Thermodynamics: An Engineering Approach 8th Edition

Published by McGraw-Hill Education
ISBN 10: 0-07339-817-9
ISBN 13: 978-0-07339-817-4

Chapter 1 - Introduction and Basic Concepts - Problems - Page 48: 1-97E

Answer

$T_{equiv}$ = 33.0 - (33.0 - $T_{ambient}$) $\times$ (0.475 - 0.0126V + 0.240$\sqrt V$)

Work Step by Step

The given equation is: $T_{equiv, °F}$= 91.4 - (91.4- $T_{ambient, °F}$) $\times$ (0.475 - 0.0203$V_{\frac{mi}{h}}$ + 0.304$\sqrt V_{\frac{mi}{h}}$ Knowing the conversion formulas to be: (1) $T_{°F}$ = $T_{°C}$ $\times$ 1.8 + 32 (2) $V_{\frac{mi}{h}}$ = $V_{\frac{km}{h}}$ $\div$ 1.6 We can plug (1) and (2) into the original equation: ($T_{equiv, °C}$ $\times$ 1.8 + 32)= 91.4 - (91.4- ($T_{ambient, °C}$ $\times$ 1.8 + 32)) $\times$ (0.475 - 0.0203($V_{\frac{km}{h}}$ $\div$ 1.6) + 0.304$\sqrt (V_{\frac{km}{h}} \div 1.6)$ $T_{equiv, °C}$ $\times$ 1.8= 91.4 -32 - (91.4 -32 - $T_{ambient, °C}$ $\times$ 1.8 ) $\times$ (0.475 - 0.0126$V_{\frac{km}{h}}$ + 0.240$\sqrt V_{\frac{km}{h}}$ Dividing by 1.8 on both sides, and since (91.4 -32) $\div$ 1.8 = 33: $T_{equiv, °C}$= 33.0 - (33.0 - $T_{ambient, °C}$ ) $\times$ (0.475 - 0.0126$V_{\frac{km}{h}}$ + 0.240$\sqrt V_{\frac{km}{h}}$
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