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3 years ago

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You just graduate from college and find a job (yah! Congrats!). You buy a new house and would like to install solar panels for t

he house (thanks for going green!). The daily energy consumption of the house is 30 kWhr. The solar energy a standard solar panel can capture varies by location. In your area, the daily energy captured by a standard 250 W panel is 1.1 kWhr. How many Ds are the daily energy per panel and daily house energy use, respectively? How many panels do you need for your home (please round to an integer)?

1 answer:

Lena [83]3 years ago

6 0

Answer: Number of panels needed is 27

Daily energy per 250 W panel is 1.1kWhr

Daily house energy used is 30kWhr

Explanation: To get 27 panel it is 250*30/1.1 =6818

6818/250= 27 panels needed daily

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A dual voltage/dual resistance AC hair dryer can be used with a 120 V or 230 V power supply. A switch allows the user to set the

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Answer:

The power is $p= 5621 W$

Explanation:

In order to obtain the power it consumes we need to first obtain the resistance of the hair dryer at 120 V

$R = \frac{V^2}{P} = \frac{(120 V)^2}{1530W} = 9.411 \Omega$

Now the power supplied in Europe at their voltage of 230V would be

$P = \frac{V^2} {R} = \frac{(230V)^2}{9.411 \ \Omega} = 5621 W$

Looking the power obtain we see that it is higher than 1530 W so this would cause the dryer to smoke.

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4 years ago

You have 6 resistors in a circuit. The voltage on each is given. Use MATLAB to calculate the total power dissipated by the resis

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Answer:

Explanation:

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4 years ago

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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

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

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4 years ago

A pressurized 2-m-diameter tank of water has a 10-cm-diameter orifice at the bottom where water discharges to the atmosphere. Th

alukav5142 [94]

Answer:

A fluid is defined as a material that deforms continuously and permanently under the

application of a shearing stress.

• The pressure at a point in a fluid is independent of the orientation of the surface

passing through the point; the pressure is isotropic.

• The force due to a pressure p acting on one side of a small element of surface dA

defined by a unit normal vector n is given by −pndA.

• Pressure is transmitted through a fluid at the speed of sound.

• The units we use depend on whatever system we have chosen, and they include quantities

like feet, seconds, newtons and pascals. In contrast, a dimension is a more

abstract notion, and it is the term used to describe concepts such as mass, length and

time.

• The specific gravity (SG) of a solid or liquid is the ratio of its density to that of water

at the same temperature.

• A Newtonian fluid is one where the viscous stress is proportional to the rate of strain

(velocity gradient). The constant of proportionality is the viscosity, µ, which is a

property of the fluid, and depends on temperature.

• At the boundary between a solid and a fluid, the fluid and solid velocities are equal;

this is called the “no-slip condition.” As a consequence, for large Reynolds numbers

(>> 1), boundary layers form close to the solid boundary. In the boundary layer,

large velocity gradients are found, and so viscous effects are important.

• At the interface between two fluids, surface tension may become important. Surface

tension leads to the formation of a meniscus, drops and bubbles, and the capillary rise

observed in small tubes, because surface tension can resist pressure differences across

the interface.

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4 years ago

Please help<br> describe the impact that a toy robot has had or could have on its intended audience

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3 years ago

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