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3 years ago

6

As the junior engineer at the Mesabi Range Hydraulic Engineering Company located in Ely, Minnesota, you have been tasked with de

signing a new irrigation canal that will be used by the English Pea Farmers Cooperative of Northern Minnesota. The canal will run from Basswood Lake to the pea fields located just south of town. The canal is to be 22 miles in length, unlined, excavated in stiff clay and must handle a flow rate of 13.15 m3/s over a slope of 0.2%. Your Boss wants you to calculate the necessary canal parameters and to also determine if the canal will fit within the 85-foot wide right-of-way established by the Lake County land-use office

1 answer:

katen-ka-za [31]3 years ago

5 0

yes it will

Explanation:

You might be interested in

Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l

sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is $0.44 \frac{W}{m^{2} \times K}$ , the average convection heat transfer coefficient over the entire plate is $0.293 \frac{W}{m^{2} \times K}$ and the total heat flux transfer to the plate is $61.6 KJ$ .

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas and surface. Further temperature boundary layer will developed on flat plate in longitudinal direction.

Hot carbon dioxide exhaust gas

physical properties

$r= 1.05 \frac{kg}{m^{3}}$

$c_p = 1.02 \frac{kJ}{Kg \times K}$

$m= 231 \times 10^{7} \frac{N \times s }{m^2}$

υ = $21.8 \times 10^{6} \frac{m^2}{s}$

$k = 32.5 \times 10^{3} \frac{W}{m \times K}$

$\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}$

$Pr = 0.725$

Apart from these other data arr given below,

$v= 3 \frac{m}{s} \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C$

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

$Nu = \frac{ h \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

where h= Average heat transfer coefficient

L= Length of a plate

k= Thermal Conductivity of carbon dioxide

Re = Reynold's Number

Pr = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

is referred as local convection heat transfer coefficient.

To find convection heat transfer coefficient at 1 m from leading edge,

$Nu = \frac{ h_local \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

Here, first we have to find Re and Pr,

$Re = \frac{r \times v \times L}{m}$

$Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}$

$Re = 20.63 \times 10^{-10}$

Pr number is take from physical property data and Pr is 0.725.

Putting value of Re and Pr in main equation,

we get

$Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 32.5 \times 10^{3} \times 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 0.44 \frac{W}{m^{2} \times K}$

(b) To find average convection heat transfer coefficient,

it can be find out as case (a), only difference is that instead of L=1 m, L=1.5 m would come,

Therefore,

$Nu = \frac{ h \times 1.5 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

Finally,

$h = \frac{0.44}{1.5}$

$h = 0.293 \frac{W}{m^{2} \times K}$

(C) Total heat flux transfer to the plate is found out by,

$Q = h \times (T_g - T_s)$

$Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140 \\ Q= 61.6 KJ$

8 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

Which of the following activities will allow an engineer to determine that a building's design is sound?

zhannawk [14.2K]

Please add more details because I don’t know if you are using a book or passage, therefore I cannot help you unless you add more detail

3 0

3 years ago

A cylinder fitted with a frictionless piston contains 2 kg of R-134a at 3.5 bar and 100 C. The cylinder is now cooled so that th

inna [77]

Answer:

The answer to the question is

The heat transferred in the process is -274.645 kJ

Explanation:

To solve the question, we list out the variables thus

R-134a = Tetrafluoroethane

Intitial Temperaturte t₁ = 100 °C

Initial pressure = 3.5 bar = 350 kPa

For closed system we have m₁ = m₂ = m

ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂

For constant pressure process we have

Work done = W = $\int\limits^a_b P \, dV$ = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)

From the tables we have

State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg

State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg

Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)

= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ

5 0

3 years ago

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any frictio

Dmitrij [34]

Complete Question

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 21 m, and all six cars have the same mass.V = -18 m What is v?X km/h

Answer:

$v=23.6m/s$

Explanation:

Velocity $v_c=18.0km/h$

Radius $r=21m$

initial velocity u $u=19=>5.27778$

Generally the equation for Angle is mathematically given by

$\theta=\frac{v_c}{2r}$

$\theta=\frac{18}{2*21}$

$\theta=0.45$

$\theta=25.7831 \textdegree$

Generally

Height of mass

$h=\frac{rsin\theta}{\theta}$

$h=\frac{21sin25.78}{0.45}$

$h=20.3m$

Generally the equation for Work Energy is mathematically given by

$0.5mv_0^2+mgh=0.5mv^2$

Therefore

$v=\sqrt{u^2+2gh}$

$v=\sqrt{=5.27778^2+2*9.81*20.3}$

$v=23.6m/s$

3 0

3 years ago