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just olya [345]
3 years ago
6

How does an airfoil create lift?

Engineering
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

An airfoil creates lift by exerting a downward force on the air as it flows past

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Which atom bond in atomic interaction combines electrons, filling its valence zone a) Van der Vaals bond; b) a covalent bond; c)
Setler79 [48]

Answer: Covalent bond

Explanation: Covalent bond is the bond that gets created when there is a sharing of electrons among atoms  and hence creating atomic interaction. The bond formed is from the shared pair because they allow the atoms or ions to achieve stability by completely filling the outer shell of the electron and thus form the covalent bond .Therefore, the correct option is the option(b) .

7 0
3 years ago
You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the
Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, \sigma /Strain, ∈

initial Strain, \epsilon_i = \frac{\sigma}{E}

\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}

\epsilon_i = 0.002

creep rate in the steady state

\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )

\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )

but Tinitial = 0

\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001

\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0
3 years ago
Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor i
Temka [501]

Answer:

Explanation:

Find the temperature at exit of compressor

T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k

Find the work done by the compressor

\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg

Find the actual workdone by the compressor

\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg

Find the temperature at exit of the turbine

T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k

Find the actual workdone by the turbine

1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg

Find the temperature of the regeneration

\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k

Find the heat supplied

Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg

Find the thermal efficiency

n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4

60.4%

Find the mass flow rate

m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42

Find the actual workdone by the compressor

\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg

Find the actual workdone by the turbine

\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg

Find the temperature of the compressor exit

\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k

Find the temperature at the turbine exit

4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k

Find the temperature of regeneration

\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k

6 0
3 years ago
Read 2 more answers
The ampere draw of a 5000 watt electric heater used on 120 volts is
Salsk061 [2.6K]

Answer:

41.67amps

Explanation:

v*a=w

120*a=5000

5000/12=41.67

8 0
3 years ago
Only respond if your the person im talkin to
bagirrra123 [75]

Answer: um wuh anyways thxs for the points!

Explanation: ....:/

5 0
3 years ago
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