Problem 2.4:
A gas with a molecular weight of
and a specific heat ratio of
flows through a variable area duct. At some point in the flow the velocity is
m/s
and the temperature is
oC.
At some other point in the flow, the temperature is
oC.
Find the velocity at this point in the flow assuming that the flow is adiabatic.
Solution:
Given:
\( \gamma = \) 1.67,
\( \hat{m} = \) 4 g/mol,
\( V_{1} = \) 180 m/s,
\( T_{1} = \) 10 oC = 283 K,
\( T_{2} = \) -10oC = 263 K,
adiabatic flow \( \implies\dot{q} = 0 \).
To calculate: \( V_{2}=? \)
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The schematic diagram of the problem description is shown in Fig. 1.
The gas constant \( R \) can be calculated using the universal gas constant, \( R_{u} = \) 8314 J/kmol-K,
and given molecular weight, \( \hat{m} = \) 4 g/mol, as
\[
R=\frac{R_{u}}{\hat{m}}=\frac{8314}{4}=2078.5\,\text{J/kg-K}
\]
The heat capacity at constant pressure can be calculated as,
\[
c_{p}=\frac{\gamma\,R}{\gamma-1}=\frac{1.67\times2078.5}{1.67-1}= 5180.7388\,\text{J/kg-K}
\]
Using the conservation of energy,
\[
c_{p}T_{1}+\frac{V_{1}^{2}}{2}+\frac{\dot{q}}{\dot{m}}=c_{p}T_{2}+\frac{V_{2}^{2}}{2}
\]
\[
5180.7388\times283+\frac{180^{2}}{2}+\cancelto{\text{adiabatic}}{\frac{\dot{q}}{\dot{m}}}=5180.7388\times263+\frac{V_{2}^{2}}{2}
\]
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The solution for \( V_{2} \) can be obtained as,
\[
V_{2}=\sqrt{180^{2}+2\times5180.7388\times\left(283-263\right)}
\]
\[
\boxed{V_{2}= 489.52\,\text{m/s}}\ .
\]