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umka21 [38]
3 years ago
11

Acme Logistics provides "Less than truck load" (LTL) services throughout the U.S. They have several hubs where they use cross-do

cking to move goods from one trailer to another. Acme built its last hub 10 years ago, and it had 36 dock doors. The cost index at that time was 140, and the total cost was $6 million. Acme plans a new hub that will have 48 dock doors. The cost index now is 195, and Acme will use a capacity factor of 0.82. What is the estimated cost of the new hub?
Engineering
1 answer:
riadik2000 [5.3K]3 years ago
6 0

Answer:

The answer is "$10.12 million"

Explanation:

To find the price of the 36 dock doors only at the current rate, just use the cost index.  

C_n=C_k \times (\frac{I_n}{I_k})\\\\

     = 6 \times \frac{195}{140} \\\\= \$ \ 8 \ million

Its current cost of 36 dock doors is, thus,\$ \ 8 \ million. To calculate the price for 48 dock doors, just use the second formula now:

\to (\frac{C_A}{C_B})=(\frac{S_A}{S_B})^{X}\\\\\to C_A=C_B \times (\frac{S_A}{S_B})^{X}\\\\

         =8 \times (\frac{48}{36})^{0.82}\\\\= \$ 10.12 \ million

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The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di
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Answer:

a) at T = 5800 k  

  band emission = 0.2261

at T = 2900 k

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b) daylight (d) = 0.50 μm

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Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

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at T = 2900 k

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attached below is a detailed solution to the problem

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3 0
3 years ago
Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 1.825 m3/s and expands adi
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Answer: (a).power developed = 776.1 kW

(b). Rate of entropy production = 1.023 kW/K

(c). efficiency = 63%

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from the question we have that

P₁ = 5 bar

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where V₁ = 0.5416 m³/Kg, S₁ = 7.5308 KJ/Kg-K and R₁ = 0.3105.6 KJ/Kg

the volumetric flow rate is given as (φ) = 1.825 m³/s

Remember that φ = ṁ V

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ṁ = φ/V = 1.825/0.5416 = 3.37 Kg/s

Also given for the Exit state;

P₂ = 1 bar

T₂ = 200°C

where V₂ = 0.5416 m³/Kg, S₂ = 7.5308 KJ/Kg-K and R₂ = 0.3105.6 KJ/Kg

(a). we are asked to determine the power developed in the Kw.

using the Flow energy equation to turbine we have;

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ш = 3.37 (3105-2815.3) = 776.1 kW

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(b). The rate of entropy production in Kw/K.

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Rate(en) = 1.023 kW/K

(c). The percent isentropic  turbine efficiency.

Πt = (R₁-R₂) / (h₁ - h₂s)

Πt = (3105.6 - 2875.3) / (3105.6 - 2740) = 63%

Πt = 63%

cheers i hope this helped!!!!!

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