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Setler [38]
3 years ago
15

IDENTIFY THE POINTER READING FOR EACH SCALE. COMPUTE THE READING USING THE ASSIGNED RANGE MULTIPLIER FOR OHMMETER SCALE.

Engineering
2 answers:
MrMuchimi3 years ago
8 0

about that one yh i dont know

daser333 [38]3 years ago
3 0

Answer:

https://gltnhs-tle.weebly.com › lesso...

Web results

Lesson 2 - T.L.E Learning Module

You might be interested in
The collapse of the magnetic field inside the ignition coil happens as a
Tanzania [10]
Primary coil


See an example below

3 0
3 years ago
A distillation column at 101 kPa is used to separate 350 kmol/h of a bubble point mixture of toluene and benzene into an overhea
AURORKA [14]

Answer:

A)

  D = 158.42 kmol/h

  B =  191.578 kmol/h

B) Rmin = 1.3095

Explanation:

<u>a) Determine the distillate and bottoms flow rates ( D and B ) </u>

F = D + B ----- ( 1 )

<em>Given data :</em>

F = 350 kmol/j

Xf = 0.45 mole

yD ( distillate comp ) = 0.97

yB ( bottom comp ) = 0.02

back to equation 1

350(0.45) = 0.97 D + 0.02 B  ----- ( 2 )

where; B = F - D

Equation 2 becomes

350( 0.45 ) = 0.97 D + 0.02 ( 350 - D )  ------ 3

solving equation 3

D = 158.42 kmol/h

resolving equation 2

B = 191.578 kmol/h

<u>B) Determine the minimum reflux ratio Rmin</u>

The minimum reflux ratio occurs when the enriching line meets the q line in the VLE curve

first we calculate the value of the enriching line

Y =( Rm / R + 1 m ) x   +  ( 0.97 / Rm + 1 )

q - line ;  y =  ( 9 / 9-1 ) x -  xf/9-1

therefore ; x = 0.45

Finally Rmin

=  (( 0.97 / (Rm + 1 ))  = 0.42

0.42 ( Rm + 1 ) = 0.97

∴ Rmin = 1.3095

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8 0
3 years ago
Somebody help me!! It’s due today
Bas_tet [7]

Answer:

1.0.7

2.6.0

3.2.7

4.1.8

5.4.2

6.3.7

5 0
3 years ago
The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
andriy [413]

Answer:

Uper Bound = 175.5 GPa, Lower Bound = 85.26 GPa

Explanation:

Rule of mixtures equations:

For a two-phase composite, modulus of elasticity upper-bound expression is,

E(c)(u) = E(m) x V(m) + E(p) x V(p)

Here, E(m) is the modulus of elasticity of matrix, E(p) is the modulus of elasticity of patriciate phase, E(c) is the modulus of elasticity of composite, V(m) is the volume fraction of matrix and V(p) is the volume fraction of composite.

For a two-phase composite, modulus of elasticity lower-bound expression is,

E(c)(l) = E(m) x E(p)/V(m) x E(p)+V(p) x E(m)

<u>Consider the expression of rule of mixtures for upper-bound and calculate the modulus of elasticity upper-bound.</u>

E(c)(u) = E(m) x V(m) + E(p) x V(p), (1)

<u>Calculate the volume fraction of matrix.</u>

V(m) + V(p) = 1

<u>Substitute 0.35 for V(p).</u>

V(m) + 0.35 = 1

V(m) = 0.65

From equation (1);

<u>Substitute 60 GPa for E(m), 390GPa for E(p), 0.65 for V(m) and 0.35 for V(p).</u>

E(c)(u )= E(m) x V(m) + E(p) x V(p)

E(c)(u) = (60 × 0.65) + (390 × 0.35)

E(c)(u) = 175.5 GPa

The modulus of elasticity upper-bound is 175.5GPa.

The modulus of elasticity of upper-bound can be calculated using the rule of mixtures expression. Since the sum of volume fraction of matrix and volume fraction of composite is equal to one V(m) + V(p) = 1. Substitute the value of volume fraction of matrix as 0.69 and obtain the volume fraction of matrix.

<u>Consider the expression of rule of mixtures for lower-bound and calculate the modulus of elasticity upper-bound.</u>

E(c)(l) = (E(m) x E(p))/ (V(m) x E(p) + V(p) x E(m))

<u>Substitute 60 GPa for E(m), 390GPa for E(p), 0.65 for V(m) and 0.35 for V(p).</u>

E(c)(l) = 60 × 390/(0.65 × 390) +(0.35 × 60)

E(c)(l) = 23400/274.5

E(c)(l) = 85.26 GPa

The modulus of elasticity lower-bound is 85.26 GPa.

8 0
4 years ago
A 0.5m diameter sphere containing pollution monitoring equipment is dragged through the Charles River at a relative velocity of
Dmitriy789 [7]

Answer:

\phi = 155.57

Explanation:

from figure

taking summation of force in x direction be zero

\sum x = 0

F_D = Tsin \theta  .....1

\frac{c_d \rho v^2 A}{2} =Tsin \theta

taking summation of force in Y direction be zero

F_B - W-  Tcos \theta

T = \frac{F_B -W}{cos \theta} .........2

putting T value in equation 1

F_D - \frac{F_B -W}{cos \theta} sin\theta

F_D = \rho g V ( 1 -Sg) tan \theta.........3

F_D = \rho g [\frac{\pi d^3}{6}] ( 1 -Sg) tan \theta

tan \theta = \frac{6 c_D \rho v&2 A}{ 2 \rho g V \pi D^3 (1- Sg)}

Water at 10 degree C  has kinetic viscosity v = 1.3 \times 10^{-6} m^2/s

Reynold number

Re = \frac{ VD}{\nu} = \frac{10\times 0.5}{1.3 \times 10^{-6}} = 3.84 \times 10^6

so for Re =3.84 \times 10^6  cd is 0.072

tan \theta = \frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}

\theta = tan^{-1} [\frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}]

\theta = - 65.57 degree

\phi = 90 - (-65.57) = 1557.57 degree

8 0
3 years ago
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