6. What living area room must be adjacent to the kitchen?

A horizontal 2-m-diameter conduit is half filled with a liquid (SG=1.6 ) and is capped at both ends with plane vertical surfaces

luda_lava [24]

Answer:

Resultant force = 639 kN and it acts at 0.99m from the bottom of the conduit

Explanation:

The pressure is given as 200 KPa and the specific gravity of the liquid is 1.6.

The resultant force acting on the vertical plate, Ft, is equivalent to the sum of the resultant force as a result of pressurized air and resultant force due to oil, which will be taken as F1 and F2 respectively.

Therefore,

Ft = F1 + F2

According to Pascal's law which states that a change in pressure at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere, the air pressure will act on the whole cap surface.

To get F1,

F1 = p x A

= p x (πr²)

Substituting values,

F1 = 200 x π x 1²

F1 = 628.32 kN

This resultant force acts at the center of the plate.

To get F2,

F2 = Π x hc x A

F2 = Π x (4r/3π) x (πr²/2)

Π - weight density of oil,

A - area on which oil pressure is acting,

hc - the distance between the axis of the conduit and the centroid of the semicircular area

Π = Specific gravity x 9.81 x 1000

Therefore

F2 = 1.6 x 9.81 x 1000 x (4(1)/3π) x (π(1)²/2)

F2 = 10.464 kN

Ft = F1 + F2

Ft = 628.32 + 10.464

Ft = 638.784 kN

The resultant force on the surface is 639 kN

Taking moments of the forces F1 and F2 about the centre,

Mo = Ft x y

Ft x y = (F1 x r) + F2(1 - 4r/3π)

Making y the subject,

y = (628.32 + 10.464(1 - 4/3π)/ 638.784

y = 0.993m

7 0

4 years ago

Plz answer all of these questions!

statuscvo [17]

Answer: all you need to to is go to help me with this question.cm

Explanation:

5 0

3 years ago

Propane is to be compressed from 0.4 MPa and 360 K to 4 MPa using a two-stage compressor. An interstage cooler returns the tempe

kow [346]

Answer:

a. 81 kj/kg

b. 420.625K

c. 101.24kj/kg

Explanation:

$\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }$

t1 = 360

p1 = 0.4mpa

p2 = 1.20

y = 1.13

substitute these values into the equation

$\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }$

$\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347$

when we cross multiply

t2 = 360 * 1.1347

= 408.5

a. the work required in the firs compressor

w=c(t2-t1)

c=1.67x10³

t1 = 360

t2 = 408.5

w = 1670(408.5-360)

= 1670*48.5

= 80995 J

= 81KJ/kg

b. $n=\frac{t2-t1}{t'2-t1}$

n = 80%

t2 = 408.5

t1 = 360

0.80 = 408.5-360 ÷ t'2-360

$0.80 =\frac{48.5}{t'2-360}$

cross multiply to get the value of t'2

0.80(t'2-360) = 48.5

0.80t'2 - 288 = 48.5

0.8t'2 = 48.5+288

0.8t'2 = 336.5

t'2 = 336.5/0.8

= 420.625

this is the temperature at the exit of the first compressor

c. cooling requirement

w = c(t2-t1)

= 1.67x10³(420.625-360)

= 1670*60.625

= 101243.75

= 101.24kj/kg

8 0

3 years ago

What scale model proves the initial concept?

Tju [1.3M]

Answer: A prototype

Explanation:

5 0

3 years ago

2. Given that a quality-control inspection can ensure that a structural ceramic part will have no flaws greater than 100 µm (100

MA_775_DIABLO [31]

Answer:

SiC=169.26 Mpa

Partially stabilized zirconia=507.77 Mpa

Explanation:

Stress intensity, K is given by

$K=\sigma Y\sqrt{\pi a}$ hence making $\sigma$ the subject where $\sigma$ is applied stress, Y is shape factor, a is crack length

$\sigma=\frac {K}{Y\sqrt{\pi a}}$ substituting the given figures and assuming shape factor, Y of 1

$\sigma=\frac {3}{1\sqrt{\pi 100*10^{-6}}}= 169.2569\approx 169.26 Mpa$

<u>Stabilized zirconia </u>

Stress intensity, K is given by

$K=\sigma Y\sqrt{\pi a}$ hence making $\sigma$ the subject where $\sigma$ is applied stress, Y is shape factor, a is crack length

$\sigma=\frac {K}{Y\sqrt{\pi a}}$ substituting the given figures and assuming shape factor, Y of 1

$\sigma=\frac {9}{1\sqrt{\pi 100*10^{-6}}}= 507.7706\approx 507.77 Mpa$

7 0

3 years ago