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3 years ago

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A team of engineers is working on a design to increase the power of a hydraulic lever. They have brainstormed several ideas. Whi

ch of the following steps should they now focus on to categorize and optimize their design choice?
The engineers should be sure the need has been identified with the client before moving forward in the design process.

The engineers should explore all of the possible solutions they have brainstormed and rank them.

The engineers should contact the client for the preventive maintenance phase and be sure the designs meet specifications.

The engineers should recognize the specifications and limitations, such as costs and materials.

1 answer:

Goshia [24]3 years ago

7 0

Answer:

Trye

Explanation:

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Three tool materials (high-speed steel, cemented carbide, and ceramic) are to be compared for the same turning operation on a ba

Tpy6a [65]

Answer:

Among all three tools, the ceramic tool is taking the least time for the production of a batch, however, machining from the HSS tool is taking the highest time.

Explanation:

The optimum cutting speed for the minimum cost

$V_{opt}= \frac{C}{\left[\left(T_c+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]^n}\;\cdots(i)$

Where,

C,n = Taylor equation parameters

$T_h$ =Tool changing time in minutes

$C_e$ =Cost per grinding per edge

$C_m$ = Machine and operator cost per minute

On comparing with the Taylor equation $VT^n=C$ ,

Tool life,

$T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)$

Given that,

Cost of operator and machine time $=\$40/hr=\$0.667/min$

Batch setting time = 2 hr

Part handling time: $T_h=2.5$ min

Part diameter: $D=73 mm$ $=73\times 10^{-3} m$

Part length: $l=250 mm=250\times 10^{-3} m$

Feed: $f=0.30 mm/rev= 0.3\times 10^{-3} m/rev$

Depth of cut: $d=3.5 mm$

For the HSS tool:

Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

So, $C_e=$ $\$20/15+2=\$3.33/edge$

Tool changing time, $T_t=3$ min.

$C= 80 m/min$

$n=0.130$

(a) From equation (i), cutting speed for the minimum cost:

$V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}$

$\Rightarrow 47.7$ m/min

(b) From equation (ii), the tool life,

$T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}$

$\Rightarrow T=53.4$ min

(c) Cycle time: $T_c=T_h+T_m+\frac{T_t}{n_p}$

where,

$T_m=$ Machining time for one part

$n_p=$ Number of pieces cut in one tool life

$T_m= \frac{l}{fN}$ min, where $N=\frac{V_{opt}}{\pi D}$ is the rpm of the spindle.

$\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc$

So, the number of parts produced in one tool life

$n_p=\frac {T}{T_m}$

$\Rightarrow n_p=\frac {53.4}{4.01}=13.3$

Round it to the lower integer

$\Rightarrow n_p=13$

So, the cycle time

$T_c=2.5+4.01+\frac{3}{13}=6.74$ min/pc

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times6.74+\frac{3.33}{13}=\$4.75/pc$

(e) Total time to complete the batch= Sum of setup time and production time for one batch

$=2\times60+ {50\times 6.74}{50}=457 min=7.62 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times4.01}{457}=0.4387=43.87\%$

Now, for the cemented carbide tool:

Cost per edge,

$C_e=$ $\$8/6=\$1.33/edge$

Tool changing time, $T_t=1min$

$C= 650 m/min$

$n=0.30$

(a) Cutting speed for the minimum cost:

$V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min$ [from(i)]

(b) Tool life,

$T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min$ [from(ii)]

(c) Cycle time:

$T_c=T_h+T_m+\frac{T_t}{n_p}$

$T_m= \frac{\pi D l}{fV_{opt}}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 363}=0.53min/pc$

$n_p=\frac {7}{0.53}=13.2$

$\Rightarrow n_p=13$ [ nearest lower integer]

So, the cycle time

$T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc$

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times3.11+\frac{1.33}{13}=\$2.18/pc$

(e) Total time to complete the batch $=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times0.53}{275.5}=0.0962=9.62\%$

Similarly, for the ceramic tool:

$C_e=$ $\$10/6=\$1.67/edge$

$T_t-1min$

$C= 3500 m/min$

$n=0.6$

(a) Cutting speed:

$V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}$

$\Rightarrow V_{opt}=2105 m/min$

(b) Tool life,

$T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min$

(c) Cycle time:

$T_c=T_h+T_m+\frac{T_t}{n_p}$

$\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc$

$n_p=\frac {2.33}{0.091}=25.6$

$\Rightarrow n_p=25 pc/tool\; life$

So,

$T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc$

(d) Cost per production unit:

$C_c= C_mT_c+\frac{C_e}{n_p}$

$\Rightarrow C_c=0.667\times2.63+\frac{1.67}{25}=$1.82/pc$

(e) Total time to complete the batch

$=2\times60+ {50\times 2.63}=251.5 min=4.19 hr$ .

(f) The proportion of time spent actually cutting metal

$=\frac{50\times0.091}{251.5}=0.0181=1.81\%$

3 0

4 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

4 years ago

Describe three differences between liquids and gases in fluid power systems.<br> Help !!!

scoundrel [369]

Gases, liquids and solids are all made up of atoms, molecules, and/or ions, but the behaviors of these particles differ in the three phases. ... gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

3 0

3 years ago

You are driving on a road where the speed limit is 35 mph. If you want to make a turn, you must start to signal at least _______

stich3 [128]

I believe it’s D) 20 feet

3 0

3 years ago

Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.

saul85 [17]

Answer:

Explanation:

Given that:

<u>At state 1:</u>

Pressure P₁ = 20 bar

Volume V₁ = 0.03 $\mathbf{m^{3}}$

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C ; $v_1 = vg_1$ = 0.0996 $\mathbf{m^{3}}$ / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

<u>At state 2:</u>

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 $\mathbf{m^{3}}$ / kg

From temperature T₂ = 200⁰ C

$v_f_2 = 0.0016 \ m^3/kg$

$vg_2 = 0.127 \ m^3/kg$

Since $vf_2 < v_2$ , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality $x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}$

$x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}$

$x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}$

$\mathsf{x_2 =0.78}$

At temperature T₂, the specific internal energy $u_f_2 = 850.6 \ kJ/kg$ , also $ug_2 = 2594.3 \ kJ/kg$

Thus,

$u_2 = uf_2 + x_2 (ug_2 -uf_2)$

$u_2 =850.6 +0.78 (2594.3 -850.6)$

$u_2 =850.6 +1360.086$

$u_2 =2210.686 \ kJ/kg$

<u>At state 3:</u>

Temperature $T_3=T_2 = 200 ^0 C ,$

$V_3 = 2V_1 = 0.06 \ m^3$

Specific volume $v_3 = 0.2 \ m^3/kg$

Thus; $vg_3 =vg_2 = 0.127 \ m^3/kg$ ,

SInce $v_3 > vg_3$ , therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at $v_3 = 0.2 \ m^3/kg$ and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 $\ m^3/kg$

The specific internal energy $u_3$ at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

$u_2-u_1$

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

$u_3-u_2$

= (2622.3 - 2210.686) kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

3 0

3 years ago