A supersonic wind tunnel is to be operated at M=2.2 in the test section. Upstream from the test section, the nozzle throat area is 0.07 m². Air is supplied at stagnation conditions of 500 K and 1.0 MPa (abs). At one flow condition, while the tunnel is being brought up to speed, a normal shock stands at the nozzle exit plane. The flow is steady. For this “starting” condition, immediately downstream from the shock find (a) the Mach number, (b) the static pressure, (c) the stagnation pressure, (d) the minimum area theoretically possible for the second throat downstream from the test section.
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To find the answers to these questions, we will need to apply the conservation equations of fluid mechanics, namely the conservation of mass, momentum, and energy.
(a) Find the Mach number downstream from the shock:
First, we need to determine the Mach number downstream from the shock. To do this, we can use the normal shock relations. One of these relations is the ratio of the Mach numbers before and after the shock, known as the Oblique Shock Relation (also called Prandtl-Meyer relation). However, since the flow is normal to the shock, the relation simplifies to the normal shock relation.
We can use the normal shock relation equation:
M2 = sqrt((2 + (γ - 1) * M1^2) / (2 * γ * M1^2 - (γ - 1)))
Given the upstream Mach number before the shock (M1 = 2.2), and assuming air as an ideal gas with a specific heat ratio (γ) of approximately 1.4, we can substitute these values into the equation to solve for M2.
(b) Find the static pressure downstream from the shock:
To find the static pressure downstream from the shock, we can use the isentropic relations for a normal shock. One of these relations is the ratio of static pressures before and after the shock:
P2/P1 = 1 + (2 * γ / (γ + 1)) * (M1^2 - 1)
Given the stagnation pressure (P1 = 1.0 MPa = 1.0 * 10^6 Pa), we can substitute the known values into the equation to solve for P2.
(c) Find the stagnation pressure downstream from the shock:
The stagnation pressure downstream from the shock is equal to the upstream stagnation pressure because normal shocks are isentropic processes. Thus, P1 = P3.
(d) Find the minimum area for the second throat downstream from the test section:
The minimum area for the second throat downstream from the test section occurs when the flow reaches choking conditions, i.e., maximum velocity at the throat. At choking conditions, the area at the throat is given by:
A* = A2 / ((γ + 1) / 2)^(γ + 1) / (2(γ - 1))^((γ - 1) / 2)
Given the area of the nozzle throat (A2 = 0.07 m^2) and the specific heat ratio (γ = 1.4), we can substitute these values into the equation to find the minimum area (A*) for the second throat downstream from the test section.
By applying these equations, you can calculate the solutions to the respective questions.