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kakasveta [241]
3 years ago
10

You are developing a mind map for measuring the viscosity of fluid. At which stage in the process will you specify the equipment

needed for the experiment?
Connecting the subtopics fo the central idea

Defining the lower-level subtopics

Creating the image

Defining the subtopics
Engineering
1 answer:
kipiarov [429]3 years ago
6 0

Answer:

Defining the subtopics

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Two plates are separated by a 1/4 in space. The lower plate is stationary; the upper plate moves at 10 ft/s. Oil (viscosity of 2
Montano1993 [528]

Answer:

τ = 0.25 lbf/in²

Explanation:

given that the oil viscosity, μ =  2.415 lb/ft-s

gap between plates = 1/4 inches = 1/4*12 = 1/48 ft

recall from newtons law of viscosity;

shear stress τ = μ du/dy =

  τ = (2.415 lb/ft-s) (10 ft/s)/(1/48) ft

τ = 1159.2 lb/ft-s²

we know that, 1 slug = 32.174 lb

lb = 1/32.174 slug

∴  τ = 1159.2/32.174 slug/ft-s² = 36 slug/ft-s²

τ = 36 slug/ft-s²

multiply both the numerator and denominator by ft, this gives

τ = 36 slug-ft/ft²-s²

τ = 36 lbf/ft² where 1 slug-ft/s² = 1lbf

since 1 ft = 12 inch = 1 ft² = 12² in² = 144 in²

∴  τ = 36/144 lbf /in² = 0.25 lbf/in²

τ = 0.25 lbf/in²

6 0
3 years ago
How meny people live in new york
nydimaria [60]

Answer:

8.419 million (2019) at 2019

4 0
3 years ago
I should over-tighten bolts so that my creations do not fall apart.​
Katarina [22]
Is this even a question ?
5 0
3 years ago
A three-phase line has a impedance of 0.4+j2.7 per phase. The line feeds 2 balanced three-phase loads that are connected in para
mamaluj [8]

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c)  54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

V_p=\frac{3810.5}{\sqrt{3} }=2200 V

The complex power S is given as:

S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA

where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA

The line current I is given as:

I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)}  =100\angle -36.87^o\ A

The phase voltage at the sending end is:

V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV

The magnitude of the line voltage at the source end of the line (V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V

b) The Total real and reactive power loss in the line is:

S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

5 0
4 years ago
What is the velocity of flow in an asphalt channel that has a hydraulic radius of 3.404 m, length of 200 m and bed slope of 0.00
yKpoI14uk [10]

Answer:

The velocity of flow is 10.0 m/s.

Explanation:

We shall use Manning's equation to calculate the velocity of flow

Velocity of flow by manning's equation is given by

V=\frac{1}{n}R^{2/3}S^{1/2}

where

n = manning's roughness coefficient

R = hydraulic radius

S = bed slope of the channel

We know that for an asphalt channel value of manning's roughness coefficient = 0.016

Applying values in the above equation we obtain velocity of flow as

V=\frac{1}{0.016}\times 3.404^{2/3}\times 0.005^{1/2}\\\\\therefore V=10.000m/s

7 0
3 years ago
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