Chapter1_B - Chapter 1 - Section B - Non-Numerical...

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Chapter 1 - Section B - Non-Numerical Solutions 1.1 This system of units is the English-system equivalent of SI. Thus, g c = 1(lb m )(ft)(poundal) 1 (s) 2 1.2 ( a ) Power is power , electrical included. Thus, Power [=] energy time [ = ] N · m s [ = ] kg · m 2 s 3 ( b ) Electric current is by definition the time rate of transfer of electrical charge. Thus Charge [=] (electric current)(time) [=] A · s ( c ) Since power is given by the product of current and electric potential, then Electric potential [=] power current [ = ] kg · m 2 A · s 3 ( d ) Since (by Ohm’s Law) current is electric potential divided by resistance, Resistance [=] electric potential current [ = ] kg · m 2 A 2 · s 3 ( e ) Since electric potential is electric charge divided by electric capacitance, Capacitance [=] charge electric potential [ = ] A 2 · s 4 kg · m 2 1.3 The following are general: ln x = ln 10 × log 10 x ( A ) P sat / kPa = P sat / torr × 100 750 . 061 kPa torr ( B ) t / C = T / K 273 . 15 ( C ) By Eqs. ( B ) and ( A ), ln P sat / kPa = ln 10 × log 10 P sat / torr + ln 100 750 . 061 The given equation for log 10 P sat / torr is: log 10 P sat / torr = a b
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This note was uploaded on 04/18/2011 for the course CEM 497 taught by Professor Ku during the Spring '11 term at Punjab Engineering College.

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Chapter1_B - Chapter 1 - Section B - Non-Numerical...

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