Thermoplastics burn upon heating. a)-True b)- false?

Problem 3.10 One/half million parts of a certain type are to be manufactured annually on dedicated production machines that run

goldenfox [79]

Answer:

4 machines
4 machines

Explanation:

(a) The production schedule is 4000 hour per year, or 240,000 minutes. In that time, 133,333 1/3 parts can be produced. The production of 500,000 parts in a year will require the use of ...

.5 million/(.1333... million/machine) = 3.75 machines

The demand can be satisfied by 4 machines.

__

(b) If availability is 95% then 3.75/0.95 ≈ 3.95 machines will be required.

The demand can be satisfied by 4 machines.

8 0

3 years ago

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Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volum

ryzh [129]

Answer:

The Question is incomplete, the complete question is as follows:

Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volume of 5,435 vehicles. If all other conditions are as described in Problem 1, how long can this grade be without the freeway LOS dropping to F?

A six-lane rural freeway (three lanes in each direction) has regular weekday users and currently operates at maximum LOS C conditions. The base free-flow speed is 65 mi/h, lanes are 11 ft wide, the right-side shoulder is 4 ft wide, and the interchange density is 0.25 per mile. The highway is one rolling terrain with 10% large trucks and buses (no recreational vehicles), and the peak-hour factor is 0.90. Determine the hourly volume for these conditions

Explanation:

Make the assumption Base continuous flow velocity (BFFS)= 65 mph. 

Pitch width= 11 ft.

Decrease in lane width pace,fLW= 1.9 mph.

Complete Lateral clearance= 4 ft. Lateral clearance speed reduction, fLC= 0.8 mph.

Complete Width of the Ramp= 0.25 mile.

Velocity reduction proportional to the ramp height, f ID= 0 mph.

Assume lane number to be = 3.

Reduction in speed corresponding to no. of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 65 – 1.9 – 0.8 – 3 – 0 = 59.3 mph

Peak Flow, V veh/hr

Peak-hour factor = 0.90

Trucks = 10%

Rolling Terrain

fHV = 1/ (1 + 0.10 (2.5-1)) = 1/1.15 = 0.8696

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = V/ (0.90*3*0.8696*1.0) = 0.426V veh/hr/ln

Average speed of vehicles, S = FFS = 59.3 mph

Level of service C

Density of LOS C lies between 18 - 25 veh/mi/ln

Maximum density = 25 veh/mi/ln

Density = Vp /S = 25

0.426V = 25 * 59.3

V = 3480 veh/hr

b) V = 5435 veh/hr

LOS dropping to F

Max density = 45 veh/mi/ln

Density = Vp/S = 45

Vp = 45 * 59.3 = 2668.5 veh/hr/ln

V/(PHF * n * fHV * fP) = 2668.5

fHV = 5435/(0.9*3*2668.5*1.0) = 0.754

1/(1+0.10 (ET -1)) = 0.754

ET = 4.26 ~ 3.5

For 4% upgrade and 10% trucks with ET = 3.5, length of the grade is Greater than 1.0 miles

6 0

3 years ago

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For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5

bonufazy [111]

Answer:

The total force resisted by the flange bolts is 163.98 N

Explanation:

Solution

The first step is to find the pipe cross section at the inlet section

Now,

A₁ = π /4 D₁²

D₁ = diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of the pipe at the inlet section

A₂ = π /4 D₂²

D₂ = diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265 * 5 = 1.963 * V₂

V₂ = 12.8 m/s

Note: Kindly find an attached copy of the part of the solution to the given question below

8 0

3 years ago

Determine the combined moment about O due to the weight of the mailbox and the cross member AB. The mailbox weighs 3.2 lb and th

koban [17]

Answer:

Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib

Answer : moment of inertia = 186.7 Ib - in

Explanation:

Given data

weight of the mailbox = 3.2 Ib

weight of the uniform cross member = 10.3 Ib

The origin is of mailbox and cross member is 0

The perpendicular distance from Y axis of centroid of the mailbox

= 4 + (25/2) = 16.5"

The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4 = 13"

therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)

= 52.8 + 133.9 = 186.7 Ib-in

8 0

3 years ago

Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the r

9966 [12]

Answer:

(a) Calculate the rod base temperature (°C). = 299.86°C

(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent. = 0.4325m

Explanation:

see attached file below

3 0

3 years ago