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larisa [96]
2 years ago
9

Hoosier Manufacturing operates a production shop that is designed to have the lowest unit production cost at an output rate of 1

15 units per hour. In the month of July, the company operated the production line for a total of 205 hours and produced 21,400 units of output. What was its capacity utilization rate for the month?
Physics
1 answer:
abruzzese [7]2 years ago
3 0

Answer:

90.77%

its capacity utilization rate for the month is 90.77%

Explanation:

The capacity utilisation rate can be expressed mathematically as;

Capacity utilisation rate = capacity used/Best operating level × 100%

Given;

Total Number of production time = 205hours

Production output/capacity used = 21400 units

Best operation rate = 115units/hour

Best operation output for the month of July( at best operation level )

=115units/hour × 205 hours = 23575 units

Capacity utilisation rate = 21400/23575 × 100%

= 90.77%

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Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is
ad-work [718]

Answer:

T_{out}=27.0000077 ºC

Explanation:

First, let's write the energy balance over the duct:

H_{out}=H_{in}+Q

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

m*Cp(T_{out}-T_{in})=Q

So, let's isolate T_{out}:

T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}

The Cp of the air at 27ºC is 1007\frac{J}{kgK} (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are T_{out} and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

P=2*H+2*A=2*0.004m+2*0.016m=0.04m

Surface area:

A=P*L=0.04m*1m=0.04m^2

Then, the heat Q is:

600\frac{W}{m^2} *0.04m^2=24W

Finally, find the exit temperature:

T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077

T_{out}=27.0000077 ºC

The temperature change so little because:

  • The mass flow is so big compared to the heat flux.
  • The transfer area is so little, a bigger length would be required.
3 0
2 years ago
Sort the phrases based on whether they represent kinetic energy or potential energy.
alexandr1967 [171]
Kinetic energy:
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Hope this helps.
8 0
2 years ago
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A cup of hot coffee can be cooled by placing a cold spoon in it. A spoon of which of the following materials would be most effec
arsen [322]

Answer: copper

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The coffee is at a higher temperature compared to the spoon, hence there is conduction of heat by the spoon from the coffee.

Heat is now transfered from the coffee to the spoon, the rate of heat transfer ( for this case coffee to spoon) is dependent on some properties but for this case, the thermal conductivity is of the most important.

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Glass (0.8 W/m.K)

It can been seen that copper has the highest value for thermal conductivity which implies that it will conduct heat faster compared to others thus cooling the coffee faster.

8 0
2 years ago
At an oceanside nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water th
Mice21 [21]
Answer is 30. just took it
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2 years ago
Select the volume units that are greater than one liter.
Andreas93 [3]
A.) kiloliter. 1 kiloliter = 1,000 liters
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hope this helps
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1 year ago
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