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MakcuM [25]
1 year ago
8

A machine has an efficiency of 15%. If the energy input is 300 joules, how much useful energy is generated?(1 point).

Engineering
1 answer:
Crazy boy [7]1 year ago
3 0

With an energy input of 300 J and an efficiency of 15%, the machine produces 45 J of useable energy.

<h3>What is energy?</h3>

Energy is characterized as a quantitative characteristic that is transmitted to a person or a physical system and can be observed in the manner in which labor is accomplished as well as in the generation of heat and light. The sun is one of the most significant energy sources.

The machine's useful energy output, or work output, can be calculated as follows:

Efficiency = work out / work in x 100

0.15 = work out / 300

Multiply by cross

Production = 0.15 x 300

Production: 45 J

Thus, with an energy input of 300 J and an efficiency of 15%, the machine produces 45 J of useable energy.

To learn more about energy, refer to the link below:

brainly.com/question/1932868

#SPJ1

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Most of our knowledge about Earth's interior comes from Seismic waves. Thus, option C is correct.

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2 years ago
Print "userNum1 is negative." if userNum1 is less than 0. End with newline. Assign userNum2 with 2 if userNum2 is greater than 1
miv72 [106K]

Answer:

From the question, we have two variables

1. userNum1

2. userNum2

And we are to print "userNum1 is negative" if userNum1 is less than 0.

Then Assign userNum2 with 2 if userNum2 is greater than 10.

Otherwise, print "userNum2 is less or equal 10.".

// Program is written in C++ Programming Language

// Comments are used for explanatory purpose

// Program starts here

#include<iostream>

using namespace std;

int main ()

{

// Declare variables

int userNum1, userNum2;

// Accept input for these variables

cin>>userNum1, userNum2;

// Condition 1

if(userNum1 < 0)

{

cout<<"userNum1 is negative"<<'\n';

}

// Condition 2

if(userNum2 > 10)

{

userNum2 = 2;

}

// If condition is less than 10

else

{

cout<<"userNum2 is less or equal to 10"<<\n;

}

return 0;

}

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3 0
3 years ago
Air at 1600 K, 30 bar enters a turbine operating at steady state and expands adiabatically to the exit, where the pressure is 2.
djyliett [7]

Solution :

The isentropic efficiency of the turbine is given as :

$\eta = \frac{\text{actual work done}}{\text{isentropic work done}}$

  $=\frac{m(h_1-h_2)}{m(h_1-h_{2s})}$

  $=\frac{h_1-h_2}{h_1-h_{2s}}$

The entropy relation for the isentropic process is given by :

$0=s^\circ_2-s^\circ_1-R \ln \left(\frac{P_2}{P_1}\right)$

$\ln \left(\frac{P_2}{P_1}\right)=\frac{s^\circ_2-s^\circ_1}{R}$

$ \frac{P_2}{P_1}=exp\left(\frac{s^\circ_2-s^\circ_1}{R}\right)$

$\left(\frac{P_2}{P_1}\right)_{s=constant}=\frac{P_{r2}}{P_{r1}}$

Now obtaining the properties from the ideal gas properties of air table :

At $T_1 = 1600 \  K,$

$P_{r1}=791.2$

$h_1=1757.57 \ kJ/kg$

Calculating the relative pressure at state 2s :

$\frac{P_{r2}}{P_{r1}}=\frac{P_2}{P_1}$

$\frac{P_{r2}}{791.2}=\frac{2.4}{30}$

$P_{r2}=63.296$

Obtaining the properties from Ideal gas properties of air table :

At $P_{r2}=63.296$,  $T_{2s}\approx 860 \ K$

Considering the isentropic relation to calculate the actual temperature at the turbine exit, we get:

  $\eta=\frac{h_1-h_2}{h_1-h_{2s}}$

$\eta=\frac{c_p(T_1-T_2)}{c_p(T_1-T_{2s})}$

$\eta=\frac{T_1-T_2}{T_1-T_{2s}}$

$0.9=\frac{1600-T_2}{1600-860}$

$T_2= 938 \ K$

So, at $T_2= 938 \ K$, $h_2=975.66 \ kJ/kg$

Now calculating the work developed per kg of air is :

$w=h_1-h_2$

  = 1757.57 - 975.66

  = 781 kJ/kg

Therefore, the temperature at the exit is 938 K and work developed is 781 kJ/kg.

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