Tech Manual for Erjavec's Automotive Technology: A Systems Approach
6th Edition
ISBN: 9781133933731
Author: ERJAVEC, Jack, Thompson, Rob
Publisher: Delmar Cengage Learning
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Chapter 18, Problem 3RQ
To determine
The term counter EMF and its relation with starting a motor.
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1- Determine the following: 1- RSHF? 2- C.C.C in tons-ref. 3- Mass
of supply air?
Fresh
Spray chilled
water
S
air
100% RH
To 34 C db & 26 wbt
S
Operation
fan
room I
Exhaust
air
Ti 22 C db & 50% RH
How do I solve this task
A weight for a lift is suspended using an adapter. The counterweight is held up with 4 screws. The weight F is 3200kg.
The screws have a strength class of 8.8. Safety factor 3
Which is the smallest bunch size that can be used?+_Sr/Fm =0,16Gr=0,71ơ=800·0.8=640 MPaAs=?Fmax= As·ơ·GrFs=ơs·AsFFm= Fs· GFSF =SF / FFm · FFm Fpreload =Fload / SF → Fload /3Fpreload per screw =Fload / SF → Fload /4As=Fpreload per screw /ơ·Gr → As= Fpreload per screw / 640· 0.71
The correct answer should be M12 with As=84.3mm²
...
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سوأل الوجه البينة
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- Find the equivalent resistance between
A and B
bellows
For the circuit shown.
• All resistances in Ohms.
2
C
2
A
4
B
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Dr. Abduljabbo
Hammade
27/12/2024
Chapter 18 Solutions
Tech Manual for Erjavec's Automotive Technology: A Systems Approach
Ch. 18 - True or False? The strength of the magnetic field...Ch. 18 - Prob. 2RQCh. 18 - Prob. 3RQCh. 18 - True or False? Many armature bearings are held in...Ch. 18 - Which of the following is not part of the starter...Ch. 18 - Which of the following tests would not be...Ch. 18 - When the starter spins but does not crank the...Ch. 18 - If the solenoid clicks while trying to crank the...Ch. 18 - Prob. 9RQCh. 18 - Prob. 10RQ
Ch. 18 - Prob. 11RQCh. 18 - The device that prevents the engine from turning...Ch. 18 - The part of the armature that the brushes ride on...Ch. 18 - Describe how to perform a starter bench test.Ch. 18 - A control circuit voltage drop test shows a 1.1...Ch. 18 - An engine cranks slowly: Technician A says that a...Ch. 18 - While discussing armature testing: Technician A...Ch. 18 - Pinion gear-to-flywheel ring gear clearance is...Ch. 18 - Technician A says that a starter no-load test is...Ch. 18 - Technician A checks a starters field coils for...Ch. 18 - Prob. 6ASRQCh. 18 - When diagnosing a no-crank no-start condition, no...Ch. 18 - While discussing a no-crank condition: Technician...Ch. 18 - Prob. 9ASRQCh. 18 - While discussing electric motors: Technician A...
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- P₂ 7+1 * P₁ ART 2 P (P₁ - P₂- Zgp) 21 / Prove that :- m² a cda A₂ == * Cde actual mip Solutionarrow_forwardQ1/ Show that (actual 02/ A simple iet == Cda Cdf х Af 2/Y - Y+1/Y 2P(P1-P2-zxgxpr)arrow_forward5. Determine the transfer function of G(s) = 01(s)/T₁(s) and 02(s)/T₁ for the mechanical system shown in Figure Q5. (Hints: assume zero initial condition) T₁(t) 01(t) 102(1) Ол N1 D1 D2 No. 1790220000 N2 Figure Q5 K2arrow_forward
- A spring package with two springs and an external force, 200N. The short spring has a loin of 35 mm. Constantly looking for spring for short spring so that total compression is 35 mm (d). Known values: Long spring: Short spring:C=3.98 N/mm Lo=65mmLo=87.4mmF=c·fTotal compression is same for both spring. 200 = (3.98(c1) × 35) + (c₂ × 35) 200 = 139.3 + 35c₂ 200 - 139.3 = 35c₂ 60.7 = 35c₂ c₂ = 60.7/35 Short spring (c₂) = 1.73 N/mm According to my study book, the correct answer is 4.82N/mm What is wrong with the calculating?arrow_forwardWhat is the reason for this composition?arrow_forwardHomework: ANOVA Table for followed design B AB Dr -1 -1 1 (15.18,12) 1 -1 -1 (45.48.51) -1 1 -1 (25,28,19) 1 1 (75.75,81)arrow_forward
- 20. [Ans. 9; 71.8 mm] A semi-elliptical laminated spring is made of 50 mm wide and 3 mm thick plates. The length between the supports is 650 mm and the width of the band is 60 mm. The spring has two full length leaves and five graduated leaves. If the spring carries a central load of 1600 N, find: 1. Maximum stress in full length and graduated leaves for an initial condition of no stress in the leaves. 2. The maximum stress if the initial stress is provided to cause equal stress when loaded. [Ans. 590 MPa ; 390 MPa ; 450 MPa ; 54 mm] 3. The deflection in parts (1) and (2).arrow_forwardQ6/ A helical square section spring is set inside another, the outer spring having a free length of 35 mm greater than the inner spring. The dimensions of each spring are as follows: Mean diameter (mm) Side of square section (mm) Active turns Outer Inner Spring Spring 120 70 8 7 20 15 Determine the (1) Maximum deflection of the two springs and (2) Equivalent spring rate of the two springs after sufficient load has been applied to deflect the outer spring 60 mm. Use G = 83 GN/m².arrow_forwardQ2/ The bumper springs of a railway carriage are to be made of rectangular section wire. The ratio of the longer side of the wire to its shorter side is 1.5, and the ratio of mean diameter of spring to the longer side of wire is nearly equal to 6. Three such springs are required to bring to rest a carriage weighing 25 kN moving with a velocity of 75 m/min with a maximum deflection of 200 mm. Determine the sides of the rectangular section of the wire and the mean diameter of coils when the shorter side is parallel to the axis of the spring. The allowable shear stress is not to exceed 300 MPa and G = 84 kN/mm². Q6/ A belicalarrow_forward
- 11. A load of 2 kN is dropped axially on a close coiled helical spring, from a height of 250 mm. The spring has 20 effective turns, and it is made of 25 mm diameter wire. The spring index is 8. Find the maximum shear stress induced in the spring and the amount of compression produced. The modulus of rigidity for the material of the spring wire is 84 kN/mm². [Ans. 287 MPa; 290 mm]arrow_forwardWhat is the reason for this composition?arrow_forwardHomework: ANOVA Table for followed design B AB Dr -1 -1 1 (15.18,12) 1 -1 -1 (45.48.51) -1 1 -1 (25,28,19) 1 1 (75.75,81)arrow_forward
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