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3 years ago

Which statement best describes a hybrid design?

Engineering

2 answers:

mario62 [17]3 years ago

4 0

Answer:

B. Hybrid

Explanation:

A hybrid desing implicate combining two different elements/methods, such a hybrid car, which works using both mechanic and electric energy.

mash [69]3 years ago

3 0

Answer:

Hybrid design - (B) It is a design based on multiple methodologies.

Explanation:

Hybrid simply means more than one. Certain times one thing can't be able to sustain the whole process for that multiple ideas are taken together and packed into one and a new design is made called hybrid design which includes properties of more than one designs.

It is simple as a child have abilities of parents as well as grandparents inherited. It is said as hybrid because it is a combination of more than one design. It helps to carry forward over what is already there and take it to the needed next level and helps to get desired successful results."

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(20 points) A 1 mm diameter tube is connected to the bottom of a container filled with water to a height of 2 cm from the bottom

zzz [600]

Solution :

Given :

h = 2 cm

Diameter of the tube , d = 1 mm

Diameter of the hose, D = 6 mm

Between 1 and 2, by applying Bernoulli's principle, we get

As point 1 is just below the free surface of liquid, so

$P_1=P_{atm} \text{ and} \ V_1=0$

$\frac{P_{atm}}{\rho g}+\frac{v_1^2}{2g} +h = \frac{P_2}{\rho g}$

$\frac{101.325}{1000 \times 9.81}+0.02 =\frac{P_2}{\rho g}$

$P_2 = 111.35 \ kPa$

Therefore, 111.325 kPa is the gas supply pressure required to keep the water from leaking back into the tube.

Velocity at point 2,

$V_2=\sqrt{\left(\frac{111.135}{\rho g}+0.02}\right)\times 2g$

= 1.617 m/s

Flow of water, $Q_2 = A_{tube} \times V_2$

$=\frac{\pi}{4} \times (10^{-3})^2 \times 1.617$

$1.2695 \times 10^{-6} \ m^3/s$

Minimum air flow rate,

$Q_2 = Q_3 = A_{hose} \times V_3$

$V_3 = \frac{Q_2}{\frac{\pi}{4}D^2}$

$V_3 = \frac{1.2695 \times10^{-6}}{\pi\times 0.25 \times 36 \times 10^{-6}}$

= 0.0449 m/s

b). Reynolds number in hose,

$Re = \frac{\rho V_3 D}{\mu} = \frac{V_3 D}{\nu}$

υ for water at 25 degree Celsius is $8.9 \times 10^{-7} \ m^2/s$

υ for air at 25 degree Celsius is $1.562 \times 10^{-5} \ m^2/s$

$Re_{hose}=\frac{0.0449 \times 6 \times 10^{-3}}{1.562 \times 10^{-5}}$

= 17.25

Therefore the flow is laminar.

Reynolds number in the pipe

$Re = \frac{V_2 d}{\nu} = \frac{1.617 \times 10^{-3}}{8.9 \times 10^{-7}}$

= 1816.85, which is less than 2000.

So the flow is laminar inside the tube.

3 0

2 years ago

simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl

Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

a) Air temperature at turbine exit

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17 for Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) $(\frac{783.05-778.18}{800.13-778.18} )$ = 764.45K

b) The net work output

first we determine the actual work input to compressor

Wc = h2 - h1 ( calculated values )

= 626.57 - 295.17 = 331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4 ( calculated values )

= 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

= 541.88 - 331.4 = 210.48 kJ/kg

c) determine thermal efficiency

лth = Wnet / qin ------ ( 1 )

where ; qin = h3 - h2

equation 1 becomes

лth = Wnet / ( h3 - h2 )

= 210.48 / ( 1324.93 - 626.57 )

= 0.3014 = 30.14%

6 0

3 years ago

Which statement about tensile stress is true? A. Forces that act perpendicular to the surface and pull an object apart exert a t

svp [43]

Answer:

A. Forces that act perpendicular to the surface and pull an object apart exert a tensile stress on the object.

Explanation:

Tensile stress is referred as a deforming force, in which force acts perpendicular to the surface and pull an object apart, attempting to elongate it.

The tensile stress is a type of normal stress, in which a perpendicular force creates the stress to an object’s surface.

Hence, the correct option is "A."

3 0

3 years ago

The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,

babunello [35]

Answer:

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × $10^6$ Pa

viscosity n = 100 Pas

angle = 18°

so Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 $\times 10^{-3}$ )²× 9 $\times 10^{-3}$ × sin18 × cos18

Qd = 94.305 × $10^{-6}$ m³/s

and

Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2

Qb = 11 × $10^{6}$ × π × 85 $\times 10^{-3}$ × ( 9 $\times 10^{-3}$ )³ × sin²18 ÷ 12 × 100 × 2

Qb = 85.2 × $10^{-6}$ m³/s

so here

volume flow rate Qx = Qd - Qb ..............3

Qx = 94.305 × $10^{-6}$ - 85.2 × $10^{-6}$

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

8 0

2 years ago

Bob and Alice are solving practice problems for CSE 2320. They look at this code: for(i = 1; i <= N; i = (i*2)+17 ) for(k = i

MissTica

Answer:

Alice is correct.

The loop are dependent.

Explanation:

for(i = 1; i <= N; i = (i*2)+17 )

for(k = i+1; k <= i+N; k = k+1) // notice i in i+1 and i+N

printf("B")

This is a nested for-loop.

After the first for-loop opening, there is no block of statement to be executed rather a for-loop is called again. And the second for-loop uses the value of i from the first for-loop. The value of N is both called from outside the loop.

So, the second for-loop depend on the first for loop to get the value of i. For clarity purpose, code indentation or use of curly brace is advised.

8 0

3 years ago

Read 2 more answers