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Kruka [31]
3 years ago
9

You find an unnamed fluid in the lab we will call Fluid A. Fluid A has a specific gravity of 1.65 and a dynamic viscosity of 210

centipoise [cP]. What is the kinematic viscosity of Fluid A, in units of stokes [St]
Engineering
1 answer:
Naily [24]3 years ago
4 0

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

<u>Calculate the kinematic viscosity of Fluid A </u>

First step : determine the density of fluid A

Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

                                            = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

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A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Po
kicyunya [14]

This question is incomplete, the missing image in uploaded along this answer below.

Answer:

The required stress is 200 Mpa

Explanation:

Given the data in the question;

diameter D = 12 mm = 12 × 10⁻³ m

Length L = 188 mm = 188 × 10⁻³ m

Poisson's ratio v = 0.34

Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m

The transverse strain will;

εˣ = Δd / D

εˣ = -0.0105 × 10⁻³ /  12 × 10⁻³ m

εˣ = -0.00088

The longitudinal strain will be;

E^z = - ( εˣ  / v )

E^z = - ( -0.00088  / 0.34 )

E^z = - ( - 0.002588 )

E^z = 0.0026

Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.

From the graph, in the Second image;

The stress is 200 Mpa

Therefore, The required stress is 200 Mpa

8 0
2 years ago
A heat pump operates on a Carnot heat pump cycle with a COP of 12.5. It keeps a space at 24°C by consuming 2.15 kW of power. Det
Vinil7 [7]

Answer:

a) T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C), b) \dot Q_{H} = 26.875\,kW

Explanation:

a) The Coefficient of Performance of the Carnot Heat Pump is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

After some algebraic handling, the temperature of the cold reservoir is determined:

T_{H}-T_{L} = \frac{T_{H}}{COP_{HP}}

T_{L} = T_{H}\cdot \left(1-\frac{1}{COP_{HP}}  \right)

T_{L} = (297.15\,K)\cdot \left(1-\frac{1}{12.5}\right)

T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)

b) The heating load provided by the heat pump is:

\dot Q_{H} = COP_{HP}\cdot \dot W

\dot Q_{H} = (12.5)\cdot (2.15\,kW)

\dot Q_{H} = 26.875\,kW

4 0
3 years ago
Who developed the process of blueprinting?
VikaD [51]
Answer: C.) John Herschel
3 0
3 years ago
You are about to perform PMCS on your M1114? What resource should you use for the procedures and instructions for performing PMC
aniked [119]

The resources and instructions that should be used for the procedures of performing PMCS are:

  1. Operator's manuals
  2. Safety cautions and warnings.
  3. Fording kit
  4. Heating and cooling systems.

<h3>What is PMCS?</h3>

PMCS is an acronym for preventive maintenance checks and services and it can be defined as the maintenance, checks, and services that are typically performed before, during, and after the use of any type of military equipment such as:

  • M1114
  • M1151
  • M1123

Basically, the resources and instructions that should be used for the procedures of performing PMCS are:

  1. Operator's manuals
  2. Safety cautions and warnings.
  3. Fording kit
  4. Heating and cooling systems.

Read more on PMCS here: brainly.com/question/15720250

#SPJ1

4 0
1 year ago
A single fixed pulley is used to lift a load of 400N by the application of an effort of 480N in 10s through a vertical height of
Allushta [10]

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

(c) The efficiency of the machine (E) = 83.3 %

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

M.A = L/E

M.A = 400 / 480

M.A = 0.833

(c) The efficiency of the machine (E);

E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%

4 0
3 years ago
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