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kiruha [24]
4 years ago
6

Consider a rectangular wing mounted in a full-scale wind tunnel. The wing model completely spans the test section so that the fl

ow sees essentially an infinite wing. The wing has a NACA 0009 airfoil section, a chord of 2.0 m, and a span of 5 m. The tunnel is operated at the following test conditions: P = 95,000 N/m2; T =25◦C; V = 70 m/s; and μ = 1.86 × 10^−5 kg/(m s).
a. Determine the operating Reynolds number.
b. Calculate the lift and drag, about the aerodynamic center for an angle of attack of 10 deg. using the most appropriate curve (indicate which curve you used).
c. Assume in the next test, the most appropriate Reynolds number is 6 × 10^6 and the wing has roughness, but the same dimensions, find the following:

i. What velocity is this wing tested at in the wind tunnel?
ii. What is the stalling angle of attack for this airfoil?
iii. What is the angle of attack for zero lift? and 3) what is the lift- curve slope?
iv. Calculate the lift and drag for case c) at the stall angle.

Engineering
1 answer:
NARA [144]4 years ago
8 0

Answer:

Explanation:

Given the parameters ;

P = 95,000 N/m2; T =25◦C; V = 70 m/s; and μ = 1.86 × 10^−5 kg/(m s).

The detailed steps and appropriate calculation is as shown in the attached files.

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Calculate the number of vacancies per cubic meter for some metal, M, at 783°C. The energy for vacancy formation is 0.95 eV/atom,
djyliett [7]

Answer:

Following are the solution to this question:

Explanation:

The number of vacancies by the cubic meter is determined.  

N_V =N exp(\frac{Q_v}{kT})

      = \frac{N_A \rho}{A} exp (\frac{Q_v}{kT})

      =  \frac{6.022 \times 10^{23} \times 6.10}{43.41} \exp(\frac{-0.95}{8.62\times 10^{-5} \times (783+273)})\\\\=  \frac{36.7342 \times 10^{23}}{43.41} \exp(\frac{-0.95}{0.0313626})\\\\=  0.846215158 \times 10^{23} \exp(-30.290856)\\\\

      =1.57 \times 10^{25} \ cm^{-3}

7 0
3 years ago
Intravenous infusions are usually driven by gravity by hanging the bottle at a sufficient height to counteract the blood pressur
Bingel [31]

Answer:

(a) BP = 11.99 KPa

(b) h = 2 m

Explanation:

(a)

Since, the fluid pressure and blood pressure balance each other. Therefore:

BP = ρgh

where,

BP = Blood Pressure

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = 1.2 m

Therefore,

BP = (1020 kg/m³)(9.8 m/s²)(1.2 m)

<u>BP = 11995.2 Pa = 11.99 KPa</u>

(b)

Again using the equation:

P = ρgh

with data:

P = Gauge Pressure = 20 KPa = 20000 Pa

ρ = density of fluid = 1020 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = height of fluid = ?

Therefore,

20000 Pa = (1020 kg/m³)(9.8 m/s²)h

<u>h = 2 m</u>

7 0
3 years ago
Write a program that uses the function isPalindrome given below. Test your program on the following strings: madam, abba, 22, 67
defon

Answer:

#include <iostream>

#include <string>

using namespace std;

bool isPalindrome(string str)

{

   int length = str.length();

   for (int i = 0; i < length / 2; i++)

   {

       if (tolower(str[i]) != tolower(str[length - 1 - i]))

           return false;

   }

   return true;

}

int main()

{

   string s[6] = {"madam", "abba", "22", "67876", "444244", "trymeuemyrt"};

   int i;

   for(i=0; i<6; i++)

   {

       //Testing function

       if(isPalindrome(s[i]))

       {

           cout << "\n " << s[i] << " is a palindrome... \n";

       }

       else

       {

           cout << "\n " << s[i] << " is not a palindrome... \n";

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5 0
3 years ago
A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o
victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width= 25 mm = 25\times 10^{-3} m

thickness = 6.5 mm = 6.5\times 10^{-3} m

crack length 2c = 0.5 mm at centre of specimen

\sigma _{applied} =  1000 N/cross sectional area

stress intensity factor  =  k  will be

\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}

                   = 6.154\times 10^{6} Pa

we know that

k =\sigma_{applied} (\sqrt{\pi C})

  =6.154\sqrt{\pi (2.5\times 10^{-04})}          [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

ifK_C = 1.15 Mpa m^{1/2} then load will be

Kc = \sigma _{frac}(\sqrt{\pi C})

1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}

\sigma _{frac} = 41.04 MPa

load = \sigma _{frac}\times Area

load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N

LAOD = 6669.86 N

3 0
3 years ago
In one study the critical stress intensity factor for human bone was calculated to be 4.05 MN/m3/2. If the value of Y in Eq. (2.
Diano4ka-milaya [45]
Where is Eq.(28) ?? You should show it to find the result
6 0
3 years ago
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