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Bingel [31]
3 years ago
12

On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced press

ure PR approaches zero. How do you explain this behavior?
Engineering
1 answer:
Alekssandra [29.7K]3 years ago
4 0

Answer:

Enthalpy is a function of pressure hence normalized enthalpy departure values will approach zero with reduced pressure approaching zero

Explanation:

On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero. this is because enthalpy is a function of pressure therefore as the Pressure is reducing towards the zero value, the gas associated with the pressure tends to behave more like an Ideal gas.

For an Ideal gas the Normalized enthalpy departure value will be approaching the zero value.

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A dual-fluid heat exchanger has 10 lbm/s water entering at 100 F, 20 psia and leaving at 50 F, 20 psia. The other fluid is glyco
kakasveta [241]

Answer:

Rate of internal heat transfer = 23.2 Btu/Ibm

mass flow rate = 21.55 Ibm/s

Explanation:

using given data to obtain values from table F7.1

Enthalpy of water at temperature of 100 F = 68.04Btu/Ibm

Enthalpy of water at temperature of 50 F = 18.05 Btu/Ibm

from table F.3

specific constant of glycerin C_{p} = 0.58 Btu/Ibm-R

<u>The rate of internal heat transfer ( change in enthalpy ) </u>

h4 - h3 = Cp ( T4 - T3 ) --------------- ( 1 )

where ; T4 = 50 F

             T3 = 10 F

             Cp = 0.58 Btu/Ibm-R

substitute given values into equation 1

change in enthalpy ( h4 - h3 ) = 23.2 Btu/Ibm

<u>Determine mass flow rate of glycol</u>

attached below is the detailed solution

mass flow rate of glycol = 21.55 Ibm/s

3 0
3 years ago
A manufacturer makes integrated circuits that each have a resistance layer with a target thickness of 200 units. A circuit won't
aleksklad [387]

Answer:

probability P = 0.32

Explanation:

this is incomplete question

i found complete A manufactures makes integrated circuits that each have a resistance layer with a target thickness of 200 units. A circuit won't work well if this thickness varies too much from the target value. These thickness measurements are approximately normally distributed with a mean of 200 units and a standard deviation of 12 units. A random sample of 17 measurements is selected for a quality inspection. We can assume that the measurements in the sample are independent. What is the probability that the mean thickness in these 16 measurements x is farther than 3 units away from the target value?

solution

we know that Standard error is expess as

Standard error = \frac{sd}{\sqrt{n}}

Standard error  = \frac{12}{\sqrt{16}}

Standard error  = 3  

so here we get Z value for 3 units away are from mean are

mean =  -1 and + 1

so here

probability P will be

probability P = P( z < -1 or z > 1)

probability P = 0.1587 + 0.1587

probability P =  0.3174

probability P = 0.32

7 0
3 years ago
simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl
Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

<u>a) Air temperature at turbine exit </u>

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17  for  Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) (\frac{783.05-778.18}{800.13-778.18} ) = 764.45K

<u>b) The net work output </u>

first we determine the actual work input to compressor

Wc = h2 - h1  ( calculated values )

     = 626.57 - 295.17 =  331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4  ( calculated values )

     = 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

         = 541.88 - 331.4  = 210.48 kJ/kg

<u>c) determine thermal efficiency </u>

лth = Wnet / qin  ------ ( 1 )

where ; qin = h3 - h2

<em>equation 1 becomes </em>

лth = Wnet / ( h3 - h2 )

      = 210.48 / ( 1324.93 - 626.57 )

      = 0.3014  =  30.14%

6 0
3 years ago
Define Plastic vs elastic deformation.
Snowcat [4.5K]

Answer:

Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.

Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.

3 0
3 years ago
2.4: Add a method called setValue(), and the description of setValue is: public int setValue(long searchKey) In this method, the
Yanka [14]

Answer:

Below is java code that must be used for the given question:

// highArray.java

// demonstrates array class with high-level interface

// to run this program: C>java HighArrayApp

////////////////////////////////////////////////////////////////

class HighArray

  {

  private long[] a;                 // ref to array a

  private int nElems;               // number of data items

  //-----------------------------------------------------------

  public HighArray(int max)         // constructor

     {

     a = new long[max];                 // create the array

     nElems = 0;                        // no items yet

     }

  //-----------------------------------------------------------

  public setValue find(long searchKey)

     {                              // find specified value

     int j;

     for(j=0; j<nElems; j++)            // for each element,

        if(a[j] == searchKey)           // found item?

           break;                       // exit loop before end

     if(j == nElems)                    // gone to end?

        return false;                   // yes, can't find it

     else

        return true;                    // no, found it

     }  // end find()

  //-----------------------------------------------------------

  public void insert(long value)    // put element into array

     {

     a[nElems] = value;             // insert it

     nElems++;                      // increment size

     }

  //-----------------------------------------------------------

  public void display()             // displays array contents

     {

     for(int j=0; j<nElems; j++)       // for each element,

        System.out.print(a[j] + " ");  // display it

     System.out.println("");

     }

  //-----------------------------------------------------------

  }  // end class HighArray

////////////////////////////////////////////////////////////////

class HighArrayApp

  {

  public static void main(String[] args)

     {

     int maxSize = 100;            // array size

     HighArray arr;                // reference to array

     arr = new HighArray(maxSize); // create the array

     arr.insert(77);               // insert 10 items

     arr.insert(99);

     arr.insert(44);

     arr.insert(55);

     arr.insert(22);

     arr.insert(88);

     arr.insert(11);

     arr.insert(00);

     arr.insert(66);

     arr.insert(33);

     arr.display();                // display items

     int searchKey = 35;           // search for item

     if( arr.find(searchKey) )

        System.out.println("Found " + searchKey);

     else

        System.out.println("Can't find " + searchKey);

     }  // end main()

  }  // end class HighArrayApp

Explanation:

6 0
2 years ago
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