Automotive Technology
7th Edition
ISBN: 9781337794213
Author: ERJAVEC, Jack.
Publisher: Cengage,
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Chapter 12, Problem 1ASRQ
While discussing the reasons for resurfacing a cylinder head; Technician A says it should be resurfaced to make it flat and very smooth so the head gasket can seal properly during engine assembly. Technician B says that it can be resurfaced to raise the compression. Who is correct?
- Technician A
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Required information
Air at 25°C (cp=1006 J/kg.K) is to be heated to 58°C by hot oil at 80°C (cp = 2150 J/kg.K) in a cross-flow heat exchanger
with air mixed and oil unmixed. The product of heat transfer surface area and the overall heat transfer coefficient is 750
W/K and the mass flow rate of air is twice that of oil.
NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part.
Air
Oil
80°C
Determine the effectiveness of the heat exchanger.
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
J/kg.K.
Cp=
4181
Water
Steam
What would be the required heat exchanger area in case of parallel-flow arrangement?
The required heat exchanger area in case of parallel-flow arrangement is
1m².
A single-pass crossflow heat exchanger is used to cool jacket water (cp = 1.0 Btu/lbm.°F) of a diesel engine from 190°F to 140°F, using
air (Cp = 0.245 Btu/lbm.°F) at inlet temperature of 90°F. Both air flow and water flow are unmixed. If the water and air mass flow rates
are 85500 lbm/h and 400,000 lbm/h, respectively, determine the log mean temperature difference for this heat exchanger. Assume
the correction factor F to be 0.92.
Air flow
(unmixed)
Water flow
(unmixed)
The log mean temperature difference of the heat exchanger is
°F.
Chapter 12 Solutions
Automotive Technology
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Ch. 12 - Many engines with WT have a phaser mounted to the...Ch. 12 - Multiple valve engines tend to be more efficient...Ch. 12 - Which of the following is not a type of valve...Ch. 12 - What type of valve lifter automatically...Ch. 12 - Prob. 8MCCh. 12 - While discussing the reasons for resurfacing a...Ch. 12 - While inspecting the timing gears of an engine:...Ch. 12 - Technician A says that positive valve stem seals...Ch. 12 - Prob. 4ASRQCh. 12 - Technician A says an engine may use one or more...Ch. 12 - Technician A says it is not necessary to measure...Ch. 12 - While discussing the Valvetronic system:...Ch. 12 - While checking a hydraulic phaser after the...Ch. 12 - While servicing a pair of aluminum cylinder heads:...Ch. 12 - While discussing camshafts Technician A says valve...
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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 explanationsarrow_forwardIn 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 isarrow_forwardStress, 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, MPaarrow_forward
- 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
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