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valkas [14]
3 years ago
9

Two heat exchangers are under consideration for purchase. A standard type of heat exchanger (code name HX1) has an initial insta

lled cost of $20,000 and a useful life of 6 years. At the end of its useful life, the standard heat exchanger has a negligible salvage value, assumed to be 0. Another type of heat exchanger (Code name HX2) with equivalent design capacity has an initial installed cost of $34,000, and a useful life of 10 years with a salvage value of $4,000. Assuming an effective interest rate of 6% per year, determine which heat exchanger will heat exchanger will result in the lower capitalized cost.

Engineering
1 answer:
Leviafan [203]3 years ago
3 0

Answer:

The lower capitalized cost is associated with HX1

Explanation:

The capitalized cost is basically the present cost. Therefore, for the Heat Exchanger HX1, the capitalized cost will be:

CC1 = First Cost

<u>CC1 = - $ 20,000</u>

Negative sign shows cash outflow.

Now, for Heat Exchanger HX2, we have:

CC2 = - $ 34,000 + ($ 4,000)(P/F, 6%, 10)

Now, we use the factor tables to calculate the present worth (P) of the salvage value, which is actually future worth, after 10 years at a compounded interest of 6% per year.

The factor table is provided in picture, with the value indicated.

CC2 = - $ 34,000 + ($ 4,000)(0.5584)

CC2 = - $ 34,000 + $ 2,233.6

<u>CC2 = - $ 31,766.4</u>

From the capitalized cost, taking look at the absolute values of both heat exchangers we can conclude that:

<u>The Heat Exchanger that will result in lower capitalized cost is HX1</u>

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Answer:

Option C

Explanation:

Tamara must choose the contractor who can provide best services among all and also has a fair pricing. Hence, option 3 is correct

If she chooses the contractor with low price, there are chances that the quality will not be good. Hence, option 2 is incorrect

Option 4 is incorrect because the contractor must take inputs of the contracting person or agency.

Option 1 is incorrect because selecting some one on the basis of reviews can be a biased selection of contractor.

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3 years ago
Chad is working on a design that uses the pressure of steam to control a valve in order to increase water pressure in showers. W
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C: Viscosity, the resistance to flow that fluids exhibit

Explanation:

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8 0
3 years ago
Estimate the design-stage uncertainty in determining the voltage drop across an electric heating element. The device has a nomin
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Answer:

Find the attachment for solution

Note: The resistance is measured in ohms (Ω). In the question Ω will replace V, where ever the resistance unit is required .

3 0
4 years ago
The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that deliv
Nataliya [291]

The question is incomplete. The complete question is :

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is $\frac{3}{8}$  in.

A) Determine the maximum shear stress developed in the shaft.

$\tau_{max}$ = ?

B) Also, what is the "wind up," or angle of twist in the shaft at full power?

$ \phi $ = ?

Solution :

Given :

Angular speed, ω = 1700 rpm

                              $ = 1700 \frac{\text{rev}}{\text{min}}\left(\frac{2 \pi \text{ rad}}{\text{rev}}\right) \frac{1 \text{ min}}{60 \ \text{s}}$

                              $= 56.67 \pi \text{ rad/s}$

Power $= 2590 \text{ hp} \left( \frac{550 \text{ ft. lb/s}}{1 \text{ hp}}\right)$

          = 1424500 ft. lb/s

Torque, $T = \frac{P}{\omega}$

                 $=\frac{1424500}{56.67 \pi}$

                 = 8001.27 lb.ft

A). Therefore, maximum shear stress is given by :

Applying the torsion formula

$\tau_{max} = \frac{T_c}{J}$

        $=\frac{8001.27 \times 12 \times 4}{\frac{\pi}{2}\left(4^2 - 3.625^4 \right)}$

      = 2.93 ksi

B). Angle of twist :

     $\phi = \frac{TL}{JG}$

         $=\frac{8001.27 \times 12 \times 100 \times 12}{\frac{\pi}{2}\left(4^4 - 3.625^4\right) \times 11 \times 10^3}$

         = 0.08002 rad

         = 4.58°

6 0
3 years ago
A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end
hjlf

Answer:

a) v=19.6 m/s

b) H=19.58 m

c) v_{f}=29.57 m/s  

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J

But we know the work is equal to change of kinetic energy, so:

W=\Delta K=\frac{1}{2}mv^{2}

v=\sqrt{\frac{2W}{m}}=19.6 m/s

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

v_{f}^{2}=v_{i}^{2}-2gh

At the maximum height the velocity is 0, so v(f) = 0.

0=v_{i}^{2}-2gH

H=\frac{19.6^{2}}{2*9.81}=19.58 m  

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

v_{f}^{2}=v_{i}^{2}-2gh

v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

6 0
3 years ago
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