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2 years ago

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An escalator handles a steady load of 26 people per minute in elevating them from the first to the second floor through a vertic

al rise of 27.5 ft. The average person weighs 124 lb. If the motor which drives the unit delivers 3.8 hp, calculate the mechanical efficiency e of the system.

1 answer:

photoshop1234 [79]2 years ago

3 0

Answer:

$\eta = 70.711\,\%$

Explanation:

The power needed to make the escalator working is obtained by means of the Work-Energy Theorem:

$\dot W = \dot U_{g}$

$\dot W = \dot n \cdot m_{p}\cdot g \cdot \Delta y$

$\dot W = \left(26\,\frac{persons}{min}\right)\cdot (124\,lbm)\cdot \left(32.174\,\frac{ft}{s^{2}}\right)\cdot \left(\frac{1\,lbf}{32.174\,\frac{lbm\cdot ft}{s^{2}} } \right)\cdot (27.5\,ft)$

$\dot W = 88660\,\frac{lbf\cdot ft}{min}\,\left(2.687\,hp\right)$

The mechanical efficiency of the escalator is:

$\eta = \frac{2.687\,hp}{3.8\,hp}\times 100\,\%$

$\eta = 70.711\,\%$

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Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required,

solong [7]

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

P₂= 80 psia

Temperature =°F Temperature is convert to °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67 x -1.817

=- 64.10Btu/lbm

The required work therefore for this isothermal compression is 64.10 Btu/lbm

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2 years ago

Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a

Elena L [17]

Answer:

$\mathbf{\tau_c =5.675 \ MPa}$

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [ $\bar 1$ 01]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

$cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]$

where;

$[d_1\ e_1 \ f_1]$ = directional indices for tensile stress

$[d_2 \ e_2 \ f_2]$ = slip direction

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_2$ = -1 , $e_2$ = 0 , $f_2$ = 1

$cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]$

$cos \ \lambda = \dfrac{1}{\sqrt{2}}$

Also, to find the angle $\phi$ between the stress [001] & normal slip plane [111]

Then;

$cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]$

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_3$ = 1 , $e_3$ = 1 , $f_3$ = 1

$cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]$

$cos \phi= \dfrac{1} {\sqrt{3} }$

However, the critical resolved SS(shear stress) $\mathbf{\tau_c}$ can be computed using the formula:

$\tau_c = (\sigma )(cos \phi )(cos \lambda)$

where;

applied tensile stress $\sigma =$ 13.9 MPa

∴

$\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})$

$\mathbf{\tau_c =5.675 \ MPa}$

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2 years ago

Modify the Rainfall Statistics program you wrote for Programming Challenge 2 of Chapter 7 . The program should display a list of

rjkz [21]

Answer:

#include<iostream>

#include <iomanip>

using namespace std;

const int NUM_MONTHS = 12;

double getTotal(double [], int);

double getAverage(double [], int);

double getLargest(double [], int, int &);

double getSmallest(double [], int, int &);

double getTotal(int rainFall,double NUM_MONTHS[])

{

double total = 0;

for (int count = 0; count < NUM_MONTH; count++)

total += NUM_MONTH[count];

return total;

}

double getAverage(int rainFall,double NUM_MONTH[])

{getTotal(rainFall,NUM_MONTH)

average= total/NUM_MONTHS;

return average;

}

double getHighest(int rainFall, double NUM_MONTHS[]) //I left out the subScript peice as I was not sure how to procede with that;

{

double largest;

largest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month++ ){

if ( values[month] > largest ){

largest = values[month];

return largest;

}

double getSmallest(int rainFall, double NUM_MONTHS[])

{

double smallest;

smallest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month){

if ( values[month] < smallest ){

smallest = values[month];

return smallest;

}

int main()

{

double rainFall[NUM_MONTHS];

for (int month = 0; month < NUM_MONTHS; month++)

{

cout << "Enter the rainfall (in inches) for month #";

cout << (month + 1) << ": ";

cin >> rainFall[month];

while (rainFall[month] < 0)

{

cout << "Rainfall must be 0 or more.\n"

<< "Please re-enter: ";

cin >> rainFall[month];

}

}

cout << fixed << showpoint << setprecision(2) << endl;

cout << "The total rainfall for the year is ";

cout << getTotal(rainFall, NUM_MONTHS)

<< " inches." << endl;

cout << "The average rainfall for the year is ";

cout << getAverage(rainFall, NUM_MONTHS)

<< " inches." << endl;

int subScript;

cout << "The largest amount of rainfall was ";

cout << getLargest(rainFall, NUM_MONTHS, subScript)

<< " inches in month ";

cout << (subScript + 1) << "." << endl;

cout << "The smallest amount of rainfall was ";

cout << getSmallest(rainFall, NUM_MONTHS, subScript)

<< " inches in month ";

cout << (subScript + 1) << "." << endl << endl;

return 0;

}

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2 years ago

Is the COP of a heat pump always larger than 1?

Liono4ka [1.6K]

Answer:

Yes

Explanation:

Yes it is true that COP of heat pump always greater than 1.But the COP of refrigeration can be greater or less than 1.

We know that

COP of heat pump= 1 + COP of refrigeration

It is clear that COP can not be negative .So from the above expression we can say that COP of heat pump is always greater than one.

3 0

2 years ago

I need help on the Coderz Challenge missions 3 part 3. PLEASE HELP!

allsm [11]

Answer:

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