A permanent-magnet dc motor has the following parameters: Ra = 0.3 Ω and kE = kT = 0.5 in MKS units. For a torque of up to 10 Nm, plot its steady state torque-speed characteristics for the following values of Va: 100 V, 75 V, and 50 V.

Answer:

Average heat transfer coefficient =  31 kw/m^2 k

Heat transfer rate per meter length of pipe =  116.808 KW

Explanation:

water temperature = 20⁰c,  

free-stream velocity = 1.5 m/s

circular pipe diameter = 2.0 cm = 0.02 m

surface temperature = 80⁰c

A) calculate average heat transfer coefficient

we apply the formula below :

m = αAv

A (area) = [tex]\pi /4 (d)^2[/tex]

m = 10^3 * [tex]\pi / 4 ( 0.02)^2[/tex] * 1.5

   = 10^3 * 0.7857( 0.0004) * 1.5

   = 0.4714 kg/s

Average heat transfer coefficient  

h = [tex]\frac{m(cp)}{A}[/tex]  ,  A = [tex]\pi DL[/tex]

L = 1 m , m = 0.4714 kgs , cp = 4.18

back to equation

h = [tex]\frac{0.4714*4.18}{\pi * 0.02 }[/tex]   = 1.970 / 0.0628 = 31.369 ≈ 31 kw/m^2 k

B) Heat transfer rate per meter length of pipe

Q = ha( ΔT ),  a = [tex]\pi DL[/tex]

   = 31 * 0.0628 * ( 80 - 20 )

  = 31 * 0.0628 * 60 = 116.808 KW

Answer: A prototype

Explanation:

The scale model that proves the initial concept is called a domain model.

A copy or depiction of something where all parts have the same dimensions as the original. A scale model is an image or copy of an object that is either larger or smaller than the object being represented's actual size.

A domain model is a type of conceptual model that is used to depict the structural elements and conceptual constraints within a domain of interest.

A domain model will include all of the entities, their attributes, and relationships, as well as the constraints that govern the conceptual integrity of the structural model elements that comprise that problem domain.

Therefore, a domain model is the scale model that proves the initial concept.

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Answer:

Both Technician A and technician B are correct.

Explanation: A transistor must have a P-N junction as that is where the positive and negative charges are connected.

A transistor also can be described as a semiconductor which acts as a switch and can be used to amplify currents. Transistors are very key and vital to electronic devices especially the mobile phones in recent times, it helps to ensure that electronic systems perform optimally.

The charges in the P-N junction is controlled by the availability of Positive and negative electrons.

Answer:

A)  ( N ) = 1.54

B)  N ( Goodman ) = 1.133,  N ( Morrow) = 1.35

Explanation:

width of steel bar = 25-mm

thickness of steel bar = 10-mm

diameter = 6-mm

load on plate = between 12 kN AND 28 kN

notch sensitivity = 0.83

A ) Fatigue factor of safety based on yielding criteria

= δa + δm = [tex]\frac{Syt}{n}[/tex]   =  91.03 + 227.58 = 490 / N

therefore Fatigue number of safety ( N ) = 1.54

δa (amplitude stress ) = kf ( Fa/A) = 2.162 * ( 8*10^3 / 190 ) = 91.03 MPa

A = area of steel bar = 190 mm^2 , Fa = amplitude load = 8 KN , kf = 2.162

δm (mean stress ) = kf ( Fm/A ) = (2.162 * 20*10^3 )/ 190 = 227.58 MPa

Fm = mean load  = 20 *10^3

B) Fatigue factor of safety based on Goodman and Morrow criteria

δa / Se + δm / Sut = 1 / N

= 91.03 / 183.15 + 227.58 / 590 = 1 /N

Hence N = 1.133 ( based on Goodman criteria )

note : Se = endurance limit (calculated) = 183.15 , Sut = 590

applying Morrow criteria

N =   1 / ( δa/Se) + (δm/ δf )

   = 1 / ( 91.03 / 183.15 ) + (227.58 / 935 )  

   = 1.35

Answer:

a. 115

b. 135

c. 220

d. 185

Explanation:

Spheriodite is microscopic constituents in some steels which is composed of spherically shaped cementide particle. It is most ductile and softest type of steel. Pearlite is two phased lamellar compose of alternating layer of ferrite and cementite. It is hard and strong but not tough. It is applied on cutting tools like chopper, blades and knives.

