REFRIGERATION+AIR COND.TECH.-W/2 ACCESS
REFRIGERATION+AIR COND.TECH.-W/2 ACCESS
8th Edition
ISBN: 9781337370042
Author: TOMCZYK
Publisher: CENGAGE L
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Chapter 2, Problem 26RQ

How many Btu of heat can be produced by 4 kWh ofelectricity?

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Design a speed warning system that receives on two lines, an indication of the speed limit on the highway. There are three possible values 45, 55, or 65 MPH. It receives from the automobile, on two other lines, an indication of the speed of the vehicle. There are four possible values under 45, between 46 and 55, between 56 and 65, and over 65 MPH. It produces two outputs. The first, F, indicates whether the car is going above the speed limit. The second, G, indicates that the car is driving at a "dangerous speed" _ defined as either over 65 MPH or more than 10 MPH above the speed limit. The inputs are coded as follows: Speed limit A B Speed C D 45 0 0 65 1 1 a) Complete the Truth table for the speed warning system Inputs Outputs A B C D F G b) Write the Minterm expressions for the Output Functions. c) Minimize the output functions d) Implement the logic e) Using LabView, verify the Correctness of the Speed Warning System.
Ex. The cantilever beam shown us, made ofrem steel with = 552 MPa is , ut subjected to fully reversed load. Neglect shear stress effect, estimate wheather the beam is safe or not safe at N= lo cycles 9 The beam is machined surface and the operating temp. is 100C. A F 200 a=0 N -200 N 1 L= 10cm D time 764 Yze.25 Gm L D= 1.3 cm d = 1 cm b= 1 cm -momend diagram AA -FL at the root of the cantilever, the bending moment is max. factor Ex. Repeat Ex. in page (24), with fluctuating load as shown below. By = 46242,041 = 552 MPa. Find the safety (NF) using Modified -Goodman, Gerber, and soderberg criterias F(N). ....400 time
Example The bar Shown Is. subsected to combined loadings as follows: P = = 20.KN F = 0 to 2 KN T= 0-5 KN.m The bar material has Se=400 mpa and out = 1379 MPa, Yiu= 1000 mpa Find the factor of safety, neglect the transverse Shear- Sol I load analysis ар F T Par = 20 KN (axial load- Pmim = -20 KN fully reversed dynamic) Emark 2km = Fmim = 0 N (bending loading -dynamic) Repeated D=20mm d=15 mm r=5mm L= =250mm

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REFRIGERATION+AIR COND.TECH.-W/2 ACCESS

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