THERMODYNAMICS LLF W/ CONNECT ACCESS
9th Edition
ISBN: 9781264446889
Author: CENGEL
Publisher: MCG
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Chapter 9.12, Problem 119P
(a)
To determine
State whether the compressor work decrease, increase or remain unchanged.
(b)
To determine
State whether the back work ratio decrease, increase or remain unchanged.
(c)
To determine
State whether the thermal efficiency decrease, increase or remain unchanged.
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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?
What is the reason for this composition?
Homework: 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)
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- 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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