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Mrac [35]
3 years ago
13

A fire hose nozzle has a diameter of 1.125 in. According to some fire codes, the nozzle must be capable of delivering at least 2

50 gal/min. If the nozzle is attached to a 3-in.-diameter hose, what pressure must be maintained just upstream of the nozzle to deliver this flowrate?
Engineering
2 answers:
Furkat [3]3 years ago
7 0

Answer:

P_{1} = 403,708\,kPa\,(58.553\,psi)

Explanation:

Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:

\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}

The initial pressure is:

P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}

The speed at outlet is:

v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}

v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }

v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )

The initial pressure is:

P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}

P_{1} = 403,708\,kPa\,(58.553\,psi)

Contact [7]3 years ago
4 0

Answer:

P1 = 42.93 psi

Explanation:

For incompressible fluid, we know that;

A1V1 = A2V2

Making V1 the subject, we obtain;

V1 = A2V2/A1

Now A2V2 is the volumetric flow rate (V') .

Thus; V1 = V'/A1

A1 = πD²/4

Thus, V1 = 4V'/πD²

V' = 250 gal/min

But the diameter is in inches, let's convert to inches³/seconds.

Thus, V' = 250 x 3.85 = 962.5 in³/s

Substituting the relevant values to obtain,

V1 = (4 x 962.5)/(π x 3²) = 136.166 in/s.

Now let's convert to ft/s;

V1 = 136.166 x 0.0833 = 11.34 ft/s

Also for V2;

V2 = (4 x 962.5)/(π x 1.125²) = 968.29 in/s.

Now let's convert to ft/s;

V2 = 968.29 x 0.0833 = 80.66 ft/s

Setting bernoulli equation between the hose and the exit, we obtain;

(p1/γ) + (V1²/2g) = V2²/2g

Where V1 and V2 are intial and final velocities and γ is specific weight of water which is 62.43 lb/ft³ and g i acceleration due to gravity which is 32.2 ft/s²

Making p1 the subject, we obtain;

p1 = (γ/2g)(V2² - V1²)

p1 = (62.43/(2x32.2))(80.66² - 11.34²)

p1 = 6182.35 lb/ft²

So Converting to psi, we have;

p1 = 6182.35/144 = 42.93 psi

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The present worth of income from an investment that follows an arithmetic gradient is projected to be $475,000. The income in ye
Nikitich [7]

Answer:

G = $37,805.65

Explanation:

I found this on another site:

475,000 = 25,000(P/A,10%,6) + G(P/G,10%,6)

475,000 = 25,000(4.3553) + G(9.6842)

9.6842G = 366,117.50

G = $37,805.65

4 0
3 years ago
A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of
OverLord2011 [107]

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

7 0
2 years ago
A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque
Helen [10]

Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

                             τ = 16*T / pi*d^3

                             T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

                             T_c = τ_c*pi*d_c^3 / 16

                             T_c = 12*1000*pi*1^3 / 16

                             T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

                             T_s = τ_s*pi*d_s^3 / 16

                             T_s = 15*1000*pi*0.25^3 / 16

                             T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

                             T = min ( 2356.2 , 46.02 )

                             T = 46.02 lb-in

6 0
3 years ago
CS3733: Homework/Practice 05 Suppose we would like to write a program called monitor which allows two other programs to communic
valina [46]

Answer:

#include<stdio.h>

#include<stdlib.h>

#include<unistd.h>

#include<sys/types.h>

#include<string.h>

#include<pthread.h>

//#include<sys/wait.h>

int main(int argc, char** argv)

{

int fd1[2];

int fd2[2];

int fd3[2];

int fd4[2];

char message[] = "abcd";

char input_str[100];

pid_t p,q;

if (pipe(fd1)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd2)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd3)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd4)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

p = fork();

if (p < 0)

{

 fprintf(stderr, "fork Failed" );

return 1;

}

// child process-1

else if (p == 0)

{

 close(fd1[0]);// Close reading end of first pipe

 char concat_str[100];

 printf("\n\tEnter meaaage:"):

 scanf("%s",concat_str);

 write(fd1[1], concat_str, strlen(concat_str)+1);

 // Concatenate a fixed string with it

 int k = strlen(concat_str);

 int i;

 for (i=0; i<strlen(fixed_str); i++)

 {

  concat_str[k++] = fixed_str[i];

 }

 concat_str[k] = '\0';//string ends with '\0'

 // Close both writting ends

 close(fd1[1]);

 wait(NULL);

//.......................................................................

 close(fd2[1]);

 read(fd2[0], concat_str, 100);

 if(strcmp(concat_str,"invalid")==0)

 {

 printf("\n\tmessage not send");

 }

 else

 {

  printf("\n\tmessage send to prog_2(child_2).");

 }

 close(fd2[0]);//close reading end of pipe 2

 exit(0);

}

else

{

 close(fd1[1]);//Close writting end of first pipe

 char concat_str[100];

 read(fd1[0], concal_str, strlen(concat_str)+1);

 close(fd1[0]);

 close(fd2[0]);//Close writing end of second pipe

 if(/*check if msg is valid or not*/)

 {

  //if not then

  write(fd2[1], "invalid",sizeof(concat_str));

  return 0;

 }

 else

 {

  //if yes then

  write(fd2[1], "valid",sizeof(concat_str));

  close(fd2[1]);

  q=fork();//create chile process 2

  if(q>0)

  {

   close(fd3[0]);/*close read head offd3[] */

   write(fd3[1],concat_str,sizeof(concat_str);//write message by monitor(main process) using fd3[1]

   close(fd3[1]);

   wait(NULL);//wait till child_process_2 send ACK

   //...........................................................

   close(fd4[1]);

   read(fd4[0],concat_str,100);

   close(fd4[0]);

   if(sctcmp(concat_str,"ack")==0)

   {

    printf("Messageof child process_1 is received by child process_2");

   }

   else

   {

    printf("Messageof child process_1 is not received by child process_2");

   }

  }

  else

  {

   if(p<0)

   {

    printf("Chiile_Procrss_2 not cheated");

   }

   else

   {

     

    close(fd3[1]);//Close writing end of first pipe

    char concat_str[100];

    read(fd3[0], concal_str, strlen(concat_str)+1);

    close(fd3[0]);

    close(fd4[0]);//Close writing end of second pipe

    write(fd4[1], "ack",sizeof(concat_str));

     

   }

  }

 }

 close(fd2[1]);

}

}

8 0
3 years ago
If you were driving the blue Prius in the situation pictured above, explain why the red Mustang should be given right-of-way at
dexar [7]

Answer:

The blue Prius, because the Mustang arrived almost in the same time. And when you arrive in an intersection at the same time of other vehicle you need yield for the car on your right if the car is on your left you have the right of way.

Explanation:

4 0
3 years ago
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