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NARA [144]
4 years ago
5

For a fluid with a Prandtl Number of 1000.0, the hydrodynamic layer is thinner than the thermal boundary layers. a) True b) Fals

e
Engineering
1 answer:
kvv77 [185]4 years ago
8 0

Answer:

(b)False

Explanation:

Given:

 Prandtl number(Pr) =1000.

We know that   Pr=\dfrac{\nu }{\alpha }

  Where \nu is the molecular diffusivity of momentum

             \alpha is the molecular diffusivity of heat.

 Prandtl number(Pr) can also be defined as

    Pr=\left (\dfrac{\delta }{\delta _t}\right )^3

Where \delta is the hydrodynamic boundary layer thickness and \delta_t is the thermal boundary layer thickness.

So if Pr>1 then hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

In given question Pr>1 so  hydrodynamic boundary layer thickness will be greater than thermal boundary layer thickness.

So hydrodynamic layer will be thicker than the thermal boundary layer.

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Technician A says that weld-on dent removal attachments may be used on a steel panel.
spin [16.1K]

Answer:

sjasfjajfhajshfdff

Explanation:

7 0
3 years ago
A particular cloud-to-ground lightning strike lasts 500 µµsec and delivers 30 kA across a potential difference of 100 MV. Assu
sasho [114]

Answer:

a) 15 C charge was delivered by the lightening bolt

b) the lightning delivered 3.0 × 10¹² W of power

c)

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is 416666.67 Wh

d)

the residential retail value of the energy delivered by the strike is $ 40.83

e)

a total of 26 lightening strikes would be required to power an average US home for a year.

Explanation:

Given that;

the lighting strike lasted for t ( time ) = 500 μsecs = 500×10⁻⁶ s

Current I = 30 kA = 30×10³ A

voltage V = 100 mV = 100×10⁶ v

a)

we know that; I = Q/t

so Q = I × t

we substitute

Q =  30×10³ × 500×10⁻⁶

Q = 15 C

Therefore 15 C charge was delivered by the lightening bolt

b)

Power P = V × I

we substitute

Power P = 100×10⁶ × 30×10³

P = 3.0 × 10¹² W

Therefore, the lightning delivered 3.0 × 10¹² W of power

c)

we know that; Power = Energy / Time

Energy = Power × Time

we substitute

E = 3.0 × 10¹²  × 500×10⁻⁶

E = 1500 × 10⁶ J

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is;

⇒ 1500×10⁶ / 3600 Wh

= 416666.67 Wh

d)

given that;  1 KWh → $ 0.098

energy delivered by the strike = 416666.67 Wh = 416.66667 KWh

so the residential retail value of the energy delivered by the strike will be;

416.66667 KWh × $ 0.098

= $ 40.83

∴ the residential retail value of the energy delivered by the strike is $ 40.83

e)

Given that; average monthly residential energy consumption is 900 kWh.

for a year; energy consumption = 12 × 900 kWh = 10,800 KWh = 10800000 Wh

Now

1 lightening strike ⇒ 416666.67 Wh

x lightening strike ⇒ 10800000 Wh

x = 10800000 / 416666.67

x = 25.9199 ≈ 26

Therefore; a total of 26 lightening strikes would be required to power an average US home for a year.  

7 0
3 years ago
6.27 LAB: Convert to binary - functions Write a program that takes in a positive integer as input, and outputs a string of 1's a
Olin [163]

Answer:

//Convert any decimal number to binary number

//Program is written in C++ Programming Language

// Comments are used for explanatory purpose

// Program starts here

#include <iostream>

using namespace std;

// Main Method declared here

int main()

{

    int x;

   cout<<"Enter any integer number: ";

    cin>>x;

   DecBin(x);

   return 0;

}

// Here a function named DecBin is declared along with an integer variable, x

void DecBin(int x)

{

   // Declare an array to store the resulting binary digits

   int bindigit[32];

   // counter for binary array

   int kount = 0;

   while (x > 0) {

       // Store the remainder of each division in the declared array

       bindigit[kount] = x % 2;

       x = x / 2;

       kount++;

   }

   // Loop to print the binary digits in reverse order

   for (int j = i - 1; j >= 0; j--)

          {

       cout << bindigit[j];

        }

}

// End of Program

4 0
3 years ago
use engineering judgement to estimate the size of cooling and heating equipment that is needed for a three story, 31,000 square
Blizzard [7]

Answer: operating cost = 22820.736 $

Energy = 32.291 KBtu/sf.

Explanation:

Total heating load is given as = 31000 * 31.25  

 = 968.75* 103 Btu

From the cooling capacity application;

If 1000ftsquare = 2.8 TR

Therefore 30000ftsquare = x

Where x is the total cooling load,

Therefore x = (30000 * 2.8) / 1000

x = 84TR.

Therefore, the total cooling load, x = 84TR

Using conversion factor;

i.e. converting Btu/hr to MBH

1 MBH = 1000 Btu per hour

we have 968.75* 103 Btu/hr = 968.75 MBH

let us proceed.

Estimating the “Annual operating cost” we first calculate the maximum operating cost (Avg/Annum)

For cooling load:

Max. operatn. Cost = 14 * 84 = 1176$

For heating load:

The average heating loading hours = 1000 hrs.

Conversion to Btu/yr gives = 968.75* 103 * 1000  

 = 968.75* 106 Btu/yr.

Conversion of 968.75* 103Btu to Kwh gives  

1Btu = 0.000293

968.75* 103 Btu = 283.84 Kwh  

Therefore, average cost gives;

(8.04/100) $ = 1Kwh

X = 283.84*103

X = 22820.736 $

Operating cost = 22820.736 $

Energy use in KBtu/sf is given as  

E = 968.75* 103 / 30000  

E = 32.291 KBtu/sf.

8 0
4 years ago
The 5 ft wide gate ABC is hinged at C and contacts a smooth surface at A. If the specific weight of the water is 62.4 lb/ft3 , f
pochemuha

Answer:

The solution is given in the attachments.

5 0
3 years ago
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