Answer:

minimum current level required =  8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = [tex]\frac{\pi }{4} * D^2 * l[/tex] = [tex]\frac{\pi }{4} * 5.5^2 * 5[/tex] = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 )  = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED  use the relation below

electrical energy =  I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore  8975.95 = I ( current )

Answer:

The Remaining Bulbs will either burn out( draw more current ) or Not burn out depending on the arrangement of the bulbs in the circuit

Explanation:

The Remaining Bulbs will either burn more brightly ( draw more current ) or Burn less brightly ( draw less current).and this depends on the arrangement of the light bulbs in the various circuit.

If the light bulbs are connected in series the remaining bulbs will burn out as soon as  any light bulb burns out and this is because bulbs connected in series receive the same amount of current ,

If the light bulbs are connected in parallel the remaining bulbs will not burn out because each bulb receives current based on its resistance.

Answer:

Explanation:

The only vertical force that can be applied at joint D is that of link CD. Since joint D is stationary, there must be no vertical force. Hence the force in link CD must be zero, as must the force in link DE.

At joint E, the only horizontal force is that applied by link EF, so it, too, must be zero.

Then link CE has 15 N tension.

The downward force in CE must be balanced by an upward force in CF. Of that force, only 1/√2 of it will be vertical, so the force in CF is a compression of 15√2 N.

In order for the horizontal forces at C to be balanced the 15 N horizontal compression in CF must be balanced by a 15 N tension in BC.

At joint F, the 15 N horizontal compression in CF must be balanced by a 15 N compression in FA. CF contributes a downward force of 15 N at joint F. Together with the external load of 10 N, the total downward force at F is 25 N. Then the tension in BF must be 25 N to balance that.

At joint B, the 25 N downward vertical force in BF must be balanced by the vertical component of the compressive force in AB. That component is 2/√5 of the total force in AB, which must be a compression of 25√5/2 N.

The horizontal forces at joint B include the 15 N tension in BC and the 25/2 N compression in AB. These are balanced by a (25/2+15) N = 55/2 N tension in BG.

In summary, the link forces are ...

_____

Note that the forces at the pins of G and A are in accordance with those that give a net torque about those point of 0, serving as a check on the above calculations.

Answer:

Clay bricks are manufactured by mining and clay moulded blocks. There are 20,000 brick klins in Pakistan.

Explanation:

In Pakistan, the clay bricks are manufactured by mining and baking the clay moulded blocks in brick kilns. According to an estimate, the baking process emits about  1.4 pounds of carbon per brick made, but in Pakistan, because the systems are outdated, brick kilns are used, which is producing more than the average amount of the pollution.

There are about 20,000 brick kilns in Pakistan. The traditional brick production in Pakistan is consists of hand-made bricks which are first baked in Fixed Chimney Bull's Trench Kilns (FCBTK), this is the most widely used brick firing technology in South Asia.

Answer:

C) a winding connected in series with the armature.

Explanation:

In a series motor, an electromagnet is used as a stator to generate its magnetic field. The field coil of this stator are connected through a commutator in series with the rotor windings. This stator which is the armature windings will conduct AC even on a DC machine, due to the  periodically reverses current direction (commutation) or due to electronic commutation (as in brushless DC motors).

Answer:

A) Flow depth = 2.46 m, Froude number after jump = 0.464

B) head loss = 0.572 m

C) dissipation ratio = 0.173

Explanation:

Given data :

Velocity before jump ( v1 ) = 7 m/s

flow depth before jump ( y1 ) = 0.8 m

g = 9.81 m/s

Esi = 3.3 m ( calculated )

attached below is a detailed solution of the problem

Answer:

the torque required to RAISE the load is Tr = 18.09 Nm

the torque required to LOWER the load is Tl = 10.069 ≈ 10.07 Nm

the Overall Efficiency e = 0.2199 ≈ 0.22

Explanation:

Given that; F = 5 kN, p = 5mm, d = 35mm

Dm = d - p/2

Dm = 35 - ( 5/2) = 35 - 2.5

DM = 32.5mm

So the torque required to RAISE the load is

Tr = ( 5 × 32.5)/2 [(5 + (π × 0.09 × 32.5)) / ( (π × 32.5) - ( 0.09 × 5))] + [( 5 × 0.06 × 45)/2]

