Question

Problem 2 The compressed-air requirements of a plant are met by a 100 hp screw compressor that runs at full load during 40 percent of the time and idles the rest of the time during operating hours. The compressor consumes 35 percent of the rated power when idling and 90 percent of the power when compressing air. The annual operating hours of the facility are 3800 hours, and the unit cost of electricity is $0.075 kWh:1 It is determined that the compressed-air requirements of the facility during 60 percent of the time can be met by a 25 hp reciprocating compressor that consumes 95 percent of the rated power when compressing air and no power when not compressing air. It is estimated that the 25 hp compressor runs 85 percent of the time. The efficiencies of the motors of the large and small compressors at or near full load are 0.90 and 0.88, respectively. The efficiency of the large motor at 35 percent load is 0.82. Determine the amount of energy and money saved as a result of switching to the 25 hp compressor during 60 percent of the time.

Answer 1

Given data:

power rating of large compressor = 100 hp

small compressor = 25 hp

Annual operating hours = 3800 hours

unit cost of electricity = $0.075/kwh

unloaded efficiency of the large motor $\eta$motor = 0.82

loaded efficiency of the large motor  $\eta$motor = 0.9

efficiency of the small motor $\eta$motor = 0.88
ANSWER:- Energy consumed of lerge compressor per year (Energy conssumed), = (power rating) (Armaal Hours) large {texto la motor arbetet (14 Tablaroheliste Substitute Hue given values in above esuation, (Energy consumed) = Cicohp') (0-746 kWhp) (3900 hours/ year) larga 5% 0.6 0.82 Tu +0. = (100) (0-746) (3800) (0-3540-41317 +0.940-4444) = 289480 [ 0-256095 +6-39996] = 283480 (0-656055] =. 185918:4714 whlyear. (Energy consumed = 185 -978 kwh/year o large

Energy consumed & small compressoy per year. ☺ (Ehergles con suped) on (POLET) (Hou15) (LF-X TF moto)+(25+772 inlocked in the - (25) (0:746) (3800) [0x0:15+0.95X085)a98. = 70840 (0+0-8075])0-88 = 70870 [0.9176) = 65 631 kwhl year i Energy Savings per year". Energy yoe savings = (energy consumed) ont (energy consumedamalk. = 185998-65037 kwhlycar. = 120947 kWh learn cost saving per yearz (Energy savings) (unte cost of electricity) = (180947) ($0.075) = 90+1.095 [yean . Cost saving 389041-085 lyear,