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jekas [21]
4 years ago
15

Consider a thermal energy reservoir at 1500 K that can supply heat at a rate of 150,000 kJ/h. Determine the exergy of this suppl

ied energy, assuming an environment temperature of 25°C.
Engineering
1 answer:
ANEK [815]4 years ago
5 0

Answer:

exergy = 33.39 kW

Explanation:

given data

thermal energy reservoir T2 = 1500 K

heat at a rate = 150,000 kJ/h = \frac{15000}{3600} kW =  41.67 kW

environment temperature T1 = 25°C = 298 K

solution

we get here maximum efficiency that is  reversible efficiency is express as

reversible efficiency = 1 - \frac{T1}{T2}    ...............1

reversible efficiency = 1 - \frac{298}{1500}  

reversible efficiency =  0.80133

and

the exergy of this supplied energy that is

exergy  = efficiency × hat supply   ................2

exergy = 0.80133 × 41.67 kW

exergy = 33.39 kW

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Iteru [2.4K]

The item that is least likely to cause an exhaust emission test failure as of a result of excessive nox emissions is; poorflow through the egr system.

<h3 /><h3>What is an exhaust emission?</h3>

Exhaust emissions simply means substances emitted into the atmosphere from the exhaust discharge nozzle of an aircraft or aircraft engine.

Now, when the test fails as a result  of excessive nox emissions, we can say that high NOx emissions can occur when an engine's air-fuel mixture is too lean. Thus, we can conclude that the reason for exhaust emission test failure as of a result of excessive nox emissions is due to poorflow through the egr system.

Read more about emissions at; brainly.com/question/14154063

7 0
3 years ago
When determining risk, it is necessary to estimate all routes of exposure in order to determine a total dose (or CDI). Recognizi
Allushta [10]

Answer:

The following are the solution to this question:

Explanation:

The Formula for calculating CDI:

\bold{CDI = \frac{C \times CR \times EF \times ED}{BW \times AT}}

_{where} \\ CDI = \text{Chronic daily Intake rate}  (\frac{mg}{kg-day})} \\\\\text{C = concentration of Toluene}\\\\\text{CR = contact rate} \frac{L}{day}\\\\\text{EF = Exposure frequency} \frac{days}{year}\\\\\text{ED = Exposure duration (in years)} = 10 \ \ years\\\\\text{BW = Body weight (kg) = 70 kg for adult}\\\\ \text{AT = average period of exposure (days) }

calculating the value of AT:

=  365 \frac{days}{year}  \times  70 \ year  \\\\ = 25550 \ days

 calculating the value of Intake based drinking:

C = 1 \ \frac{mg}{L}

CR = 2 \frac{L}{day} Considering that adult females eat 2 L of water a day,

EF = 350 \frac{days}{year} for drink

calculating the CDI value:

\to CDI = \frac{(1 \times 2 \times 350 \times 10)}{(70 \times  25550)}\\\\

             = \frac{(2 \times 3500)}{(70 \times  25550)}\\\\ = \frac{(7000)}{(70 \times  25550)}\\\\ = \frac{(100)}{(25550)}\\\\=0.00391 \frac{mg}{ kg-day}

Centered on inhalation, intake:

C = \frac{1 \mu g} { m^3} \ \ \  or \ \ \ \ 0.001  \ \ \frac{mg}{m^3}\\\\CR = 20  \frac{m^3}{day}\\\\EF = 15 \frac{min}{day}  \ \ or\ \  5475 \frac{min}{yr} \ \ \  or \ \  3.80 \frac{days}{year}\\

calculating the value of CDI:

\to CDI = \frac{(0.001 \times 20 \times 3.80 \times 10)}{(70 \times 25550)}

             = \frac{(0.76)}{(1788500)}\\\\= 4.24 \times 10^{-7} \ \ \frac{mg}{kg-day}

7 0
3 years ago
A single-degree-of-freedom mass-spring-damper system is observed during its free vibration and the displacement amplitude decays
AleksandrR [38]

Answer:

Logarithmic decrement is equal to 0.182

Explanation:

given,

amplitude decay = 9 dB          

number of cycles = 12 cycles        

mass of the system = 7 kg        

spring stiffness = 3000 N/m            

logarithmic decrement = ?                  

now,                                                      

logarithmic decreament = ln\ D^{\frac{1}{n}}

                                        = ln\ 9^{\frac{1}{12}}

                                        =ln (1.2)

                                        = 0.182

Hence, Logarithmic decrement is equal to 0.182

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4 years ago
1.-¿Cómo actuaba el virus Miguel Ángel?
Black_prince [1.1K]
I don’t know so sorry about that
4 0
3 years ago
This question is in C programming.Write a loop that sets newScores to oldScores shifted once left, with element 0 copied to the
andriy [413]

Answer:

Replace

/* Your solution goes here */

with the following;

for (i = 0; i < SCORES_SIZE - 1; i++) {

newScores[i] = oldScores[i + 1];

}

newScores[SCORES_SIZE - 1] = oldScores[0];

The full program becomes

#include

int main(void) {

const int SCORES_SIZE = 4;

int lowerScores[SCORES_SIZE];

int i;

for (i = 0; i < SCORES_SIZE; ++i) {

scanf("%d", &(lowerScores[i]));

}

for (i = 0; i < SCORES_SIZE - 1; i++) {

newScores[i] = oldScores[i + 1];

}

newScores[SCORES_SIZE - 1] = oldScores[0];

for (i = 0; i < SCORES_SIZE; ++i) {

printf("%d ", lowerScores[i]);

}

printf("\n");return 0;

}

What the included code segment does it that;

for (i = 0; i < SCORES_SIZE - 1; i++) {

newScores[i] = oldScores[i + 1];

}

The above assigns the values of array oldScores to newScores starting with the second element of array oldScores till the last.

newScores[SCORES_SIZE - 1] = oldScores[0];

The above line then assigns the first element of oldScores to the first element of newScores array

7 0
4 years ago
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