Is cross flow more efficient or counter flow

An inductor is energized as in the circuit of Fig. 2-4a. The circuit has L 10 mH and VCC 14 V. (a) Determine the required on tim

Savatey [412]

Answer:

A) 11.1 ms

B) 5.62 Ω

Explanation:

L ( inductance ) = 10 mH

Vcc = 14V

<u>A) determine the required on time of the switch such that the peak energy stored in the inductor is 1.2J </u>

first calculate for the current ( i ) using the equation for energy stored in an inductor hence

i = $\sqrt{\frac{2W}{L} }$ ----- ( 1 )

where : W = 1.2j , L = 10 mH

Input values into equation 1

i = 15.49 A

Now determine the time required with expression below

i( t ) = 15.49 A

L = 10 mH, Vcc = 14

hence the time required ( T-on ) = 11.1 ms

attached below is detailed solution

B) <u>select the value of R such that switching cycle can be repeated every 20 ms </u>

using the expression below

τ = $\frac{L}{R}$ ---- ( 2 )

but first we will determine the value of τ

τ = t-off / 5 time constants

= (20 - 11.1 ) / 5 = 1.78 ms

Back to equation 2

R = L / τ

= (10 * 10^-3) / (1.78 * 10^-3)

= 5.62 Ω

3 0

3 years ago

According to the video, what are some decisions that Architects make? Check all that apply. how much money a building will be wo

Rudiy27

Answer: what traffic patterns people will follow,

how people will feel when they enter or leave a building,

where to put doors, walls, and windows,

what building materials to use.

Explanation:

so the answers ar 2,3,5, and 6

6 0

3 years ago

Read 2 more answers

An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression wh

Kamila [148]

Answer:

hello your question lacks some information attached is the complete question

A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in

(ii) maximum bending stress in compression = 0.7413*10^6 Ib-in

B) (i) The average shear stress at the neutral axis = 0.7904 *10 ^5 psi

(ii) Average shear stress at the web = 18.289 * 10^5 psi

(iii) Average shear stress at the Flange = 1.143 *10^5 psi

Explanation:

First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)

A) Calculate the maximum bending stress in tension and compression

lintel load = 10000 Ib

simple span = 6 ft

( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS

I = 53.54

i) The maximum bending stress (fb) in tension=

= $\frac{M_{mm}Y }{I}$ = $\frac{6.48 * 10^6 * 2.375}{53.54}$ = 0.287 * 10^6 Ib-in

ii) The maximum bending stress (fb) in compression

= $\frac{M_{mm}Y }{I}$ = $\frac{6.48 *10^6*(8.5-2.375)}{53.54}$ = 0.7413*10^6 Ib-in

B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange

i) The average shear stress at the neutral axis

V = $\frac{wL}{2}$ = $\frac{1000*6*12}{2}$ = 3.6*10^5 Ib

Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - $\frac{0.5}{2}$ ) * $\frac{(2.375 - (\frac{0.5}{2} ))}{2}$

= 5.878 in^3

t = VQ / Ib = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi

ii) Average shear stress at the web ( value gotten from the shear stress at the flange )

t = 1.143 * 10^5 * (8 / 0.5 ) psi

= 18.289 * 10^5 psi

iii) Average shear stress at the Flange

t = VQ / Ib = $\frac{3.6*10^5 * 8*0.5*(2.375*(0.5/2))}{53.54 *0.5}$

= 1.143 *10^5

4 0

4 years ago

A 3-oz serving of roasted, skinless chicken breast contains 140 Cal, 27 g of protein, 3 g of fat, 13 mg of calcium, and 64 mg of

Art [367]

Answer:

Explanation:

a) Given C [ cal pro fat calc sod]

[140 27 3 13 64]

P = [cal pro fat calc sod]

[180 4 11 24 662]

B = [cal pro fat calc sod]

[50 5 1 82 20]

To find C+2P+3B = [140 27 3 13 64] + 2[180 4 11 24 662] + 3[50 5 1 82 20]

= [650 54 28 307 1448]

The entries represent skinless chicken breast , One-half cup of potato salad and One broccoli spear.

7 0

3 years ago

A trapezoidal section has a 5.0-ft bed width, 2.5-ft depth, and 1:1 side slope. Evaluate its geometric elements (Area, water dep

aleksley [76]

Answer:

a) 18.75 ft^2

b) 2.5 ft

c) 12.07 ft

d) 10 ft

e) 1.553 ft

f) 1.875 ft

Explanation:

Given data :

5.0-ft bed width, ( b )

2.5-ft depth ( y )

1 : 1 side slope

Evaluate

a) Area of trapezoidal section

A = by + my^2

we assume m = 1

A = [5 + (1 * 2.5 ) ] *2.5

= ( 7.5 ) 2.5 = 18.75 ft^2

b) Calculate water depth

water depth = 2.5 ft

c) Calculate wetted perimeter

P = b + 2y √ 1 + m^2

= 5 + (2.5*2) √ 1 + 1 ^2 = 12.07 ft

d) calculate top width

T = b + 2my

= 5 + 2 ( 1 * 2.5 ) = 10 ft

e) calculate hydraulic radius

R = A / P = 18.75 / 12.07

= 1.553 ft

f) calculate hydraulic depth

D = A / T = 18.75 / 10 = 1.875 ft

4 0

3 years ago