JNTU
JNTU B.Tech III AERODYNAMICS II Previous Question Paper (February - 2022 - R13)
Answer any five questions. All questions carry equal marks.
Main
Derive an expression for One-dimensional steady flow momentum equation.
Answer
The one-dimensional steady flow momentum equation is derived from Newton's Second Law of Motion, which states that the net force acting on a fluid control volume is equal to the rate of change of momentum of the fluid passing through it. Consider a steady flow through a stream tube between two cross-sections 1 and 2 with areas A1 and A2, velocities u1 and u2, and pressures p1 and p2.
Applying the control volume approach, the net force F acting on the fluid in the x-direction is the sum of pressure forces and body forces (such as gravity). For a horizontal stream tube, the momentum balance is given by:
Where ṁ is the mass flow rate, defined as ρAu. The pressure forces acting on the control volume are p₁A₁ at the inlet and p₂A₂ at the outlet, with an average pressure p_avg acting on the side walls of the stream tube. The force balance is expressed as:
In the case of a constant area duct where A₁ = A₂ = A, the equation simplifies significantly as the side wall pressure force term vanishes:
Rearranging this yields the final form of the one-dimensional steady flow momentum equation:
This expression demonstrates that for steady, one-dimensional, incompressible flow in a constant area duct, the sum of pressure and momentum flux per unit area remains constant.
Derive an expression of entropy change in terms of stagnation pressure.
Answer
For an ideal gas undergoing an adiabatic process, the Gibbs equation relates entropy change to pressure and temperature changes. Starting from the relation for a change in specific entropy, ds = cp(dT/T) - R(dp/p), and using the definition of stagnation temperature T0 and stagnation pressure p0, we consider an isentropic process from a state (p, T) to the stagnation state (p0, T0).
The entropy change between two states 1 and 2 is given by:
For a flow process where the stagnation state is reached isentropically, the entropy remains constant. When considering the change in entropy due to non-isentropic effects (such as shock waves or friction), we relate the stagnation pressure ratio to the entropy change. Using the relation T0/T = 1 + ((γ-1)/2)M2 and the isentropic relation p0/p = (1 + ((γ-1)/2)M2)γ/(γ-1), the entropy change is expressed as:
This shows that the change in entropy is directly proportional to the natural logarithm of the ratio of stagnation pressures, indicating that a decrease in stagnation pressure in an adiabatic flow corresponds to an increase in entropy.
Show that the condition of incompressibility for gases is
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Derive an expression for pressure rise across a normal shock.
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Develop the relation
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Describe flow through a convergent-divergent diffuser, making use of oblique shock.
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Show that (i) Tangential component of the flow velocity is constant across an oblique shock, and (ii) Changes across an oblique shock wave are governed only by the components of velocity normal to the wave.
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Derive an expression for Prandtl-Meyer function.
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Explain over expanded and under expanded flows.
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Derive expression for pressure coefficient for a supersonic flow.
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