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3 years ago

Which measuring tool will be used to determine the diameter of a crankshaft journal?

Engineering

1 answer:

guapka [62]3 years ago

4 0

Answer:

The dial bore gauge measures the inside of round holes, such as the bearing journals. This one tool can measure 2″ up to 6″ diameter holes. Both tools are needed in order to check the interior and exterior dimensions of the crankshaft, rods and engine block journals, as well as the thickness of the bearings themselves.

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Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.

Nutka1998 [239]

Answer:

β = $\frac{c}{\sqrt{km} }$ = 0.7071 ≈ 1 ( damping condition )

closed-form expression for the response is attached below

Explanation:

Given : x + 2x + 2x = 0 for Xo = 0 mm and Vo = 1 mm/s

computing a solution :

M = 1,

c = 2,

k = 2,

Wn = $\sqrt{\frac{k}{m} }$ = $\sqrt{2}$

next we determine the damping condition using the damping formula

β = $\frac{c}{\sqrt{km} }$ = 0.7071 ≈ 1

from the condition above it can be said that the damping condition indicates underdamping

attached below is the closed form expression for the response

6 0

2 years ago

How do I do this? Blueprints, complete the missing view.

Ymorist [56]

Explanation:

Look at the drawings and decide which view is missing. Front? Side? Top? Then draw it

7 0

3 years ago

Suppose we use radix sort to sort the English-language strings below using standard lexicographic ordering (i.e. sort in alphabe

nlexa [21]

Answer:

a) 4 passes are required to sort the string.

b) 4

c) i) TARP

ii) CHIP

iii) PART

iv) TARP

v) TARP

d) O(k+n), n is no. of strings, k is largest no. of character in among the string

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Explanation:

7 0

3 years ago

A spring-mass-damper instrument is employed for acceleration measurements. The spring constant is 12000 N/m. The mass is 5 g. Th

shepuryov [24]

Answer:

a) 246.56 Hz

b) 203.313 Hz

c) Add more springs

Explanation:

Spring constant = 12000 N/m

mass = 5g = 5 * 10^-3 kg

damping ratio = 0.4

a) Calculate Natural frequency

Wn = √k/m = $\sqrt{12000 / 5*10^{-3} }$

= 1549.19 rad/s ≈ 246.56 Hz

b) Bandwidth of instrument

W / Wn = $\sqrt{1-2(0.4)^2}$

W / Wn = 0.8246

therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz

C ) To increase the bandwidth we have to add more springs

5 0

3 years ago

Problem 10.012 SI A vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. S

pychu [463]

Answer:

a) 4.1 kw

b) 4.68 tons

c) 4.02

Explanation:

Saturated vapor enters compressor at ( p1 ) = 2.6 bar

Saturated liquid exits the condenser at ( p2 ) = 12 bar

Isentropic compressor efficiency = 80%

Mass flow rate = 7 kg/min

A) Determine compressor power in KW

compressor power = m ( h2 - h1 )

= 7 / 60 ( 283.71 - 248.545 )

= 4.1 kw

B) Determine refrigeration capacity in tons = m ( h1 - h4 )

= 7/60 ( 248.545 - 107.34 )

= 16.47 kw = 4.68 tons

C) coefficient of performance ( COP )

= Refrigeration capacity / compressor power

= 16.47 / 4.1 = 4.02

Attached below is the beginning part of the solution

7 0

2 years ago