Quelles sont les types de carburant utilisés en aviation

A liquid drug, with the viscosity and density of water, is to be administered through a hypodermic needle. The inside diameter o

vredina [299]

Answer:

(a) The maximum volume flow rate for which the flow will be laminar is 0.0190 cubic meter per second

(b) The pressure drop required to deliver the maximum flow rate is 148962.96 Pascal

(c) The corresponding wall shear stress is 7600 Pascal

Explanation:

Reynolds number = 2299, density of water = 1000kg/m^3, diameter of needle = 0.27mm = 0.00027m, Length of needle = 50mm = 0.05m, viscosity of water = 0.00089kg/ms, area = 0.05m × 0.05m = 0.0025m^2, coefficient of friction = 64 ÷ Reynolds number = 64 ÷ 2299 = 0.028

Velocity = (Reynolds number × viscosity) ÷ (density × diameter) = (2299 × 0.00089) ÷ (1000 × 0.00027) = 2.046 ÷ 0.27 = 7.58m/s

(a) Maximum volume flow rate = velocity × area of needle = 7.58 × 0.0025 = 0.0190 cubic meter per second

(b) Pressure drop = ( coefficient of friction × length × density × velocity^2) ÷ (2 × diameter) = (0.028 × 0.05 × 1000 × 7.58^2) ÷ (2 × 0.00027) = 80.44 ÷ 0.00054 = 148962.96 Pascal

(c) Wall shear stress = (density × volume flow rate) ÷ area = (1000 × 0.0190) ÷ 0.0025 = 7600 Pascal

7 0

3 years ago

Route Choice The cost of roadway improvements to the developer is a function of the amount of traffic being generated by the the

aksik [14]

Row choice the cost of roadway improvements to the developer and functional the amount of trafficBeing generated by the theater as well as the Ralph’s ladies

7 0

3 years ago

Two advantages of deforming steel at room temperature rather than at elevated temperatures are: (select 2 answers from the optio

Aleks [24]

Answer:

A AND C

Explanation:

Two advantages of deforming steel at room temperature rather than at elevated temperatures are: (select 2 answers from the options below)

A. better dimensional accuracy to allow form complex parts with complex geometries.

C. a smoother surface finish to allow elimination of finish machining and grinding operations.

7 0

3 years ago

After reading through the code, what will happen when you click run?* 1 point when run move forward while there is a pile do rem

KATRIN_1 [288]

Explanation:

I'm not exactly a master at coding, but I'm pretty sure that:

The farmer will remove dirt as long as there is a pile, then stop when the pile is done.

5 0

3 years ago

Two cars are traveling on level terrain at 55 mi/h on a road with a coefficient of adhesion of 0.75. The driver of car 1 has a 2

Vlad1618 [11]

Answer:

0.981

Explanation:

velocity of cars ( v1 , v2 ) = 55 mi/h

coefficient of adhesion ( u ) = 0.75

Reaction time of driver of car 1 = 2.3 -s

Reaction time of driver of car 2 = 1.9 -s

breaking efficiency of car 2 ( n2 ) = 0.80

Determine the braking efficiency of car 1

First determine the distance travelled during reaction time ( dr )

dr = v * tr ------- ( 1 )

tr ( reaction time )

v = velocity

note : 1 mile = 1609 m , I hour = 60 * 60 secs

back to equation 1

for car 1

dr1 =( 55 * 2.3 * 1609 ) / ( 60 * 60 )

= 56.53 m

for car 2

dr2 = ( 55 * 1.9 * 1609 ) / ( 60 * 60 )

= 46.70 m

next we calculate the stopping distances ( d ) using the relation below

ds = d + dr

d = distance travelled during break

dr = distance travelled during reaction time

where : d = $\frac{v^2intial}{2ugn}$

for car 1

d1 = $\frac{(55)^2}{2*0.75*9.81* n1} * (\frac{1609}{3600} )^2$

∴ d1 = $\frac{41.10}{n1}$

for car 2

d2 = $\frac{(55)^2}{2*0.75*9.8*0.8} * (\frac{1609}{3600} )^2$

∴ d2 = 51.38

since the stopping distance for both cars are the same

d1 + dr1 = d2 + dr2

( 41.10 /n1 ) + 56.53 = 51.38 + 46.70

solve for n1

hence n1 = 0.981 ( braking efficiency of car 1 )

4 0

3 years ago