(20pts) Air T[infinity] = 10 °C and u[infinity] = 100 m/s flows over a flat plate. Assume that the density of air is 1.0 kg/m3 a
nd the dynamics viscosity is 1.8 ×10-6 kg/m∙s. (a) Determine the velocity boundary layer thickness at, δ at x = 2 m (x indicates the distance from the leading edge of the plate). (b) Determine local friction coefficient at x = 2 m. (c) Determine the Nusselt number at x = 2 m. Assume that the velocity boundary layer thickness is the same as the thermal boundary layer thickness (δ = δT).
1 answer:
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Answer:
Explanation:
Hello!
To solve this problem you must follow the following steps, which are fully registered in the attached image.
1. Draw the complete outline of the problem.
2. 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.
3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.
3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.
4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values and finds the work done.
5. draw the pV diagram using the 300F isothermal line
Answer:
The correct answer is option (A) 0.060 uF
Note: Kindly find an attached image of the complete question below
Sources: The complete question was well researched from Quizlet.
Explanation:
Solution
Given that:
C₁ = 0.1 μF
C₂ =0.22 μF
C₃ = 0.47 μF
In this case, C₁, C₂ and C₃ are in series
Thus,
Their equivalent becomes:
1/Ceq = (1/C₁ + 1/C₂ +1/C₃
1/Ceq =[ (1/0.1 + 1/0.22 +1/0.47)]
1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]
1/Ceq =[(0.1034 + 0.047 + 0.022)/(0.01034)
1/Ceq =[(0.1724)/(0.01034)]
1/Ceq = [(16.67)]
1/Ceq =(1/16.67) = 0.059μf
Ceq = 0.059μf ≈ 0.060μf
Therefore the equivalent capacitance of the three series capacitors is 0.060μf
Answer:
eccentrcity of orbit is 0.22
Explanation:
GIVEN DATA:
Initial velocity of satellite = 8333.3 m/s
distance from the sun is 600 km
radius of earth is 6378 km
as satellite is acting parallel to the earth therefore
and radial component of given velocity is zero
we have
h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s
we know that
so
solvingt for
therefore eccentrcity of orbit is 0.22