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3 years ago

What happens to the electrolyte, during discharging?

Engineering

1 answer:

lisov135 [29]3 years ago

3 0

During the discharge ions combine with anode to form a compound and release one or more electron.

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The speed of sound is 1150 ft/s convert to mile/h

Sonbull [250]

Answer:

784.090909mph

Explanation:

1ft/s=0.681818 mph

1150ft/s=0.681818 x 1150=784.090909

6 0

3 years ago

Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage cha

alexira [117]

Answer:

Δr=20.45 %

Explanation:

Given that

Rake angle α = 15°

coefficient of friction ,μ = 0.15

The friction angle β

tanβ = μ

tanβ = 0.15

β=8.83°

2φ + β - α = 90°

φ=Shear angle

2φ + 8.833° - 15° = 90°

φ = 48.08°

Chip thickness r given as

$r=\dfrac{tan\phi}{cos\alpha +sin\alpha\ tan\phi}$

$r=\dfrac{tan48.08^{\circ}}{cos15^{\circ} +sin15^{\circ}\ tan48.08^{\circ}}$

r=0.88

New coefficient of friction ,μ' = 0.3

tanβ' = μ'

tanβ' = 0.3

β'=16.69°

2φ' + β' - α = 90°

φ'=Shear angle

2φ' + 16.69° - 25° = 90°

φ' = 49.15°

Chip thickness r' given as

$r'=\dfrac{tan\phi'}{cos\alpha +sin\alpha\ tan\phi'}$

$r'=\dfrac{tan44.15^{\circ}}{cos49.15^{\circ} +sin49.15^{\circ}\ tan44.15^{\circ}}$

r'=0.70

Percentage change

$\Delta r=\dfrac{r-r'}{r}\times 100$

$\Delta r=\dfrac{0.88-0.70}{0.88}\times 100$

Δr=20.45 %

8 0

3 years ago

Technician A says that weld-through primer can be removed from the immediate weld area to improve weld quality. Technician B say

lisabon 2012 [21]

Answer:

83737373777473737373738388383838

4 0

3 years ago

What is the mass in both slugs and kilograms of a 1000-lb beam?

Likurg_2 [28]

Answer:

31.1 slug, 453.4 Kg

Explanation:

given,

mass of the beam is 1000 lb

to convert mass of beam into slugs and kilograms.

1 lb is equal to 0.0311 slug

1000 lb = 1000 × 0.0311

= 31.1 slug

now, conversion of lb into kg

1 lb is equal to 0.4534 kg

so,

1000 lb = 1000 × 0.4534

= 453.4 Kg

hence, 1000 lb of beam in slugs is equal to 31.1 slugs and in kilo gram is 453.4 Kg.

7 0

3 years ago

The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me

masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

$\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0$ Equation 1

Similarly, the equation for outer node “o” will be

$\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0$ Equation 2

The conventive thermal resistance in i-node will be

$R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w$ Equation 3

The conventive hermal resistance per unit area is

$R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w$ Equation 4

The conductive thermal resistance per unit area is

$R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w$ Equation 5

Since $q_{rad}$ is given as 100, $T_{o}$ is 40 $T_ \infty$ is 300 $T_{\infty, o}$ is 25

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

$\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0$ Equation 6

$\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0$

$T_{i}-40= \frac {L}{0.05}*150$

$T_{i}-40=3000L$

$T_{i}=3000L+40$ Equation 7

From equation 6 we can substitute wherever there’s $T_{i}$ with 3000L+40 as seen in equation 7 hence we obtain

$\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0$

The above can be simplified to be

$\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0$

$\frac {260-3000L}{0.033}=50$

-3000L=1.665-260

$L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm$

Therefore, insulation thickness is 86mm

8 0

3 years ago