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2 years ago

The radiation meter is showing radiation 2x as much as background. Is this a hot zone? If so why or why not?

Engineering

1 answer:

kakasveta [241]2 years ago

5 0

Answer:

This is not a clear indication of the hot zone as the information of the radioactivity of the background is not provided clearly.

Explanation:

According to IAEA as well as NRCP, the hot area is defined on the basis of the radioactivity reading it shows instead of contrast or comparative reading from the background. The value of radiation activity which will be required to declare an area as hot zone is if it is greater than 0.1 mSv/h or $1.5091\times 10^{29} kg^{-1} s^{-1}$ .

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What are some common ways of converting fuel to useful energy? Check all that apply.

Rainbow [258]

Answer: Burning it.

Explanation: When your car moves its due to the burning of fuel, works like a turbine the molecules evaporate and turn a turbine like object.

3 0

3 years ago

Read 2 more answers

A meter stick can be read to the nearest millimeter and a travelling microscope can be read to the nearest 0.1 mm. Suppose you w

german

Answer: No

Explanation:

Length= 2cm= 20mm

Now meter stick can read to nearest millimeter.

It is given that length is to be measured with a precision of 1% of 20mm= 1/100 * 20= 0.2mm

Since the least count is 1mm of meter stick and precision required is less than that. So, meter stick cannot be used for this, travelling microscope can be used for this as it can read to 0.1mm.

3 0

3 years ago

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 well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape unti

natima [27]

Answer:

Final mass of Argon= 2.46 kg

Explanation:

Initial mass of Argon gas ( M1 ) = 4 kg

P1 = 450 kPa

T1 = 30°C = 303 K

P2 = 200 kPa

k ( specific heat ratio of Argon ) = 1.667

assuming a reversible adiabatic process

<u>Calculate the value of the M2 </u>

Applying ideal gas equation ( PV = mRT )

P₁V / P₂V = m₁ RT₁ / m₂ RT₂

hence : m2 = P₂T₁ / P₁T₂ * m₁

= (200 * 303 ) / (450 * 219 ) * 4

= 2.46 kg

<em>Note: Calculation for T2 is attached below</em>

5 0

2 years ago

In an experiment, one selected two samples of copper-silver alloy. One sample has 40 wt% of silver and 60wt% of copper and the o

mote1985 [20]

I belive it’s 3 sorry if it’s wrong

6 0

3 years ago