Tr = 81.25 × (14.1892 / 101.6518) + 6.75

Tr = 11.3414 + 6.75

Tr = 18.09 Nm

the torque required to LOWER the load is

Tl =  ( 5 × 32.5)/2 [(π × 0.09 × 32.5) - 5) / ( (π × 32.5) + ( 0.09 × 5))] + [( 5 × 0.06 × 45)/2]

Tl = 81.25 × 4.1892 / 102.5518 + 6.75

Tl = 3.3190 + 6.75

Tl = 10.069 ≈ 10.07 Nm

So since torque required to LOWER the load is positive

that is, the thread is self locking

Therefore the efficiency is

e = ( 5 × 5 ) / ( 2π × 18.09 )

e = 25 / 113.6628

e = 0.2199 ≈ 0.22

Answer:

  resistance

Explanation:

A strain gauge changes resistance with applied strain.

Answer:

There are five (5) unknown support reactions in this problem.

Explanation:

A roller joint rotates and translates along the surface on which the roller rests. The resulting reaction force is always a single force that is perpendicular to, and away from, the surface. This allows the roller to move in a single plane along the surface where it rests.

A cable support provides support in one direction, parallel, and in opposite direction to the load on it. There exists a single reaction from the cable pointed upwards.

A ball-and-socket joint have  reaction forces in all 3 cardinal  directions. This allows it to move in the x-y-z plane.

The total unknown reactions on the member are five in number.

Answer:

a. myMethod() has access to and can use myField.

Explanation:

Logic programming is a kind of programming which is largely based on formal logic.  The statement are written in logical forms which express rules about the domain. In the given scenario the my method will have access to my field which is private field. My method non static public field can also use my field class.

Answer:

30 feet

Explanation:

Given data :

design speed = 30 miles/h

super elevation = 0.08

determine the width of the turning roadway

calculate the value of R = V^2 / 15( e + p)

e = 0.08 , p = 0.2 , v = 30

R = (30)^2 / 15 ( 0.08 + 0.2 )

  = 900 / 15 ( 0.28 )

  ≈ 215 ft

pavement width from the calculation above = 28 ft

width of the turning roadway = pavement width + 2 = 30 feet ( because there are two vertical widths joining up the main road at the T junction )

Answer:

It's easier to buy one, but you can search for a tutorial on how to make one on Yuotube.

There are quite a few videos on how to make one.

Answer:

D.  Resistance = 30 ohms

Explanation:

Using Ohm's law

V = I times R

Given:

V = 60 V

I = 2 A

Resistance = V / I = 60 V / 20 A

Resistance = 30 ohms

Answer:

Electrical conductivity or specific conductance is the reciprocal of electrical resistivity. It represents a material's ability to conduct electric current. It is commonly signified by the Greek letter σ (sigma), but κ (kappa) (especially in electrical engineering) and γ (gamma) are sometimes used.

Answer:

thread depth = 2.5 mm

thread width = 2.5 mm

mean diameter = 27.5 mm

root diameter = 25 mm

lead of screw = 5 mm

Explanation:

given data

power screw diameter D = 30 mm

thread pitch  P = 5 mm

solution

First, we get here thread depth fr square thread

thread depth = [tex]\frac{P}{2}[/tex]   ......................1

thread depth = [tex]\frac{5}{2}[/tex]

thread depth = 2.5 mm

and

thread width for square thread

thread width = [tex]\frac{P}{2}[/tex]   ......................2

thread width = [tex]\frac{5}{2}[/tex]

thread width = 2.5 mm

and

mean diameter is

mean diameter = D - [tex]\frac{P}{2}[/tex]    ................3

mean diameter = 30 - [tex]\frac{5}{2}[/tex]

mean diameter = 27.5 mm

and

root diameter is

root diameter = D - P   ....................4

root diameter = 30 - 5

root diameter = 25 mm

and

lead of screw for single thread so n = 1

so lead of screw = 1 × 5

lead of screw = 5 mm

Answer:

As the asteroid falls closer to the Earth's surface its Gravitational Potential energy decreases and its Kinetic energy increases.

Answer:

C) Decreases with an increase in temperature

Explanation:

As the temperature of a transistor increases, the thermal runaway property of the transistor becomes more significant and the transistors, conducting more freely as a result of the rise in temperature, causes an increase in the collector current or leakage current. The transistor base-emitter voltage decreases as a result.

