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professor190 [17]
4 years ago
14

1. What is the linear distance traveled in one revolution of a 9 in diameter wheel?

Engineering
1 answer:
zmey [24]4 years ago
5 0

Explanation:

1. Circumference is pi times diameter.

C = πd

C = 9π inches

C ≈ 28.3 inches

2. Mechanical advantage is ratio of output force to input force.

MA = Fout / Fin

MA = 155 lb / 15 lb

MA ≈ 10.3

3. Mechanical advantage is ratio of input distance over output distance.

MA = din / dout

MA = 10.5 in / 0.85 in

MA ≈ 12.4

4. Mechanical advantage of pulley system is the number of strands.

MA = 8

5. Mechanical advantage is ratio of output force to input force.

MA = Fout / Fin

8 = Fout / 200 lb

Fout = 1600 lb

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If it is desired to lay off a distance of 10,000' with a total error of no more than ± 0.30 ft. If a 100' tape is used and the
Ira Lisetskai [31]

Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

3 0
3 years ago
Air enters the compressor of a simple gas turbine at 100 kPa, 300 K, with a volumetric flow rate of 5 m3/s. The compressor press
Zepler [3.9K]

Answer:

a) 3581.15067 kw

b) 95.4%

Explanation:

<u>Given data:</u>

compressor efficiency = 85%

compressor pressure ratio = 10

Air enters at:    flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K

At turbine inlet : pressure = 950 kPa, temperature = 1400k

Turbine efficiency = 88% , exit pressure of turbine = 100 kPa

A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW

attached below is a detailed solution to the given question

6 0
3 years ago
A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?
Ierofanga [76]

Answer: (b)

Explanation:

Given

Original length of the rod is L=100\ cm

Strain experienced is \epsilon=82\%=0.82

Strain is the ratio of the change in length to the original length

\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm

Therefore, new length is given by (Considering the load is tensile in nature)

\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm

Thus, option (b) is correct.

8 0
3 years ago
The operating sequence to light the main burners on an intermittent pilot system is:______.
11111nata11111 [884]
The pilot valve and spark igniter are energized, the pilot flame is proved, and then the main gas valve is energized.
7 0
2 years ago
(a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and pr
Gnoma [55]

Answer:

See explanation below

Explanation:

Hypo-eutectoid steel has less than 0,8% of C in its composition.

It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.

Ferrite has a higher tensile strength than cementite but cementite is harder.

Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:

Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel

Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel

Hyper-eutectoid steel is harder than Hyper-eutectoid steel

Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.

When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because

1. It is harder

2. It has low cost

3. It is lighter

When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because

1. It is ductile

2. It has high tensile strength

3. It is durable

7 0
3 years ago
Read 2 more answers
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