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erastova [34]
3 years ago
14

The velocity profile for a thin film of a Newtonian fluid that is confined between the plate and a fixed surface is defined by u

= (5y-0.5y2) mm/s where y is in mm. Determine the force P that must be applied to the plate to cause this motion. The plate has a surface area of 4000mm2 in contact with the fluid. Take µ= 0.55 N.s/m2
Engineering
1 answer:
zimovet [89]3 years ago
8 0

Answer:

F = 0.0022N

Explanation:

Given:

Surface area (A) = 4,000mm² = 0.004m²

Viscosity = µ = 0.55 N.s/m²

u = (5y-0.5y²) mm/s

Assume y = 4

Computation:

F/A = µ(du/dy)

F = µA(du/dy)

F = µA[(d/dy)(5y-0.5y²)]

F = (0.55)(0.004)[(5-1(4))]

F = 0.0022N

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The following is a list of metals and alloys:
viktelen [127]

Answer:

A) Gray cast iron

B) Aluminum

C) Titanium alloy

D) Tool steel

E) Titanium alloy

F) magnesium

G) Tungsten

5 0
3 years ago
Do all websites use the same coding to create?
Sonbull [250]

Answer:

yes.

Explanation:

because all websites use coding

6 0
3 years ago
Two technicians are discussing the intake air temperature (IAT) sensor. Technician A says that the computer uses the IAT sensor
mart [117]

Both the technicians are correct.

Explanation

Intake air temperature sensor is used in engines of vehicles to monitor the temperature of air entering the engine.

They are basically made of thermistors whose electrical resistance changes according to temperature.

Depending upon the reading and accuracy of intake air temperature sensor, the power-train control module (PCM) will decide about the air and fuel mixture ratio in the engine.

The hot air in engine requires less fuel to operate the engine parts while cold air requires more fuel to operate the engine.

The ratio of air and fuel mixture should be maintained in the engine and it is done by PCM only after getting the input from IAT. So technician B is saying correct.

Also the IAT works as a backup to support the engine coolant temperature sensor by the computer.

As the IAT checks the temperature of outside air, it will help to change the coolant temperature of the engine based on the environment.

Thus technician A is also correct. So both the technicians are correct.

6 0
3 years ago
Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.
Nutka1998 [239]

Answer:

β = \frac{c}{\sqrt{km} } =  0.7071 ≈ 1 ( damping condition )

closed-form expression for the response is attached below

Explanation:

Given :  x + 2x + 2x = 0   for Xo = 0 mm and Vo = 1 mm/s

computing a solution :

M = 1,

c = 2,

k = 2,

Wn = \sqrt{\frac{k}{m} }  = \sqrt{2}  

next we determine the damping condition using the damping formula

β = \frac{c}{\sqrt{km} } =  0.7071 ≈ 1

from the condition above it can be said that the damping condition indicates underdamping

attached below is the closed form expression for the response

6 0
2 years ago
Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0
3 years ago
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