Chapter 7, Problem 7.1P

To determine

The type of gas from given properties and flow conditions.

Answer to Problem 7.1P

The gas is Methane (CH₄).

Explanation of Solution

Given information:

For the ideal gas,

Pressure p₁ = 1 atm

Temperature T₁ = 20° C

Velocity V₁ = 12 m/s

Downstream, at a distance of 60 cm,

The thickness of boundary layer = 5 mm

Calculation:

For the given conditions of flow,

The velocity is 12 m/s which is sufficiently low, the distance from the leading edge is 60 cm which is also very short distance.

For low velocity and sufficiently small distance we can assume that the flow is laminar.

From the relation of thickness of laminar sub-layer:

$\frac{\delta }{x}=\frac{5.0}{\sqrt{{\mathrm{Re}}_x}}$

$\begin{array}{l}\frac{0.005}{0.6}=\frac{5.0}{\sqrt{\frac{V\times x}{\upsilon }}}\\ \frac{0.005}{0.6}=\frac{5.0}{\sqrt{\frac{12\times 0.6}{\upsilon }}}\end{array}$

$\upsilon =2\times {10}^{-5}{\text{m}}^{\text{2}}\text{/s}$

From, the calculated viscosity, the type of gas can easily be identified, the gas with this kinematic viscosity is Methane (CH₄),

Now, from this kinematic viscosity, the type of flow can also be identified by calculating the Reynold’s number.

$\begin{array}{l}\mathrm{Re}=\frac{Vx}{\upsilon }\\ \mathrm{Re}=\frac{12\times 0.6}{2\times {10}^{-5}}\\ \mathrm{Re}=3,60,000<500,000\end{array}$

The Reynold’s number is lesser than 5,00,000, that means the flow is Laminar and hence, the assumed laminar flow condition is also correct.

Conclusion:

Thus, the gas is Methane (CH₄).