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SCORPION-xisa [38]
2 years ago
13

A stainless steel ball (=8055 kg/m3, Cp= 480 J/kgK) of diameter D =15 cm is removed from theoven at a uniform temperature of 3

50oC. The ball is then subjectedto the flow of air at 1 atm pressure and 30oC with a velocityof 6 m/s. The surface temperature of the ball eventuallydrops to 250oC. Determine the average convection heat transfercoefficient during this cooling process and estimate howlong this process has taken.
Engineering
1 answer:
aleksandrvk [35]2 years ago
3 0

Answer:

i) 25.04 W/m^2 .k

ii) 23.82 minutes = 1429.2 secs

Explanation:

Given data:

Diameter of steel ball = 15 cm

uniform temperature = 350°C

p = 8055 kg/m^3

Cp = 480 J/kg.k

surface temp of ball drops to  250°C

average surface temperature = ( 350 + 250 ) / 2 = 300°C

<u>i) Determine the average convection heat transfer coefficient during the cooling process</u>

<em>Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table  " properties of air "  contained in your textbook</em>

average convection heat transfer coefficient = 25.04 W/m^2 .k

<u>ii) Determine how long this process has taken </u>

Time taken by the process = 23.82 minutes = 1429.2 seconds

Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs

attached below is the detailed solution of the given question

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g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What
earnstyle [38]

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity (v), measured in meters per second, is determined by the following expression:

v = \frac{4\cdot \dot V}{\pi \cdot D^{2}} (1)

Where:

\dot V - Flow rate, measured in cubic meters per second.

D - Diameter, measured in meters.

If we know that \dot V = 0.01\,\frac{m^{3}}{s} and D = 0.05\,m, then the flow velocity is:

v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}

v \approx 5.093\,\frac{m}{s}

The density and dinamic viscosity of the glycerin at 20 ºC are \rho = 1260\,\frac{kg}{m^{3}} and \mu = 1.5\,\frac{kg}{m\cdot s}, then the Reynolds number (Re), dimensionless, which is used to define the flow regime of the fluid, is used:

Re = \frac{\rho\cdot v \cdot D}{\mu} (2)

If we know that \rho = 1260\,\frac{kg}{m^{3}}, \mu = 1.519\,\frac{kg}{m\cdot s}, v \approx 5.093\,\frac{m}{s} and D = 0.05\,m, then the Reynolds number is:

Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }

Re = 211.230

A pipeline is in turbulent flow when Re > 4000, otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor (f), dimensionless, is determined by the following expression:

f = \frac{64}{Re}

If we get that  Re = 211.230, then the friction factor is:

f = \frac{64}{211.230}

f = 0.303

The friction factor is 0.303.

4 0
2 years ago
Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity i
RoseWind [281]

To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.

The stagnation temperature can be defined as

T_0 = T+\frac{V^2}{2c_p}

Where

T = Static temperature

V = Velocity of Fluid

c_p = Specific Heat

Re-arrange to find the static temperature we have that

T = T_0 - \frac{V^2}{2c_p}

T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})

T = 672.88K

Now the pressure of helium by using the Adiabatic pressure temperature is

P = P_0 (\frac{T}{T_0})^{k/(k-1)}

Where,

P_0= Stagnation pressure of the fluid

k = Specific heat ratio

Replacing we have that

P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}

P = 0.399Mpa

Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa

<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>

3 0
3 years ago
Twenty-five wooden beams were ordered or a construction project. The sample mean and he sample standard deviation were measured
aksik [14]

Answer:

Correct option: B. 90%

Explanation:

The confidence interval is given by:

CI = [\bar{x} - z\sigma_{\bar{x}} , \bar{x}+z\sigma_{\bar{x}} ]

If \bar{x} is 190, we can find the value of z\sigma_{\bar{x}}:

\bar{x} - z\sigma_{\bar{x}}  = 188.29

190 - z\sigma_{\bar{x}}  = 188.29

z\sigma_{\bar{x}}  = 1.71

Now we need to find the value of \sigma_{\bar{x}}:

\sigma_{\bar{x}} = s / \sqrt{n}

\sigma_{\bar{x}} = 5/ \sqrt{25}

\sigma_{\bar{x}} = 1

So the value of z is 1.71.

Looking at the z-table, the z value that gives a z-score of 1.71 is 0.0436

This value will occur in both sides of the normal curve, so the confidence level is:

CI = 1 - 2*0.0436 = 0.9128 = 91.28\%

The nearest CI in the options is 90%, so the correct option is B.

4 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
A metal bar has a 0.6 in. x 0.6 in. cross section and a gauge length of 2 in. The bar is loaded with a tensile force of 50,000 l
Aleks [24]

Answer:

modulus =3.97X10^6 Ib/in^2, Poisson's ratio = 0.048

Explanation:

Modulus is the ratio of tensile stress to tensile strain

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And contraction occur from 0.6 in x 0.6 in to 0.599 in x 0.599 in while 2 in extended to 2.007, with extension of 0.007 in

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