Ask question

Ask question

All categories

Genrish500 [490]

3 years ago

15

An injection-molding machine has a first cost of $1,050,000 and a salvage value of $240,000 in any year. The maintenance and ope

rating cost is $235,000 with an annual gradient of $75,000. The MARR is 10%. What is the most economic life?
a. years (EUAC - $759.376)
b. 7 years (EUAC - $672,616)
c. 3 years (EUAC - $904,384)
d. 5 years (EUAC - $608,428)

1 answer:

Ilya [14]3 years ago

4 0

Explanation:

First cost =$ 1,050,000

Salvage Value =$ 225,000 Maintenance & Operating cost =$ 235,000 Maintenance & Operating Gradient =$ 75,000 MARR =10 %

EUAB - EAUC $=\$ 1,050,000(\mathrm{~A} / \mathrm{P}, 10 \%, \mathrm{n})+\$ 225,000(\mathrm{~A} / \mathrm{F}, 10 \%, \mathrm{n})$

$$

-\$ 235,000-\$ 75,000(\mathrm{~A} / \mathrm{G}, 10 \%, \mathrm{n})

$$

Try $\mathrm{n}=4$ years:

$$

\text { EUAB - EAUC }=\$ 331,275+\$ 48,488-\$ 235,000-\$ 103,575=-\$ 621,362

$$

Try $\mathrm{n}=5$ years:

EUAB - EUAC=-\$ 276,990+\$ 36,855-\$ 235,000-\$ 135,750=-$ 610,885

Try n=6 years

{ EUAB - EUAC }=-$ 241,080+$ 29,160-$ 235,000-$ 166,800=-$ 613,720

Thus, year 5 has the minimum EUAB - EUAC, hence the most economic life is 5 years.

You might be interested in

Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l

sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is $0.44 \frac{W}{m^{2} \times K}$ , the average convection heat transfer coefficient over the entire plate is $0.293 \frac{W}{m^{2} \times K}$ and the total heat flux transfer to the plate is $61.6 KJ$ .

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas and surface. Further temperature boundary layer will developed on flat plate in longitudinal direction.

Hot carbon dioxide exhaust gas

physical properties

$r= 1.05 \frac{kg}{m^{3}}$

$c_p = 1.02 \frac{kJ}{Kg \times K}$

$m= 231 \times 10^{7} \frac{N \times s }{m^2}$

υ = $21.8 \times 10^{6} \frac{m^2}{s}$

$k = 32.5 \times 10^{3} \frac{W}{m \times K}$

$\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}$

$Pr = 0.725$

Apart from these other data arr given below,

$v= 3 \frac{m}{s} \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C$

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

$Nu = \frac{ h \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

where h= Average heat transfer coefficient

L= Length of a plate

k= Thermal Conductivity of carbon dioxide

Re = Reynold's Number

Pr = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

is referred as local convection heat transfer coefficient.

To find convection heat transfer coefficient at 1 m from leading edge,

$Nu = \frac{ h_local \times L }{k} = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })$

Here, first we have to find Re and Pr,

$Re = \frac{r \times v \times L}{m}$

$Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}$

$Re = 20.63 \times 10^{-10}$

Pr number is take from physical property data and Pr is 0.725.

Putting value of Re and Pr in main equation,

we get

$Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 32.5 \times 10^{3} \times 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

$h_local = 0.44 \frac{W}{m^{2} \times K}$

(b) To find average convection heat transfer coefficient,

it can be find out as case (a), only difference is that instead of L=1 m, L=1.5 m would come,

Therefore,

$Nu = \frac{ h \times 1.5 }{32.5 \times 10^{3}} = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })$

Finally,

$h = \frac{0.44}{1.5}$

$h = 0.293 \frac{W}{m^{2} \times K}$

(C) Total heat flux transfer to the plate is found out by,

$Q = h \times (T_g - T_s)$

$Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140 \\ Q= 61.6 KJ$

8 0

3 years ago

Rough Riders Inc. manufactures jeans in the cutting and sewing process. Jeans are manufactured in 50-jean batch sizes. The cutti

miskamm [114]

Answer: me and ten points now

Explanation:

3 0

3 years ago

I will mark as brainliest !

Sliva [168]

Answer:

7.8 Mph

Explanation:

Rate of cycling = 1.1 rev/s

Rear wheel diameter = 26 inches

Diameter of sprocket on pedal = 6 inches

Diameter of sprocket on rear wheel = 4 inches

Circumference of rear wheel = \pi d=26\piπd=26π

Speed would be

\begin{gathered}\text{Rate of cycling}\times \frac{\text{Diameter of sprocket on pedal}}{\text{Diameter of sprocket on rear wheel}}\times{\text{Circumference of rear wheel}}\\ =1.1\times \frac{6}{4}\times 26\pi\\ =134.77432\ inches/s\end{gathered}Rate of cycling×Diameter of sprocket on rear wheelDiameter of sprocket on pedal×Circumference of rear wheel=1.1×46×26π=134.77432 inches/s

Converting to mph

1\ inch/s=\frac{1}{63360}\times 3600\ mph1 inch/s=633601×3600 mph

134.77432\ inches/s=134.77432\times \frac{1}{63360}\times 3600\ mph=7.65763\ mph134.77432 inches/s=134.77432×633601×3600 mph=7.65763 mph

The Speed of the bicycle is 7.8 mph

3 0

3 years ago

Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the approp

Anvisha [2.4K]

Answer:

The question has some details missing : The 35-kg block A is released from rest. Determine the velocity of the 13-kgkg block BB in 4 ss . Express your answer to three significant figures and include the appropriate units. Enter positive value if the velocity is upward and negative value if the velocity is downward.

Explanation:

The detailed steps and appropriate calculation is as shown in the attached file.

6 0

3 years ago

Which system of linear inequalities is represented by the graph? y > x – 2 and y x + 1 y x + 1 y > x – 2 and y < x + 1

KatRina [158]

Answer:

The graph representing the linear inequalities is attached below.

Explanation:

The inequalities given are :

y>x-2 and y<x+1

For tables for values of x and y and get coordinates to plot for both equation.

In the first equation;

y>x-2

y=x-2

y-x = -2

The table will be :

x y

-2 -4

-1 -3

0 -2

1 -1

2 0

The coordinates to plot are : (-2,-4) , (-1,-3), (0,-2), (1,-1) ,(2,0)

Use a dotted line and shade the part right hand side of the line.

Do the same for the second inequality equation and plot then shade the part satisfying the inequality.

The graph attached shows results.

5 0

4 years ago

Read 2 more answers