AUTOMATIC TRANS...-CLASSRM.+SHOP MANUAL
7th Edition
ISBN: 9781337792158
Author: ERJAVEC
Publisher: CENGAGE L
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Textbook Question
Chapter 24, Problem 8SA
Type B engine misfires are excessive if the misfiring exceeds
to
percent in a
rpm period.
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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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۲۷
- 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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5
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bar K.
Dr. Abduljabbo
Hammade
27/12/2024
Chapter 24 Solutions
AUTOMATIC TRANS...-CLASSRM.+SHOP MANUAL
Ch. 24 - Describe the difference between an open- and a...Ch. 24 - Explain the use and importance of system strategy...Ch. 24 - Describe an OBD II warm-up cycle.Ch. 24 - Explain the trip and drive cycle in an OBD II...Ch. 24 - Describe how engine misfire is detected in an OBD...Ch. 24 - Describe the purpose of having both upstream and...Ch. 24 - Briefly describe the monitors in an OBD II system.Ch. 24 - Type B engine misfires are excessive if the...Ch. 24 - The monitor system checks the action of the can...Ch. 24 - The monitor system has a(n) and test to check the...
Ch. 24 - The fuel monitor checks fuel trim and fuel trim.Ch. 24 - Which sensor is used for misfire monitoring on OBD...Ch. 24 - Which of the following statements is not true?...Ch. 24 - A computer is capable of doing all of the...Ch. 24 - Which of the following memory circuits is used to...Ch. 24 - Technician A says that the oxygen sensor provides...Ch. 24 - While discussing OBD II systems: Technician A says...Ch. 24 - While discussing the catalyst efficiency monitor:...Ch. 24 - While discussing monitoring systems: Technician A...Ch. 24 - While discussing the comprehensive monitoring...Ch. 24 - Prob. 6ASRQCh. 24 - While discussing diagnostic procedures: Technician...Ch. 24 - Technician A says that the enable criteria are the...Ch. 24 - While discussing PCM monitor tests: Technician A...Ch. 24 - While discussing the misfire monitor: 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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