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kotykmax [81]
3 years ago
7

Problem 3.10 One/half million parts of a certain type are to be manufactured annually on dedicated production machines that run

16 hours per day, five days per week, 50 weeks per year. (a) The cycle time to produce one part is 1.8 min. If availability and utilization are each 100%, and setup time is neglected, how many of the machines will be required to satisfy demand? (b) Solve part (a) except that availability = 95%.
Engineering
2 answers:
goldenfox [79]3 years ago
8 0

Answer:

  • 4 machines
  • 4 machines

Explanation:

(a) The production schedule is 4000 hour per year, or 240,000 minutes. In that time, 133,333 1/3 parts can be produced. The production of 500,000 parts in a year will require the use of ...

  .5 million/(.1333... million/machine) = 3.75 machines

The demand can be satisfied by 4 machines.

__

(b) If availability is 95% then 3.75/0.95 ≈ 3.95 machines will be required.

The demand can be satisfied by 4 machines.

sesenic [268]3 years ago
7 0

Answer:

  • 4 machines
  • 4 machines

A.

B

➡ are both in 4 machines used in

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vazorg [7]

Answer:

(M_t)_{rated}=61.11lb-in

Explanation:

speed of motor (N)=1500 rpm

power=4 hp = 4 \times 0.7457 =2.9828 KW

service factor(k)= 2.75

now,

KW=\frac{2\pi n M_t}{60 \times 10^6} \\2.9828=\frac{2\pi \times 1500 M_t}{60 \times 10^6}\\M_t=\frac{2.9828\times 60 \times 10^6}{2\pi \times 1500 }

M_t= 18,989.09 \ N-mm= 168.06 lb-in

torque rating

(M_t)_{design}=k_s\times (M_t)_{rated}\\168.06= 2.75\times (M_t)_{rated}\\(M_t)_{rated}=\frac{168.06}{2.75} \\(M_t)_{rated}=61.11lb-in

4 0
4 years ago
The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
UkoKoshka [18]
Maybe it willl be I don’t know
7 0
3 years ago
An electric field is expressed in rectangular coordinates by E = 6x2ax + 6y ay +4az V/m.Find:a) VMN if point M and N are specifi
Fittoniya [83]

Answer:

a.) -147V

b.) -120V

c.) 51V

Explanation:

a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).

b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.

c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.

Honestly, these things take practice to get used to. It's really hard to explain this.

3 0
3 years ago
A hydraulic cylinder is to be used to move a workpiece in a manufacturing operation through a distance of 50 mm in 10 s. A force
svetlana [45]

Answer:

The answer to this question is 1273885.3 ∅

Explanation:

<em>The first step is to determine the required  hydraulic flow rate liquid if working pressure and  if a cylinder with a piston diameter of 100 mm is available.</em>

<em>Given that,</em>

<em>The distance = 50mm</em>

<em>The time t =10 seconds</em>

<em>The force F = 10kN</em>

<em>The piston diameter is = 100mm</em>

<em>The pressure = F/A</em>

<em> 10 * 10^3/Δ/Δ </em>

<em> P = 1273885.3503 pa</em>

<em>Then</em>

<em>Power = work/time  = Force * distance /time</em>

<em> = 10 * 1000 * 0.050/10</em>

<em>which is  =50 watt</em>

<em>Power =∅ΔP</em>

<em>50 = 1273885.3 ∅</em>

5 0
3 years ago
Which bulb has the lowest total cost of operation? (a) Incandescent (b) Fluorescent (c) LED
Finger [1]

Answer: LED have the lowest cost of operation.

Explanation:

If we ignore the initial procurement cost of the items the operational cost of any device consuming electricity is given by

Cost=Energy\times cost/unit

Among the three item's LED consumes the lowest power to give the same level of brightness as compared to the other 2 item's thus LED's shall have the lowest operational cost.

6 0
4 years ago
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