REFRIGERATION+AIR COND.TECH.-W/2 ACCESS
8th Edition
ISBN: 9781337370042
Author: TOMCZYK
Publisher: CENGAGE L
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Chapter 25, Problem 14RQ
What is the function of the current sensing relay in an oil safety control scheme?
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Chapter 25 Solutions
REFRIGERATION+AIR COND.TECH.-W/2 ACCESS
Ch. 25 - Which is not a basic component of the compression...Ch. 25 - Describe modulating fluid flow.Ch. 25 - The purpose of the crankcase pressure-regulating...Ch. 25 - Describe the function of an electric evaporator...Ch. 25 - Prob. 5RQCh. 25 - Briefly describe the function of a...Ch. 25 - Two types of relief valves are_______and ______.Ch. 25 - Why is low-ambient control necessary?Ch. 25 - Which is not a method of low-ambient control on...Ch. 25 - Low-pressure switch
Ch. 25 - High-pressure switchCh. 25 - Low-ambient(switch)Ch. 25 - Oil safety switchCh. 25 - What is the function of the current sensing relay...Ch. 25 - Why is there a time delay in an oil safety...Ch. 25 - What is off-cycle defrost?Ch. 25 - Two types of off-cycle defrost are______and...Ch. 25 - Two methods for accomplishing defrost on...Ch. 25 - Filter drierCh. 25 - Heat exchangerCh. 25 - Suction accumulatorCh. 25 - ReceiverCh. 25 - Prob. 23RQCh. 25 - Prob. 24RQCh. 25 - What will happen if the float in an oil separator...Ch. 25 - The function of the evaporator pressure regulator...Ch. 25 - What is the function of the oil return hole or...Ch. 25 - What is the function of the pressure equalization...
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- 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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