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Zigmanuir [339]
3 years ago
11

The USAF is thinking of purchasing a improved automatic controller (no human required) at a cost of $100,000 that can make 4,000

target contacts per hour to replace USAF specialists that can, on average, make 1,000 target contacts per hour. Note that a USAF Specialist is paid a relatively low salary of $8/hour (but with good veteran benefits) and that the controller will require $1/hour maintenance costs. How many hours of target detection would it take to pay off this new controller?
Engineering
1 answer:
blsea [12.9K]3 years ago
8 0

Answer:

It would take 3225.8 hours of target detection to pay off the new controller

Explanation:

First, to find out how many hours it would take, we calculate what is the cost per contact for both options:

\mbox{cost per contact}=\frac{\mbox{cost per hour}}{\mbox{contacts per hour}} \\\\\mbox{cost per contact}_{specialist} =\frac{8 \frac{\$}{h} }{1000\frac{contacts}{hour} } =8*10^{-3} \frac{\$}{contact}\\\\\mbox{cost per contact}_{controller} =\frac{1 \frac{\$}{h} }{4000\frac{contacts}{hour} } =2.5*10^{-4} \frac{\$}{contact}

Knowing the cost per contact, we calculate savings per contact when using the controller:

Savings=\mbox{cost per contact}_{specialist}-\mbox{cost per contact}_{controller}\\Savings=8*10^{-3} \frac{\$}{contact}-2.5*10^{-4} \frac{\$}{contact}=7.75*10^{-3} \frac{\$}{contact}

Now that we know how much the automatic controller saves, we calculate how many contacts it would take to pay it off, and then, taking into account contacts per hour, how many hours:

Hours=\frac{\mbox{cost of the controller}}{\mbox{savings per contact}}*\frac{\mbox{1}}{\mbox{contacts per hour}}  \\\\Hours=\frac{\$100000}{7.75*10^{-3} \frac{\$}{contact}} *\frac{\mbox{1}}{4000\frac{contacts}{hour} }}=3225,8 \mbox{ hours}

That means, without taking into account veteran benefits (that raise the cost of the USAF specialist), it would take around 134 days of target detection to pay off the new controller.

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The electric motor exerts a torque of 800 N·m on the steel shaft ABCD when it is rotating at a constant speed. Design specificat
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Answer:

d= 4.079m ≈ 4.1m

Explanation:

calculate the shaft diameter from the torque,    \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}

Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).

r = Radius of the shaft.

T = Twisting Moment or Torque.

J = Polar moment of inertia.

C = Modulus of rigidity for the shaft material.

l = Length of the shaft.

θ = Angle of twist in radians on a length.  

Maximum Torque, ζ= τ ×  \frac{ π}{16} × d³

τ= 60 MPa

ζ= 800 N·m

800 = 60 ×  \frac{ π}{16} × d³

800= 11.78 ×  d³

d³= 800 ÷ 11.78

d³= 67.9

d= \sqrt[3]{} 67.9

d= 4.079m ≈ 4.1m

3 0
3 years ago
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A heat engine that rejects waste heat to a sink at 520 R has a thermal efficiency of 35 percent and a second- law efficiency of
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Answer:

The source temperature is 1248 R.

Explanation:

Second law efficiency of the engine is the ratio of actual efficiency to the maximum possible efficiency that is reversible efficiency.

Given:  

Temperature of the heat sink is 520 R.

Second law efficiency is 60%.

Actual thermal efficiency is 35%.

Calculation:  

Step1

Reversible efficiency is calculated as follows:

\eta_{II}=\frac{\eta_{a}}{\eta_{rev}}

0.6=\frac{0.35}{\eta_{rev}}

\eta_{rev}=0.5834

Step2

Source temperature is calculated as follows:

\eta_{rev}=1-\frac{T_{L}}{T}

\eta_{rev}=1-\frac{520}{T}

0.5834=1-\frac{520}{T}

T = 1248 R.

The heat engine is shown below:

Thus, the source temperature is 1248 R.

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3 years ago
1. An air standard cycle is executed within a closed piston-cylinder system and consists of three processes as follows:1-2 = con
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Answer:

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3 years ago
The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200K and 400K, re
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Answer:

a) W_cycle = 200 KW , n_th = 33.33 %  , Irreversible

b) W_cycle = 600 KW , n_th = 100 %     , Impossible

c) W_cycle = 400 KW , n_th = 66.67 %  , Reversible

Explanation:

Given:

- The temperatures for hot and cold reservoirs are as follows:

  TL = 400 K

  TH = 1200 K

Find:

For each case W_cycle , n_th ( Thermal Efficiency ) :

(a) QH = 600 kW, QC = 400 kW

(b) QH = 600 kW, QC = 0 kW

(c) QH = 600 kW, QC = 200kW

- Determine whether the cycle operates reversibly, operates irreversibly, or is impossible.

Solution:

- The work done by the cycle is given by first law of thermodynamics:

                                 W_cycle = QH - QC

- For categorization of cycle is given by second law of thermodynamics which states that:

                                 n_th < n_max     ...... irreversible

                                 n_th = n_max     ...... reversible

                                 n_th > n_max     ...... impossible

- Where n_max is the maximum efficiency that could be achieved by a cycle with Hot and cold reservoirs as follows:

                                n_max = 1 - TL / TH = 1 - 400/1200 = 66.67 %

And,                         n_th = W_cycle / QH

a) QH = 600 kW, QC = 400 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 400 = 200 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 200 / 600 = 33.33 %

   - The type of process according to second Law of thermodynamics:

               n_th = 33.333 %                n_max = 66.67 %

                                       n_th < n_max  

      Hence,                Irreversible Process  

b) QH = 600 kW, QC = 0 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 0 = 600 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 600 / 600 = 100 %

   - The type of process according to second Law of thermodynamics:

                 n_th = 100 %                 n_max = 66.67 %

                                     n_th > n_max  

      Hence,               Impossible Process              

c) QH = 600 kW, QC = 200 kW

   - The work done by cycle according to First Law is:

                                W_cycle = 600 - 200 = 400 KW

   - The thermal efficiency of the cycle is given by n_th:

                                n_th = W_cycle / QH

                                n_th = 400 / 600 = 66.67 %

   - The type of process according to second Law of thermodynamics:

               n_th = 66.67 %                 n_max = 66.67 %

                                     n_th = n_max  

      Hence,                Reversible Process

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