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2 years ago

10

A manufacturer has been asked to produce 100 customized metal discs with a particular pattern engraved on them. Which production

process should be selected?

2 answers:

topjm [15]2 years ago

8 0

I’m just here for points because I have test and I need them lol

saul85 [17]2 years ago

3 0

Answer:

Job Shop production

Explanation:

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At a certain location, wind is blowing steadily at 7 m/s. Determine the mechanical energy of air per unit mass and the power gen

Kaylis [27]

Answer:

Explanation:

From the information given;

The velocity of the wind blow V = 7 m/s

The diameter of the blades (d) = 80 m

Percentage of the overall efficiency $\eta_{overall} = 30\%$

The density of air $\rho = 1.25 kg/km^3$

Then, we can use the concept of the kinetic energy of the wind blowing to estimate the mechanic energy of air per unit mass by using the formula:

$e_{mechanic} = \dfrac{mV^2}{2}$

here;

m = $\rho AV$

= $1.25 \times \dfrac{\pi}{4}(80)^2 \times 7$

= 43982.29 kg/s

∴

$W = e_{mechanic} = \dfrac{mV^2}{2}$

$= \dfrac{43982.29 \times 7^2}{2}$

$= 1077566.105 \ W$

$\mathbf{ =1077.566 \ kW}$

The actual electric power is:

$W_{electric} = \eta_{overall} \times W$

$W_{electric} = 0.3 \times 1077.566$

$\mathbf{W_{electric} =323.26 \ kW}$

8 0

2 years ago

Calculate the availability of a system where the mean time between failures is 900 hours and the mean time to repair is 100 hour

Debora [2.8K]

Answer:

The availability of system will be 0.9

Explanation:

We have given mean time of failure = 900 hours

Mean time [to repair = 100 hour

We have to find availability of system

Availability of system is given by $\frac{mean\time\ of\ failure}{mean\ time\ of\ failure+mean\ time\ to\ repair}$

So availability of system $=\frac{900}{900+100}=\frac{900}{1000}=0.9$

So the availability of system will be 0.9

7 0

3 years ago

A heat pump designer claims to have an air-source heat pump whose coefficient of performance is 1.8 when heating a building whos

Anit [1.1K]

Answer:

The claim is valid.

Explanation:

Let assume that heat pump is reversible. The coefficient of performance for the heat pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{300\,K}{300\,K-260\,K}$

$COP_{HP} = 7.5$

The claim is valid as real heat pumps have lower coefficients of performance.

3 0

3 years ago

How should you move your board through the planer? (Pick two choices.)

iragen [17]

Answer:

I would say do it at an even pace

Explanation:

Doing it a slow pace takes time quickly will probably not to good gor you and doing it at an irregular pace is just way to fast

4 0

3 years ago

A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg

lakkis [162]

Answer:

$L=107.6m$

Explanation:

Cold water in: $m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C$

Hot water in: $m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C$

$D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m$

Step 1: Determine the rate of heat transfer in the heat exchanger

$Q=m_{c}C_{c}(T_{c,out}-T_{c,in})$

$Q=1.2*4.18*(80-20)$

$Q=1.2*4.18*(80-20)$

$Q=300.96kW$

Step 2: Determine outlet temperature of hot water

$Q=m_{h}C_{h}(T_{h,in}-T_{h,out})$

$300.96=2*4.18*(160-T_{h,out})$

$T_{h,out}=124\°C$

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

$dT_{1}=T_{h,in}-T_{c,out}$

$dT_{1}=160-80$

$dT_{1}=80\°C$

$dT_{2}=T_{h,out}-T_{c,in}$

$dT_{2}=124-20$

$dT_{2}=104\°C$

$LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}$

$LMTD = \frac{104-80}{ln(\frac{104}{80})}$

$LMTD = \frac{24}{ln(1.3)}$

$LMTD = 91.48\°C$

Step 4: Determine required surface area of heat exchanger

$Q=UA_{s}LMTD$

$300.96*10^{3}=649*A_{s}*91.48$

$A_{s}=5.07m^{2}$

Step 5: Determine length of heat exchanger

$A_{s}=piDL$

$5.07=pi*0.015*L$

$L=107.57m$

7 0

2 years ago