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ddd [48]
2 years ago
13

1. A company has an annual demand for a product of 2000 units, a carrying cost of $20per unit per year, and a setup cost of $100

. Through a program of setup reduction, thesetup cost is reduced to $20. Run costs are $2 per unit. Calculate:
A. The EOQ before setup reduction.
B. The EOQ after setup reduction.
C. The total and unit cost before and after setup reduction.
Engineering
1 answer:
Xelga [282]2 years ago
5 0

Answer:

B the EOQ after setup reduction.

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den301095 [7]

Answer:

See explaination

Explanation:

Let's define tuple as an immutable list of Python objects which means it can not be changed in any way once it has been created.

Take a look at the attached file for a further detailed and step by step solution of the given problem.

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A(n) ____ is an object setting used to control the visible display of objects.
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A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the
Tanya [424]

Answer:

a) 23.89 < -25.84 Ω

b) 31.38 < 25.84 A

c) 0.9323 leading

Explanation:

A) Calculate the load Impedance

current on load side = 0.75 p.u

power factor angle = 25.84

I_{load} = 0.75 < 25.84°

attached below is the remaining part of the solution

<u>B) Find the input current on the primary side in real units </u>

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<u>C) find the input power factor </u>

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<em></em>

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8 0
3 years ago
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8 0
3 years ago
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When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner
ipn [44]

Answer:

\dot W = 1.667\, kW

Explanation:

A well-sealed house means that there is no mass interaction between air indoors and outdoors. Hence, cooling process is isochoric. The heat removed by the air conditioner is:

\dot Q_{L} = \frac{m_{air}}{\Delta t} \cdot c_{v, air} \cdot (T_{o}-T_{f})\\\dot Q_{L} = \frac{800\, kg}{(30\, min)\cdot (\frac{60\, s}{1 \, min} )}\cdot (0.7 \frac{kJ}{kg \cdot K}) \cdot (15\, K)\\\dot Q_{L} = 4.667\, kW

The power drawn by the air conditioner is:

\dot W = \frac{\dot Q_{L}}{COP_{R}} \\\dot W = \frac {4.667\, kW}{2.8}\\\dot W = 1.667\, kW

3 0
3 years ago
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