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Aneli [31]
3 years ago
12

An example of a transient analysis involving the 1st law of thermodynamics and conservation of mass is the filling of a compress

ed air tank. Assume that an air tank is being filled using a compressor to a pressure of 5 atm, and that it is being fed with air at a temperature of 25°C and 1 atm pressure. The compression process is adiabatic. Will the temperature of the air in the tank when it is done being filled i.e. once the pressure in the tank reaches 5 atm), be greater than, equal to, or less that the temperature of the 25°C air feeding the compressor?
A. Greater than 25°C
B. Unable to determine
C. Same as 25°C
D. Less than 25°C
Engineering
1 answer:
pickupchik [31]3 years ago
8 0

Answer:

<em>The temperature will be greater than 25°C</em>

Explanation:

In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.

mathematically

Change in the internal energy of a system ΔU = ΔQ + ΔW

in an adiabatic process, ΔQ = 0

therefore

ΔU = ΔW

where ΔQ is the change in heat into the system

ΔW is the work done by or done on the system

when work is done on the system, it is conventionally negative, and vice versa.

also W = pΔv

where p is the pressure, and

Δv = change in volume of the system.

In this case,<em> work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C </em>

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13. Write a function which is passed two strings. The function creates a new string from the two original strings by copying one
attashe74 [19]

Answer:

I am writing the code in C++. Let me know if you want the program in some other programming language.

#include <iostream>  // includes header file for input output functions

using namespace std;     //to identify objects like cin cout

string CopyStrings(string string1, string string2)  

{   string newString = "";    

   for (int loop = 0; loop < string1.length() ||  

                   loop < string2.length(); loop++)      {      

       if (loop < string1.length())  

           newString += string1[loop];          

       if (loop < string2.length())  

           newString += string2[loop];      }  

   return newString;   }  

int main()  

{   string stringA = "ace";  

   string stringB = "bdf";  

   cout << CopyStrings(stringA, stringB);   }

Output:

abcdef

Explanation:

The function CopyStrings() function takes two strings i.e. string1 and string2 as parameters to copy characters from both the string one character from each.

The newString variable stores the new string after copying characters from both strings string1 and string2.

Then the for loop starts which has a variable loop which is an index variable that traverses through both the strings stored in string1 and string2. The loop continues to execute until it moves through entire length of string1 and string2 which means it copies all the characters from both string1 and string2. length() is used here which returns length of the string1 and string2.

If statement in the for loop checks the character that loop (index) variable is pointing to is less than the string1 length which means it checks each character stored in string1. For example if string1 contains "ace" and loop variable is moving through the string and is currently at "a" then this condition is true. If the condition evaluates to true then the body of if statement is executed. The next statement stores that character a into the newString variable.

Next If statement checks character that loop variable is pointing to is less than the string2 length which means it checks each character stored in string2. For example if string2 contains "bdf" and loop variable is moving through the string and is currently at "b" then this condition is true. If the condition evaluates to true then the body of if statement is executed. The next statement stores that character b into the newString variable.

Then the second iteration starts which again first stores the next character i.e. c from string1 into newString and then stores next character i.e d from string2 into newString.

Then the third iteration starts which again first stores the next character i.e. e from string1 into newString and then stores next character i.e f from string2 into newString.

Then the loop breaks as the loop variable reaches end of both the string1 and string2.

return newString will return the copied string into the output screen which is abcdef.

The screenshot of code along with output is attached.

3 0
3 years ago
A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to
Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 =  (16/3) * 10^2 V

c)  E = 64/15 J

d)  dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the final potential difference across each capacitor

(c) the final energy of the system

(d) the decrease in energy when the capacitors are connected.

Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

                           Q_initial = C_1*V

                           Q_initial = 20*10^-6 * 800

                           Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

                          Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

                          V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

                          Q_2 = Q_1*C_2/C_1

                          Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

                          Q_initial = 1.5*Q_1

                          Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

                          V_2 = V_1 = Q_1 / C_1  

                                            = 10.67*10^-3 / 20*10^-6

                                            = (16/3) * 10^2 V

c)

- The energy in the system is:

                          E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

                           E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

                          E = 64/15 J

d)

- The decrease in energy of the capacitors is:

                           dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

                          dE = 0.5*10^-6*20*800^2 - (64/15)

                          dE = 32/5 - 64/15 = 32/15 J

5 0
3 years ago
Please help me with this. Plzzz.
Drupady [299]

Answer:

450,000m = 450km = 4.5E5

32,600,000W = 32.6MW = 3.26E7

59,700,000,000cal = 59.7Gcal = 5.97E10

0.000000083s = 83ns = 8.3E-8

35,000Ω = 35kΩ = 3.5E4

Explanation:

Giga   = 1,000,000,000

Mega = 1,000,000

kilo     = 1,000

unit    = 1

deci   = .1

centi  = .01

milli    = .001

micro = .000001

nano = .0000000001

pico  = .000000000001

You should be able to look at these and convert between them in seconds if you want to pursue anything in engineering.

7 0
3 years ago
A 35 ft long solid steel rod is subjected to a load of 8,000 lb. This load causes the rod to stretch 0.266 in. The modulus of el
solong [7]

Answer:

53.67

Explanation:

3 0
2 years ago
Expalin the application of diesel cycle in detail.
mars1129 [50]

Explanation:

Diesel cycle:

        All diesel engine work on diesel cycle .In diesel cycle there are four process .These processes are as follows

1. Adiabatic reversible compression

2.Heat addition at constant pressure

3.Adiabatic reversible expansion

4.Constant volume heat rejection

In general compression ratio in diesel engine is high as compare to petrol engine.But the efficiency of diesel cycle is less as compare to petrol cycle for same compression ratio.

Applications of diesel cycle:

Generally diesel cycle used for heavy vehicle or equipment because heavy vehicle or equipment is required high initial torque.So this cycle have lots of applications such as in industrial machining,in trucks,power plant,in mining ,in defense or military,large motors ,compressor and pump etc.

   

5 0
3 years ago
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