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Which of the following was an effect of world war 2 on agricultural industry

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Answer 1

Answer:

Answer:

Option C..Farmers saught new technology to help with the workload

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8. 15 A manual arc welding cell uses a welder and a fitter. The cell operates 2,000 hriyr. The welder is paid $30/hr and the fitter is paid $25/hr. Both rates include applicable overheads. The cycle time to complete one welded assembly is 15. 4 min. Of this time, the arc-on time is 25%, and the fitter's participation in the cycle is 30% of the cycle time. A robotic arc welding cell is being considered to replace this manual cell. The new cell would have one robot, one fitter, and two workstations, so that while the robot is working at the first sta tion, the fitter is unloading the other station and loading it with new components. The fitter's rate would remain at $25/hr. For the new cell, the production rate would be eight welded assemblies per hour. The arc-on time would increase to almost 52%, and the fitter's participation in the cycle would be about 62%. The installed cost of the robot and worksta tions is $158,000. Power and other utilities to operate the robot and arc welding equipment will be $3. 80/hr, and annual maintenance costs are $3,500. Given a 3-year service life, 15% rate of return, and no salvage value, (a) determine the annual quantity of welded assem blies that would have to be produced to reach the break-even point for the two methods. (b) What is the annual quantity of welded assemblies produced by the two methods work. Ing 2,000 hryr?

Answers

The annual quantity of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.

To determine the break-even point between the manual arc welding cell and the robotic cell, we need to calculate the total costs for each method and then equate them.

For the manual arc welding cell:

Labor cost per hour = (welder's hourly rate x arc-on time) + (fitter's hourly rate x fitter's participation in the cycle) = ($30 x 0.25) + ($25 x 0.3) = $11.25

Labor cost per welded assembly = labor cost per hour x cycle time per assembly / 60 = $11.25 x 15.4 / 60 = $2.89

Overhead cost per welded assembly = (labor cost per hour x (1 - arc-on time - fitter's participation in the cycle)) x cycle time per assembly / 60 = ($30 x 0.45) x 15.4 / 60 = $4.68

Total cost per welded assembly = labor cost per welded assembly + overhead cost per welded assembly = $2.89 + $4.68 = $7.57

Total cost per hour = total cost per welded assembly x production rate = $7.57 x 8 = $60.56

Total cost per year = total cost per hour x hours of operation per year = $60.56 x 2,000 = $121,120

For the robotic arc welding cell:

Labor cost per hour = fitter's hourly rate x fitter's participation in the cycle = $25 x 0.62 = $15.50

Labor cost per welded assembly = labor cost per hour x cycle time per assembly / 60 = $15.50 x 15.4 / 60 = $3.97

Overhead cost per welded assembly = power and utility cost per hour + annual maintenance cost / production rate = $3.80 + $3,500 / (8 x 2,000) = $3.80 + $0.22 = $4.02

Total cost per welded assembly = labor cost per welded assembly + overhead cost per welded assembly + (installed cost / (production rate x service life)) = $3.97 + $4.02 + ($158,000 / (8 x 3)) = $3.97 + $4.02 + $6,208.33 = $14.19

Total cost per hour = total cost per welded assembly x production rate = $14.19 x 8 = $113.52

Total cost per year = total cost per hour x hours of operation per year = $113.52 x 2,000 = $227,040

To find the break-even point, we set the total cost of the manual arc welding cell equal to the total cost of the robotic arc welding cell and solve for the annualquantity of welded assemblies:

$121,120 + x($7.57) = $227,040 + x($14.19)

$7.57x - $14.19x = $227,040 - $121,120

$-6.62x = $105,920

x = $105,920 / $6.62

x = 15,982.7

Therefore, the annualquantity of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.

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Given the latent heat of fusion (melting) and the latent heat of vaporisation for water are Δhs = 333.2 kJ/kg and Δhv = 2257 kJ/kg, respectively. Use these values to estimate the total energy required to melt 100 kg of ice at 0 °C and boil off 40 kg of water at 100 °C. a) 239,028 kJ b) 95,250 kJ c) 185,500 kJ d) 362,628 kJ e) 123,600 kJ

Answers

Answer:

C)185,500 KJ

Explanation:

Given that

Latent heat fusion = 333.23 KJ/kg

Latent heat vaporisation = 333.23 KJ/kg

Mass of ice = 100 kg

Mass of water = 40 kg

Mass of vapor=60 kg

Ice at 0°C ,first it will take latent heat of vaporisation and remain at constant temperature 0°C and it will convert in to water.After this water which at 0°C will take sensible heat and gets heat up to 100°C.After that at 100°C vapor will take heat as heat of vaporisation .

Sensible heat for water Q

$Q=mC_p\Delta T$

For water

$C_p=4.178\ KJ/Kg.K$

Q=4.178 x 40 x 100 KJ

Q=16,712 KJ

So total heat

Total heat =100 x 333.23+16,712 + 60 x 2257 KJ

Total heat =185,455 KJ

Approx Total heat = 185,500 KJ

So the answer C is correct.

