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gayaneshka [121]
3 years ago
12

In an air compressor the compression takes place at a constant internal energy and 50KJ of heat are rejected to the cooling wate

r for every Kilogram of air. Calculate the work input for the compression stroke per kilogram of air?
Engineering
1 answer:
lozanna [386]3 years ago
8 0

Answer:

work is 50 kj

Explanation:

Given  data

heat (Q) = 50 kj

To find out

work input for the compression stroke per kilogram of air

Solution

we will apply here "first law of thermodynamics" i.e.

The First Law of Thermodynamics states that heat is a form of energy, subject to the principle of conservation of energy, that heat energy cannot be created or destroyed. It can be transferred from one location to another  location. i.e.

ΔU = Q – W                        ................1

here ΔU is change in internal energy, Q is heat and W is work done

here U = 0 because air compressor the compression takes place at a constant internal energy in question

so that by equation 1

Q = W

and Q = 50

so work will be 50 kj

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A Michelson interferometer operating at a 500 nm wavelength has a 3.73-cm-long glass cell in one arm. To begin, the air is pumpe
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Answer:

The number of bright-dark fringe is 42

Solution:

As per the question:

Wavelength of light, \lambda = 500\ nm = 500\times 10^{- 9}\ m

Length of the glass cell, x = 3.73 cm = 0.0373 m

Refractive index, \mu = 1.00028

Now,

To calculate the bright-dark fringe shifts, we use the formula given below:

d_{m} = \frac{2x}{\lambda }\times (\mu - 1)

Now, substituting the appropriate values in the above formula:

d_{m} = \frac{2\times 0.0373}{500\times 10^{- 9}}\times (1.00028 - 1)

d_{m} = 41.77 ≈ 42

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Answer:

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What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

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MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

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Tm = (3.5 + 1) / (0.02 × 229.18  )

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3 years ago
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