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Liono4ka [1.6K]
3 years ago
7

A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of

dimensions 5.00 cm by 6.11 cm that rotates in an adjustable magnetic field. What is the field strength needed to produce a 24.0 V peak emf?
Engineering
1 answer:
larisa86 [58]3 years ago
4 0

Answer:

The field strength needed is 0.625 T

Explanation:

Given;

angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s

area of the rectangular coil, A =  L x B = 0.0611 x 0.05 = 0.003055 m²

number of tuns of the coil, N = 300 turns

peak emf = 24 V

The peak emf is given by;

emf₀ = NABω

B = (emf₀ ) / (NA ω)

B = (24) / (300 x 0.003055 x 41.893)

B = 0.625 T

Therefore, the field strength needed is 0.625 T

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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
3 years ago
A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of
lyudmila [28]

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

P_1V_1=P_2V_2  

where,

P_1 = initial pressure of gas  = 150 kPa

P_2 = final pressure of gas  = ?

V_1 = initial volume of gas   = v L

V_2 = final volume of gas  = \frac{v}{3}L

150\times v=P_2\times \frac{v}{3}  

P_2=450kPa

Therefore, the new pressure of the gas will be 450 kPa.

7 0
3 years ago
Which of the following is a common use for commas?
andreyandreev [35.5K]

Answer:

connecting two independent clauses

4 0
3 years ago
Read 2 more answers
A colorful design for a product meant to appeal to a preschool audience reflects which of the following points in industrial des
lord [1]
Appeal to emotion

the product doesn’t need to be colorful to function but it would be more appealing to the intended audience if it was
6 0
2 years ago
A work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R1
ella [17]

Answer:

a)

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b) 3.1 secs

Explanation:

a) Determine the normal times in TMUs for these motion elements

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b ) Determine the total time for this work element in seconds

first we have to determine the total TMU = ∑ TMU = 86.4 TMU

note ; 1 TMU = 0.036 seconds

hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds

7 0
2 years ago
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