For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value of 2.1. If, after 146 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 86% completion?

Answers

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

$Y = 1 - e^(-kt^n)$

here Y is percentage of completion of reaction = 50%

t is duration of reaction = 146 sec

so,

$0.50 = 1 - e^(-k^146^2.1)$

$0.50 = e^(-k306.6)$

taking natural log on both side

ln(0.5) = -k(306.6)

$k = 2.26* 10^(-3)$

for 86 % completion

$0.86 = 1 - e^{-2.26* 10^(-3) * t^(2.1)}$

$e^{-2.26* 10^(-3) * t^(2.1)} = 0.14$

$-2.26* 10^(-3) * t^(2.1) = ln(0.14)$

$t^(2.1) = 869.96$

t = 25.10 sec

Determine the boundary work done by a gas during an expansion process if the pressure and volume values at various states are measured to be 300 kPa, 1 L; 290 kPa, 1.1 L; 270 kPa, 1.2 L; 250 kPa, 1.4 L; 220 kPa, 1.7 L; and 200 kPa, 2 L

Answers

By Riemann sums, the boundarywork done by a gas during an expansionprocess based on the information given by the statement is approximately 0.243 joules.

How to determine the boundary work done by a gas during an expansion process

A process is a consecution of states of a system. The boundary work (W), in kilojoules, is the work done by the system on surroundings and in a P-Vdiagram this kind of work is equal to the area below the curve, which can be approximated by Riemann sums:

$W = \sum\limits_(i=1)^(n-1) p_(i)\cdot (V_(i+1)-V_(i)) + (1)/(2)\sum\limits_(i=1)^(n-1) (p_(i+1)-p_(i))\cdot (V_(i+1)-V_(i))$ (1)

Where:

p - Pressure, in kilopascals.
V - Volume, in cubic meters.

$W = (1)/(2) \sum\limits_(i=1)^(n-1) (p_(i+1)+p_(i))\cdot (V_(i+1)-V_(i))$

Now we proceed to calculate the boundary work:

W = 0.5 · [(300 kPa + 290 kPa) · (1.1 × 10⁻³ m³ - 1 × 10⁻³ m³) + (270 kPa + 290 kPa) · (1.2 × 10⁻³ m³ - 1.1 × 10⁻³ m³) + (250 kPa + 270 kPa) · (1.4 × 10⁻³ m³ - 1.2 × 10⁻³ m³) + (220 kPa + 250 kPa) · (1.7 × 10⁻³ m³ - 1.4 × 10⁻³ m³) + (200 kPa + 220 kPa) · (2 × 10⁻³ m³ - 1.7 × 10⁻³ m³)]

W = 0.243 kJ

By Riemann sums, the boundarywork done by a gas during an expansionprocess based on the information given by the statement is approximately 0.243 joules.

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Answer:

attached below

Explanation:

An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a lagging power factor of 0.77. Determine the size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging.

Answers

Answer:

$Q=41.33 KVAR\ \nat\n\ 480 Vrms$

Explanation:

From the question we are told that:

Voltage $V=480/0 \textdegree V$

Power $P=120kW$

Initial Power factor $p.f_1=0.77 lagging$

Final Power factor $p.f_2=0.9 lagging$

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

$p.f_1=0.77$

$cos \theta_1 =0.77$

$\theta_1=cos^(-1)0.77$

$\theta_1=39.65 \textdegree$

And

$p.f_2=0.9$

$cos \theta_2 =0.9$

$\theta_2=cos^(-1)0.9$

$\theta_2=25.84 \textdegree$

Therefore

$Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=-41.33VAR$

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

$Q=41.33 KVAR\ \nat\n\ 480 Vrms$

An activated sludge plant is being designed to handle a feed rate of 0.438 m3 /sec. The influent BOD concentration is 150 mg/L and the cell concentration (MLVSS) is 2,200 mg/L. If you wish to operate the plant with a food-to-microorganism ratio of 0.20 day-1 , what volume of aeration tank should you use? Please give your answer in m3 .

Answers

Answer:

Volume of aeration tank = 1.29 x 10^4 m³

Explanation:

Food/Micro- organism Ratio = 0.2/day

Feed Rate (Q) = 0.438 m³/s

Influent BOD = 150 mg/L

MLVSS = 2200 mg/L

The above mentioned parameters are related by the equation

F/M = QS₀/VX

where S₀ is the influent BOD and X is the cell concentration, with V being the Volume of the tank. Re- arranging the above equation for V and putting the values in, we get

V = 0.4380 x 150/0.2 x 2200

V = 0.1493 (m³/s) x day

V = 0.1493 x 24 x 60 x 60

V = 1.29 x 10^4 m³

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A Coca Cola can with diameter 62 mm and wall thickness 300 um has an internal pressure of 100 kPa. Calculate the principal stresses at a point on the cylindrical surface of the can far from its ends.