Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, Binder Ready Version 8th Edition

Published by Wiley
ISBN 10: 1119080703
ISBN 13: 978-1-11908-070-1

Chapter 1 - Problems - Page 33: 1.53

Answer

Calculate the density of Helium by using the following equation: $$ \rho=\frac{p}{R T} \ldots \ldots(1) $$ Here, density of ideal gas is $\rho$, Absolute pressure is $p$, and gas constant is $R$ absolute temperature is $T$ Sample calculation: Gas constant for Helium is $2077 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$ Substitute $50 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2}$ for $p, 2077 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$ for $R$, and $50{ }^{\circ} \mathrm{C}$ for $T$ in the equation $(1)$ $$ \begin{aligned} \rho &=\frac{50 \times 10^{3}}{2077 \times(50+273)} \\ &=\frac{50 \times 10^{3}}{670871} \\ &=0.0745 \mathrm{~kg} / \mathrm{m}^{3} \end{aligned} $$ Calculate remaining density values by varying pressure and temperature values and tabulate the values: $$ \begin{array}{|l|l|l|l|} \hline \begin{array}{l} \text { Pressure in } \\ \mathrm{kPa} \end{array} & \begin{array}{l} \text { Temperature in degree } \\ \text { Celsius } \end{array} & \begin{array}{l} \text { Gas constant for Helium in } \\ \mathrm{J} / \mathrm{kg}-\mathrm{K} \end{array} & \begin{array}{l} \text { Density in } \\ \mathrm{kg} / \mathrm{m}^{3} \end{array} \\ \hline 50 & 50 & 2077 & 0.0745 \\ \hline 100 & 100 & 2077 & 0.129 \\ \hline 200 & 200 & 2077 & 0.1707 \\ \hline \end{array} $$

Work Step by Step

Calculate the density of Helium by using the following equation: $$ \rho=\frac{p}{R T} \ldots \ldots(1) $$ Here, density of ideal gas is $\rho$, Absolute pressure is $p$, and gas constant is $R$ absolute temperature is $T$ Sample calculation: Gas constant for Helium is $2077 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$ Substitute $50 \times 10^{3} \mathrm{~N} / \mathrm{m}^{2}$ for $p, 2077 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}$ for $R$, and $50{ }^{\circ} \mathrm{C}$ for $T$ in the equation $(1)$ $$ \begin{aligned} \rho &=\frac{50 \times 10^{3}}{2077 \times(50+273)} \\ &=\frac{50 \times 10^{3}}{670871} \\ &=0.0745 \mathrm{~kg} / \mathrm{m}^{3} \end{aligned} $$ Calculate remaining density values by varying pressure and temperature values and tabulate the values: $$ \begin{array}{|l|l|l|l|} \hline \begin{array}{l} \text { Pressure in } \\ \mathrm{kPa} \end{array} & \begin{array}{l} \text { Temperature in degree } \\ \text { Celsius } \end{array} & \begin{array}{l} \text { Gas constant for Helium in } \\ \mathrm{J} / \mathrm{kg}-\mathrm{K} \end{array} & \begin{array}{l} \text { Density in } \\ \mathrm{kg} / \mathrm{m}^{3} \end{array} \\ \hline 50 & 50 & 2077 & 0.0745 \\ \hline 100 & 100 & 2077 & 0.129 \\ \hline 200 & 200 & 2077 & 0.1707 \\ \hline \end{array} $$
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