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mestny [16]
3 years ago
13

From the information generated in Prob. 6.4 (from your previous Aero HW#1), calculate the maximum rate of climb for the single-e

ngine light plane at sea level and at 12,000-ft altitude.
Engineering
1 answer:
soldi70 [24.7K]3 years ago
5 0

Answer:

R/C @ sea Level = 42.5 ft/s

R/C @ 12,000 ft = 24.6 ft/s

Explanation:

Given:

- From problem in 6.4:

         excess power @ sea Level = 232 hp

         excess power @ 12,000 ft = 134 hp

- Thrust delivered by the engine P = 3000 lb

Find:

Maximum rate of climb R/C @ sea Level and @12,000 ft altitude.

Solution:

- Maximum rate of climb (i.e no drag + maximum lift) @ sea level:

                  R/C = excess [email protected] level / Thrust delivered by engine P

                  R/C = 232 hp*(550) / 3000 = 42.5 ft / s

- Maximum rate of climb (i.e no drag + maximum lift) @ 12,000 ft:

                  R/C = excess [email protected] level / Thrust delivered by engine P

                  R/C = 134 hp*(550) / 3000 = 24.6 ft / s              

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For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration will raise the carbon concentratio
diamong [38]

This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.

Answer:

the required time to achieve the same concentration at a 4.9 is 83.733 hrs

Explanation:

Given the data in the question;

treatment time t₁ = 11.3 hours

Carbon concentration = 0.444 wt%

thickness at surface x₁ = 1.8 mm = 0.0018 m

thickness at identical steel x₂ = 4.9 mm = 0.0049 m

Now, Using Fick's second law inform of diffusion

x^2 / Dt = constant

where D is constant

then

x^2 / t = constant

x^2_1 / t₁ = x^2_2 / t₂

x^2_1 t₂ = t₁x^2_2

t₂ = t₁x^2_2 / x^2_1

t₂ = (x^2_2 / x^2_1)t₁

t₂ = ( x_2 / x_1 )^2 × t₁

so we substitute

t₂ = ( 0.0049  / 0.0018  )^2 × 11.3 hrs

t₂ = 7.41 × 11.3 hrs

t₂ = 83.733 hrs

Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs

8 0
2 years ago
The following electrical characteristics have been determined for both intrinsic and p-type extrinsic gallium antimonide (GaSb)
xxTIMURxx [149]

Answer:

0.5m^2/Vs and 0.14m^2/Vs

Explanation:

To calculate the mobility of electron and mobility of hole for gallium antimonide we have,

\sigma = n|e|\mu_e+p|e|\mu_h (S)

Where

e= charge of electron

n= number of electrons

p= number of holes

\mu_e= mobility of electron

\mu_h=mobility of holes

\sigma = electrical conductivity

Making the substitution in (S)

Mobility of electron

8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)

0.639=\mu_e+\mu_h

Mobility of hole in (S)

2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))

0.1436 = 7.6*10^{-3}\mu_e+\mu_h

Then, solving the equation:

0.639=\mu_e+\mu_h (1)

0.1436 = 7.6*10^{-3}\mu_e+\mu_h (2)

We have,

Mobility of electron \mu_e = 0.5m^2/V.s

Mobility of hole is \mu_h = 0.14m^2/V.s

6 0
3 years ago
Ma poate ajuta cineva?
kari74 [83]
Da, sigur. cu ce ai nevoie de ajutor?
5 0
3 years ago
What are the inputs, outputs and side effects of a PSP?
Kipish [7]

Answer:

Side effects - sudden loss of balance/ repeated falls

Outputs - sever sickness and could me factual

Inputs/corrections of this- medications and experimental treatments to help slow the process of deterioration

5 0
3 years ago
Why is the back-work ratio much higher in the brayton cycle than in the rankine cycle?
zloy xaker [14]

The back-work ratio much higher in the Brayton cycle than in the Rankine cycle because a gas cycle is the Brayton cycle, while a steam cycle is the Rankine cycle. Particularly, the creation of water droplets will be a constraint on the steam turbine's efficiency. Since gas has a bigger specific volume than steam, the compressor will have to work harder while using gas.

<h3>What are modern Brayton engines?</h3>

Even originally Brayton exclusively produced piston engines, modern Brayton engines are virtually invariably of the turbine variety. Brayton engines are also gas turbines.

<h3>What is the ranking cycle?</h3>

A gas cycle is the Brayton cycle, while the Ranking cycle is a steam cycle. The production of water droplets will especially decrease the steam turbine's performance. Gas-powered compressors will have to do more work since gas's specific volume is greater than steam's.

Th

To know more about Rankine cycle, visit: brainly.com/question/13040242

#SPJ4

4 0
1 year ago
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