TD πD Stress in Spring at F = Fo: To psi…arrow_forward
- CHAIN DRIVE DESIGN Initial Input Data: Application: Garage Door Opener Drive type: AC Motor Driven machine Chain and Sprocket to pull the door up Degrees per Radian: 57.2958 degrees Sprocket Diameter: D = 1.690 in Number of strands: Chain number: 1 40 Service factor: 1.3 Table 7-10 No. of teeth Computed Data: Actual Motor Power Input: 0.000 hp Sprocket Speed (for sprocket attached to gear shaft) Design power: 0.00 rpm 0 hp 11 12 0.06 0.15 0.29 0.56 0.99 1.09 1.61 2.64 TABLE 7-7 Horsepower Ratings-Single Strand Roller Chain No. 40 0.500 inch pitch 10 25 50 100 180 200 300 500 700 900 1000 1: 0.06 0.14 0.27 0.52 0.91 1.00 1.48 2.42 3.34 4.25 4.70 ! 3.64 4.64 5.13 13 0.07 0.16 0.31 0.61 1.07 1.19 1.75 2.86 3.95 5.02 5.56 Design Decisions-Chain Type and Teeth Numbers: 14 Chain number: Use Table 7-7 Chain pitch: p = in 15 Number of Teeth: N = Per Table 7-7 16 0.08 0.20 0.39 0.75 1.32 1.46 2.15 3.52 0.07 0.17 0.34 0.66 1.15 1.28 1.88 3.08 0.08 0.19 0.36 0.70 1.24 1.37 2.02 3.30 4.55 5.80…arrow_forwardInput Data: Torque needed to overcome rolling friction in rollers, slides and other moving parts, except for Motor and Worm Gear the worm gear T₁ = Length of travel of door: Time for door to open or close: LD = 50 lb-in. 90 in t= 12.5 seconds Pitch diameter for chain sprocket: DPC 1.690 in Weight of Door: P = No. of worm threads: Nw = Worm Pitch diameter: Dw Diametral pitch: Pd Normal pressure angle: Degrees per Radian: Number of gear teeth: Calculated Data: Linear velocity of door and chain (in/sec): Linear velocity of door and chain (ft/min): Output Speed of Gear and Sprocket: Upward Force due to Weight of Door: Фо = = NG= 240 lb 2 1.250 in 12 14.5 degrees 57.2958 degrees 28 Vα= in/sec VC= ft/min NG = rpm FD lb Net Upward Force on Door: Fou lb Torque on gear ignoring rolling friction: TG = lb-in. Formula = FDU FD-2 x Fo (note: Fo is the Max Operating load of the extension springs). This is also the initial tension in the chain. TG = FDU X DPC/2 This is the also the torque on the…arrow_forwardQ5/A: A car with a track of 1.5 m and a wheelbase of 2.9 m has a steering gear mechanism of the Ackermann type. The distance between the front stub axle pivots is 1.3 m. The length of each track arm is 150 mm, and the length of the track rod is 1.2 m. Find the angle turned through by the outer wheel if the angle turned through by the inner wheel is 30°. (6 Marks) Q5/B: Write True on the correct sentences and False on the wrong sentences listed below:- 1- In automobiles, the power is transmitted from the gearbox to the differential through bevel gears. 2- The minimum radius circle drawn to the cam profile is called the base circle. 3- The Proell governor, compared to the Porter governor, has less lift at the same speed. 4- The balancing of rotating and reciprocating parts of an engine is necessary when it runs at a slow speed. (6.5 Marks) ***Best of Luck *** جامعة بابل UNIVERSITY OF BABYLON Examiner: Mohanad R. Hameed Head of Department: Dr. Dhyai H. Jawadarrow_forward
- University of Babylon Collage of Engineering/ Al-Musayab Department of Automobiles Mid Examination/ Stage: 3rd Subject: Theory of Vehicles Date: 14 \ 4 \2025 Time: 1.5 Hours 2025-2024 Q1: The arms of a Porter governor are 250 mm long. The upper arms are pivoted on the axis of revolution, but the lower arms are attached to a sleeve at a distance of 50 mm from the axis of rotation. The weight on the sleeve is 600 N and the weight of each ball is 80 N. Determine the equilibrium speed when the radius of rotation of the balls is 150 mm. If the friction is equivalent to a load of 25 N at the sleeve, determine the range of speed for this position. Q2: In a loaded Proell governor shown in Figure below each ball weighs 3 kg and the central sleeve weighs 25 kg. The arms are of 200 mm length and pivoted about axis displaced from the central axis of rotation by 38.5 mm, y=238 mm, x=303.5 mm, CE 85 mm, MD 142.5 mm. Determine the equilibrium speed. Fe mg E M 2 Q3: In a spring loaded Hartnell type…arrow_forwardusing the theorem of three moments, find all the reactions and supports, I need the calculations onlyarrow_forwardQ.5: (10 Marks) Select the correct answer (choose 10 only) 1. The forward whirling speed is ......... the static structure tilting speed. (a) Less than (b) Higher than (c) equal to 2. The divergence between the forward and backward whirling speeds increases as: (a) The rotating speed increase (b) the polar moment of inertia increases (c) Both (a) and (b) (d) do not change 3. Increasing the system natural frequency can be done by: (a) add masses (b) adding braces and supports (c) increase damping 4. The amplitude of vibration due to external force can be reduced by: (a) Increasing damping (b) Decreasing damping (c) Increasing mass 5. Tuned absorbers are used to: (a) Shift the natural frequency (b) increase damping (c) Increase stiffness 6. Accelerometers sensors contains: г (a) Piezoelectric materials (b) Magnet and coil (c) coil only 7. Increasing the stiffness of the system causes: (a) Less transmitted force (b) more transmitted force (c) Transmitted force does not change 8. The…arrow_forward
- Q.1: (15 Marks) Find the first three natural frequencies and mode shapes of the axial and torsional vibration for a steel shaft free at both ends, having a length of 3 m. Find the subsequent axil motion if the shaft is subjected to the following initial conditions, given that E = 210 GPa, G=80 GPa, p = 7800 kg/m³: f(x)=0 v(x) = {1 2.8arrow_forwardQ.4: (15 Marks) A uniform rotor of mass 500 kg and diametral moment of inertia of 20 kg.m², is supported by identical short bearings of stiffness 1 MN/m in the horizontal and vertical directions. If the distance between the bearings is 0.6 m: (a) What is the corresponding polar moment of inertia if the backward whirling speed is 80% of the static structure tilting natural frequency? (b) Determine the forward whirling speed. 45.27arrow_forwardUniversity of Babylon Collage of Engineering/ Al-Musayab Department of Automobiles Mid Examination/ Stage: 3rd Subject: Theory of Vehicles Date: 14 \ 4 \2025 Time: 1.5 Hours 2025-2024 Q1: The arms of a Porter governor are 250 mm long. The upper arms are pivoted on the axis of revolution, but the lower arms are attached to a sleeve at a distance of 50 mm from the axis of rotation. The weight on the sleeve is 600 N and the weight of each ball is 80 N. Determine the equilibrium speed when the radius of rotation of the balls is 150 mm. If the friction is equivalent to a load of 25 N at the sleeve, determine the range of speed for this position. Q2: In a loaded Proell governor shown in Figure below each ball weighs 3 kg and the central sleeve weighs 25 kg. The arms are of 200 mm length and pivoted about axis displaced from the central axis of rotation by 38.5 mm, y=238 mm, x=303.5 mm, CE 85 mm, MD 142.5 mm. Determine the equilibrium speed. Fe mg E M 2 Q3: In a spring loaded Hartnell type…arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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