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mario62 [17]
3 years ago
15

An old refrigerator consumes 247 W of power. Assuming that the refrigerator operates for 19 hours everyday, what is the annual o

perating cost of the refrigerator, if the cost of electricity is $0.06 per kWh
Engineering
2 answers:
german3 years ago
6 0

Answer:

The annual operating cost of the refrigerator is $102.78.

Explanation:

Power consumed by the refrigerator = 247 W = 247/1000 = 0.247 kW

Daily operation of the refrigerator = 19 hours

Annual operation of the refrigerator = 365 × 19 = 6,935 hours

Annual energy consumed = 0.247 kW × 6,935 hours = 1712.945 kWh

1 kWh of electricity cost $0.06

1712.945 kWh will cost 1712.945 × $0.06 = $102.78

Annual operating cost = $102.78

BabaBlast [244]3 years ago
3 0
<h2>Answer:</h2>

$102.7767

<h2>Explanation:</h2>

Energy, E, consumed by a device is the product of the power, P, consumed by the device and the duration (time), t, of consumption. i.e;

E = p x t         ----------------(i)

The refrigerator consumes 247W of power and operates 19 hours per day.

This implies that the refrigerator consumes 0.247kW of power and operates 19 hours per day.

This means that the energy used per day is given by substituting p = 0.247kW and t = 19h into equation (i) as follows;

E = 0.247kW x 19h

E = 4.693kWh

From the question, it is given that the cost of electricity is $0.06 per kWh. i.e

$0.06 = 1 kWh

Then;

4.693kWh = 4.693kWh x $0.06 / 1kWh = $0.28158

Therefore, the cost of electricity per day due to the refrigerator is $0.28158.

Now, to get the annual (365 days) cost of electricity, we multiply the result by 365 as follows;

Annual cost = 365 x $0.28158

Annual cost = $102.7767

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Dust, dirt, or metal chips can pose a potential ____ injury risk in a shop.
Liono4ka [1.6K]

Answer: Eye injury

Explanation: small material such as dust, dirt, and metal shards can harm your eyes with potential blindness or infection.

7 0
1 year ago
Help please if you don't know don't give wrong answer please​
mrs_skeptik [129]

4.75cm * 5.22cm is 24.795cm²

but let's do this with m² instead:

0.0475m * 0.0522m = 0.0024795

now we can compare it with the 49780 much easier.

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so the representative fraction is 1:4,480.

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6 0
2 years ago
The modulus of elasticity for a ceramic material having 6.0 vol% porosity is 303 GPa. (a) Calculate the modulus of elasticity (i
Phantasy [73]

Answer:

modulus of elasticity for the nonporous material is 340.74 GPa

Explanation:

given data

porosity = 303 GPa

modulus of elasticity = 6.0

solution

we get here  modulus of elasticity for the nonporous material Eo that is

E = Eo (1 - 1.9P + 0.9P²)    ...............1

put here value and we get Eo

303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )  

solve it we get

Eo = 340.74 GPa

8 0
2 years ago
A pipe produces successive harmonics at 300 Hz and 350 Hz. Calculate the length of the pipe and state whether it is closed at on
MAXImum [283]

Answer:

The pipe is open ended and the length of pipe is 3.4 m.

Explanation:

For identification of the type of pipe checking the successive frequencies in both the open pipe and closed pipe as below

Equation for nth frequency for open end pipe is given as

f_n=\frac{nv}{2L}

For (n+1)th value the frequency is

f_{n+1}=\frac{(n+1)v}{2L}

Taking a ratio of both equation and solving for n such that the value of n is a whole number

\frac{f_{n+1}}{f_n}=\frac{\frac{(n+1)v}{2L}}{\frac{nv}{2L}}\\\frac{350}{300}=\frac{(n+1)}{n}\\350n =300n+300\\50n =300\\n =6\\

So n is a whole number this means that the pipe is open ended.

For confirmation the  nth frequency for a closed ended pipe is given as

f_n=\frac{(2n+1)v}{4L}

For (n+1)th value the frequency is

f_{n+1}=\frac{(2n+3)v}{4L}

Taking a ratio of both equation and solving for n such that the value of n is a whole number

\frac{f_{n+1}}{f_n}=\frac{\frac{(2n+3)v}{2L}}{\frac{(2n+1)v}{2L}}\\\frac{350}{300}=\frac{(2n+3)}{(2n+1)}\\700n+350 =600n+900\\100n =550\\n =5.5\\

As n is not a whole number so this is further confirmed that the pipe is open ended.

Now from the equation of, with n=6, v=340 m/s and f=300 Hz

f_n=\frac{nv}{2L}\\300=\frac{6 \times 340}{2L}\\L=\frac{2040}{600}\\L=3.4 m

The value of length is 3.4m.

5 0
2 years ago
A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pip
JulsSmile [24]

Answer:

Part A: (d/D=0.1)

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV percent=21.7%

Explanation:

We are going to use the following volume flow rate equation:

DeltaV=\frac{\pi * DeltaP}{8*u*l}(R^{4}-r^{4} -\frac{(R^{2}-r^{2})}{ln\frac{R}{r}}^{2})

Above equation can be written as:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{d}{D} )^{4}+\frac{(1-(\frac{d}{D})^{2})}{ln\frac{d}{D}}^{2})

First Consider no wire i.e d/D=0

Above expression will become:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0)^{4}+\frac{(1-(0)^{2})}{ln0}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}

Part A: (d/D=0.1)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.1)^{4}+\frac{(1-(0.1)^{2})}{ln0.1}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.574

DeltaV percent=\frac{(\frac{\pi*R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.574}{\frac{\pi*R^{4}*DeltaP}{8*u*l} }*100

DeltaV percent=\frac{1-0.574}{1}*100

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.783

DeltaV percent=\frac{(\frac{\pi *R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.783}{\frac{\pi *R^{4}*DeltaP}{8*u*l} }*100

DeltaV percent=\frac{1-0.783}{1}*100

DeltaV percent=21.7%

5 0
2 years ago
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