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2 years ago

An open tank contain oil of specific gravity 0.75 on top of

Engineering

1 answer:

n200080 [17]2 years ago

8 0

Let's not take into account atmospheric pressure.

Pressure P = Pressure of oil column + Pressure of water column

= h1 . d1.g + h2.d2. g

=( 0.5 x 800 x 9.8) + ( 1x 1000x 9.8) = 3920+ 9800 = 13720

Hence guage pressure at bottom is

P= 13720 N/ m^2

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what do you expect the future trends of an operating system in terms of (a) cost (b) size (c) multitasking (d) portability (e) s

tino4ka555 [31]

Answer:

plz follow in titkok

Explanation:

3 0

3 years ago

Please answer fast. With full step by step solution.

lina2011 [118]

Let f(z) = (4z ² + 2z) / (2z ² - 3z + 1).

First, carry out the division:

f(z) = 2 + (8z - 2) / (2z ² - 3z + 1)

Observe that

2z ² - 3z + 1 = (2z - 1) (z - 1)

so you can separate the rational part of f(z) into partial fractions. We have

(8z - 2) / (2z ² - 3z + 1) = a / (2z - 1) + b / (z - 1)

8z - 2 = a (z - 1) + b (2z - 1)

8z - 2 = (a + 2b) z - (a + b)

so that a + 2b = 8 and a + b = 2, yielding a = -4 and b = 6.

So we have

f(z) = 2 - 4 / (2z - 1) + 6 / (z - 1)

f(z) = 2 - (2/z) (1 / (1 - 1/(2z))) + (6/z) (1 / (1 - 1/z))

Recall that for |z| < 1, we have

$\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n$

Replace z with 1/z to get

$\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}$

so that by substitution, we can write

$\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}$

Now condense f(z) into one series:

$\displaystyle f(z) = 2 - \sum_{n=0}^\infty 2^{-n+1} z^{-(n+1)} + 6 \sum_{n=0}^\infty z^{-n-1}$

$\displaystyle f(z) = 2 - \sum_{n=0}^\infty \left(6+2^{-n+1}\right) z^{-(n+1)}$

$\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{-(n-1)+1}\right) z^{-n}$

$\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{2-n}\right) z^{-n}$

So, the inverse Z transform of f(z) is $\boxed{6+2^{2-n}}$ .

4 0

3 years ago

The inlet and exhaust flow processes are not included in the analysis of the Otto cycle. How do these processes affect the Otto

lara31 [8.8K]

Answer:

Suction and exhaust processes do not affect the performance of Otto cycle.

Explanation:

Step1

Inlet and exhaust flow processes are not including in the Otto cycle because the effect and nature of both the process are same in opposite direction.

Step2

Inlet process or the suction process is the process of suction of working fluid inside the cylinder. The suction process is the constant pressure process. The exhaust process is the process of exhaust out at constant pressure.

Step3

The suction and exhaust process have same work and heat in opposite direction. So, net effect of suction and exhaust processes cancels out. The suction and exhaust processes are shown below in P-V diagram of Otto cycle:

Process 0-1 is suction process and process 1-0 is exhaust process.

7 0

3 years ago

A manometer containing a fluid with a density of 60 lbm/ft3 is attached to a tank filled with air. If the gage pressure of the a

8090 [49]

Answer:

The fluid level difference in the manometer arm = 22.56 ft.

Explanation:

Assumption: The fluid in the manometer is incompressible, that is, its density is constant.

The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.

And P(gage) = ρgh

ρ = density of the manometer fluid = 60 lbm/ft³

g = acceleration due to gravity = 32.2 ft/s²

ρg = 60 × 32.2 = 1932 lbm/ft²s²

ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³

h = fluid level difference between the two arms of the manometer = ?

P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²

1353.6 = ρg × h = 60 lbf/ft³ × h

h = 1353.6/60 = 22.56 ft

A diagrammatic representation of this setup is presented in the attached image.

Hope this helps!

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4 years ago

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cricket20 [7]

Answer:

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3 years ago