Show that the solution of thin airfoil theory of a symmetric airfoil given below satisfies the Kutta condition. What angle of at
tack is required to develop a sectional lift coefficient of 0.5. Using the circulation distribution for the symmetric thin airfoil, calculate the sectional pitching moment about a point 0.5 chord from the leading edge.gamma(theta) = 2 alpha U_ infinity 1 + cos theta/sin theta
1 answer:
Answer:
The kutta condition is satisfied
Explanation:
Thin airfoil theory: This can be described as a simple theory of airfoils that relates angle of attack to lift for incompressible, inviscid flows.
The theory simply puts the idea that the flow around an airfoil is as two-dimensional flow around a thin airfoil. It can be imagined as addressing an airfoil of zero thickness and infinite wingspan.
kindly check attachment for the calculation of the sectional pitching moment about a point 0.5 chord from the leading edge.gamma(theta) = 2 alpha U_ infinity 1 + cos theta/sin theta.
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Answer:
a)temperature=69.1C
b)3054Kw
Explanation:
Hello!
To solve this problem follow the steps below, the complete procedure is in the attached image
1. draw a complete outline of the problem
2. to find the temperature at the turbine exit use termodinamic tables to find the saturation temperature at 30kPa
note=Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
3. Using thermodynamic tables find the enthalpy and entropy at the turbine inlet, then find the ideal enthalpy using the entropy of state 1 and the outlet pressure = 30kPa
4. The efficiency of the turbine is defined as the ratio between the real power and the ideal power, with this we find the real enthalpy.
Note: Remember that for a turbine with a single input and output, the power is calculated as the product of the mass flow and the difference in enthalpies.
5. Find the real power of the turbine
Solution :
Given :
External diameter of the hemispherical shell, D = 500 mm
Thickness, t = 20 mm
Internal diameter, d = D - 2t
= 500 - 2(20)
= 460 mm
So, internal radius, r = 230 mm
= 0.23 m
Density of molten metal, ρ =
=
The height of pouring cavity above parting surface is h = 300 mm
= 0.3 m
So, the metallostatic thrust on the upper mold at the end of casting is :
Area, A
= 7043.42 N
Answer:
investment 10 years from now is $1,238,000 .
Explanation:
given data
sum = $500,000
rate = 12% =0.12
total time = 10 year
solution
as present value After 2 years from now is $500,000
so time period is now = 8 year ( 10 - 2 )
so we apply future value formula that is
Future value = present value × ............1
put here value we get
Future value = $500,000 ×
Future value = $500,000 × 2.476
Future value = $1,238,000
so investment 10 years from now is $1,238,000 .