Engineering > QUESTIONS & ANSWERS > Fundamentals of instrumentation and control questions,. Graded A (All)
Fundamentals of instrumentation and control questions,. Graded A
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8.1 convert the following temperatures to Fahrenheit: 1150C, 456K and 4230R a) 1150C Tc = 5/9 (Tf - 32) Tf = 9/5 x Tc +32 = 9/5 x 115 +32 =239 F b) 456K to Fahrenheit Tf = (9/5 x (Tk-273.15)) +... 32 = 9/5 x (456-273.15) +32 = 9/5 x 182.85 + 32 = 361.13 0F c) 4230R to Fahrenheit Tf = TR- 459.69 = 423- 459.69 =-36.690F 8.2 Convert the following temperatures to Rankine: -130C, 645K, -1230F TR = (9/5 x TC + 32) + 459.69 = (9/5 x -13 +32) + 459.69 = 8.6 + 459.69 = 468.290R b) 645 K TR= (9/5 x (Tk – 273.16) + 459.69 = (9/5 x (645 – 273.16) + 32) + 459.69 = 1161.0020R c) -1230F TR= Tf + 459.69 = -123 + 459.69 = 336.690R 8.3 Convert the following temperatures to centigrade a) 1150F Tc= 5/9 (Tf – 32) = 5/9 (115-32) =46.110 C b) 356K Tc =TK -273.16 = 356 – 273.16 = 82.840C I also do class attendance and help with class work, for enquiries reach me through my Google account Wamaesymon [Show More]
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