UPSC 2017 Maths Optional Paper 2 Q7c — Step-by-Step Solution

15 marks · Section B

Euler's equation of motion for inviscid flow · Mechanics & Fluid Dynamics · asked 2× in 13 yrs · Read the full method →

Question

A stream is rushing from a boiler through a conical pipe, the diameters of the ends of which are $D$ and $d$ . If $V$ and $v$ be the corresponding velocities of the stream and if the motion is assumed to be steady and diverging from the vertex of the cone, then prove that $\dfrac{v}{V}=\dfrac{D^2}{d^2}e^{(v^2-V^2)/2K}$ , where $K$ is the pressure divided by the density and is constant.

Technique

Steady-flow continuity $\rho A q=$ const with $A\propto(\text{diameter})^2$ ; compressible Bernoulli $\tfrac{q^2}{2}+\int\frac{dp}{\rho}=$ const; isothermal closure $p=K\rho$ giving $\int dp/\rho=K\ln\rho$ .

Solution

Setup. Steady, compressible flow of gas down a conical (slowly varying) pipe; no body force. Let $\rho$ be the density, $p$ the pressure, and $q$ the speed. The condition ” $K=p/\rho$ is constant” means the flow is isothermal, $p=K\rho$ . Let subscripts denote the two ends:

end of diameter $D$ : speed $V$ , density $\rho_1$ ;
end of diameter $d$ : speed $v$ , density $\rho_2$ .

Step 1 — Continuity (conservation of mass)

For steady flow the mass flux $\rho\,A\,q$ is the same across every cross-section, where $A$ is the area. The pipe is circular so $A\propto(\text{diameter})^2$ ; thus $A_1\propto D^2$ , $A_2\propto d^2$ :

\rho_1\,D^2\,V=\rho_2\,d^2\,v.

Hence

\frac{v}{V}=\frac{\rho_1}{\rho_2}\cdot\frac{D^2}{d^2}. \tag{1}

Step 2 — Bernoulli’s equation for compressible flow

Along a streamline (no gravity, steady), the compressible Bernoulli integral is

\frac{q^2}{2}+\int\frac{dp}{\rho}=\text{const}.

For the isothermal relation $p=K\rho$ we have $dp=K\,d\rho$ , so

\int\frac{dp}{\rho}=\int\frac{K\,d\rho}{\rho}=K\ln\rho.

Therefore

\frac{q^2}{2}+K\ln\rho=\text{const}.

Applying this at the two ends:

\frac{V^2}{2}+K\ln\rho_1=\frac{v^2}{2}+K\ln\rho_2\ \Longrightarrow\ K\ln\frac{\rho_1}{\rho_2}=\frac{v^2-V^2}{2}.

Hence

\frac{\rho_1}{\rho_2}=e^{(v^2-V^2)/2K}. \tag{2}

Step 3 — Combine

Substituting (2) into (1):

Answer

\boxed{\;\frac{v}{V}=\frac{D^2}{d^2}\,e^{(v^2-V^2)/2K}.\;}