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3 years ago

Some easy points points you

Engineering

2 answers:

Ira Lisetskai [31]3 years ago

8 0

Answer:

thanks appreciate it. also can u mark this as brainliest and thank it and 5 star rate it :)

Explanation:

Mila [183]3 years ago

5 0

Answer:

thanks -_-

Explanation:

be cuz it nice

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An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are a

DedPeter [7]

Answer:

a) he rate of heat loss from the steam line is 18.413588 kW

b) the annual cost of heat loss from line is $12904.25

Explanation:

first we find the area

A = πdL

d is the diameter (0.1m) and L is the length (25m)

A = π × 0.1 × 25

A = 7.85 m²

Now rate of heat loss through convection

qconv = hA(Ts -Ta)

h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)

so we substitute

qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)

qconv = 9817.477 J/s

Now heat lost through radiation

qrad = ∈Aα ( Ts⁴ - Ta⁴)

∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),

first we shall covert our temperatures from Celsius to kelvin scale

Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)

so we substitute

qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )

qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰

qrad = 8596.112 J/s

Now to get the total rate of heat loss through convection and radiation, we say

q = qconv + qrad

q = 9817.477 + 8596.112

q = 18413.588 J/s ≈ 18.413588 kW

Therefore the rate of heat loss from the steam line is 18.413588 kW

annual cost of heat lost rate

A = C × q/n × ( 3600 × 24 × 365 )

C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)

so we substitute

A = 0.02/10⁶ × 18413.588/0.9 × ( 3600 × 24 × 365 )

A = $12904.25

Therefore the annual cost of heat loss from line is $12904.25

4 0

4 years ago

Technician A says that a 12 Volt light bulb that draws 12 amps has a power output of 1 watt. Technician B says that a motor that

Artemon [7]

Answer:

Technician B

Explanation:

Resistance, $R=\frac {V}{I}$ where V is the voltage and I is the current in amps

Therefore, $R=\frac {12}{12}=1 ohm$

Power=VI=12*12=144 W

Therefore, the power is 144 W and resistance is 1 Ohm. This implies that technician A is wrong while technician B is correct

6 0

4 years ago

Quadrature encoders operate with two photodetectors offset by _______ degrees. (a) 45 (b) 90 (c) 180 (d) 270

Lemur [1.5K]

Answer:

i think it should be 90 degree

5 0

3 years ago

Listed below are snippets from a prgram to perform input validation for

leva [86]

Answer:

#include <iostream>

#include <string>

#include "user.h"

#include "password.h"

using namespace Authenticate;

using namespace std;

int main()

{

inputUserName();

inputPassword();

cout << "Your username is " << getUserName() <<

" and your password is: " <<

getPassword() << endl;

return 0;

}

user.h:

#ifndef USER_H

#define USER_H

#include <string>

using namespace std;

namespace Authenticate

{

namespace

{

bool isvalid();

}

void inputUserName();

string getUserName();

}

#endif

user.cpp:

#include <iostream>

#include "user.h"

namespace Authenticate

{

string username="";

namespace

{

bool isvalid()

{

if(username.length() == 8)

return true;

else

return false;

}

void inputUserName(){

{

cout << "Enter your username (8 letters only)" << endl;

cin >> username;

}

while(!isvalid());

}

string getUserName()

{

return username;

}

password.h:

#ifndef PASSWORD_h

#define PASSWORD_h

#include <string>

using namespace std;

namespace Authenticate

{

namespace

{

bool isValid();

}

void inputPassword();

string getPassword();

}

#endif

password.cpp:

#include <iostream>

#include <string>

using namespace std;

namespace Authenticate

{

string password="";

namespace

{

bool isValid()

{

if(password.length() >= 8)

{

for(int i=0; i<password.length(); i++)

if(password[i] >= '0' && password[i] <= '9')

return true;

return false;

}

else

return false;

}

void inputPassword(){

{

cout << "Enter your password (at least 8 characters " <<

"and at leat one non-letter)" << endl;

cin >> password;

}

while(!isValid());

}

string getPassword()

{

return password;

}

3 0

3 years ago

2. Given that a quality-control inspection can ensure that a structural ceramic part will have no flaws greater than 100 µm (100

MA_775_DIABLO [31]

Answer:

SiC=169.26 Mpa

Partially stabilized zirconia=507.77 Mpa

Explanation:

Stress intensity, K is given by

$K=\sigma Y\sqrt{\pi a}$ hence making $\sigma$ the subject where $\sigma$ is applied stress, Y is shape factor, a is crack length

$\sigma=\frac {K}{Y\sqrt{\pi a}}$ substituting the given figures and assuming shape factor, Y of 1

$\sigma=\frac {3}{1\sqrt{\pi 100*10^{-6}}}= 169.2569\approx 169.26 Mpa$

<u>Stabilized zirconia </u>

Stress intensity, K is given by

$K=\sigma Y\sqrt{\pi a}$ hence making $\sigma$ the subject where $\sigma$ is applied stress, Y is shape factor, a is crack length

$\sigma=\frac {K}{Y\sqrt{\pi a}}$ substituting the given figures and assuming shape factor, Y of 1

$\sigma=\frac {9}{1\sqrt{\pi 100*10^{-6}}}= 507.7706\approx 507.77 Mpa$

7 0

3 years ago

Some easy points points you ​

Some easy points points you