With increased heating due to heavy current flow, the transistor is damaged.

Answer:

moment of inertia = 4.662 * 10^6 [tex]mm^4[/tex]

Explanation:

Given data :

Mass of machine = 400 kg = 400 * 9.81 = 3924 N

length of span = 3.2 m

E = 200 * 10^9 N/m^2

frequency = 9.3 Hz

Wm ( angular frequency ) = 2 [tex]\pi f[/tex] = 58.434 rad/secs

also Wm = [tex]\sqrt{\frac{g}{t} }[/tex]  ------- EQUATION 1

g = 9.81

deflection of simply supported beam

t = [tex]\frac{wl^3}{48EI}[/tex]

insert the value of t into equation 1

W[tex]m^2[/tex] = [tex]\frac{g*48*E*I}{WL^3}[/tex]   make I the subject of the equation

I ( Moment of inertia about the neutral axis ) = [tex]\frac{WL^3* Wn^2}{48*g*E}[/tex]

I = [tex]\frac{3924*3.2^3*58.434^2}{48*9.81*200*10^9}[/tex]  = 4.662 * 10^6 [tex]mm^4[/tex]

Answer: hello attached below is the diagram which is part of your question

Total entropy change  = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k  it violates Clausius increase of entropy which is Sgen > 0

Explanation:

Clausius statement states that it is impossible to transfer heat energy from a cooler body to a hotter body in a cycle or region without any other external factors affecting it .  

applying the increase in entropy principle to prove this

temp of cold reservoir (t hot)= 600 k

temp of hot reservoir(t cold) = 1220 k

energy (q) = 100 kj

total entropy change  = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k

entropy change in cold reservoir = Q/t cold = 100 / 600 = -0.166 kj/k

entropy change in hot reservoir = Q / t hot = 100 / 1220 = 0.083 kj/k

hence it violates  Clausius inequality of increase of entropy principle which is states that generated entropy has to be > 0

Answer:

the rate of heat loss from the duct to the attic space = 1315.44 W

the pressure difference between the inlet and outlet sections of the duct  = 7.0045 N/m²

Explanation:

We know that properties of air 80⁰C  and 1atm  (from appendix table) are;

density p = 0.9994 kg/m³, Specifice heat Cp = 1008 J/kg.⁰C

Thermal conductivity k = 0.02953 W/m.⁰C, Prandtl number Pr = 0.7154,

Kinematic viscosity v = 2.097 × 10⁻⁵ m²/s

haven gotten that, we calculate the hydraulic diameter of square duct

Dh = 4Ac / P      { Ac = is cross sectional area of duct and P = perimeter}

now we substitute a² for Ac and 4a for P ( we know from the question that a = 0.2 m)

Dh = 4a² / 4a

Dh = 4(0.2)² / 4(0.2)

Dh = 0.2 m

Now we calculate the average velocity of air

Vₐ = Vˣ / Ac        { vˣ = volume flow rate of air}

Vₐ = Vˣ / a²      { Ac = a² }, we know that a = 0.2m₂, Vˣ = 0.15 m³

Vₐ = 0.15 / (0.2)²

Vₐ = 3.75 m/s

Next we calculate the Reynolds number

Re = Vₐ Dh / V

Re = (3.75 × 0.2) / 2.097× 10⁻⁵

Re = 35765.379

The  Reynolds number IS GREATER than 10,000

so the flow is turbulent and entry length in this case is nearly 10 times the hydraulic diameter

Lh ≈ Lt ≈ 10D

= 10 × 0.2

= 2m

As this length is quite small when compared to the total of tube, we assume fully developed flow for the entire tube length.

Now we calculate the Nusselt number from this relation;

Nu =  0.023 Re⁰'⁸ Pr⁰'³

so we substitute for Re and Pr

Nu = 0.023(35765.379)⁰'⁸ (0.7154)⁰'³

Nu = 91.4

Now calculate the convective heat transfer coefficient

h = Nu × K/ Dh

we substitute

h = 91.4 × 0.02953 W/m.°C / 0.2 m

h = 13.5 W/m².°C

We calculate the surface area of the square duct

Aₓ = 4aL       { L= length of duct}

we substitute

Aₓ = 4 × 0.2 × 8

Aₓ = 6.4 m²

Mass flow rate of air

m = pVˣ

we substitute again ( from our initials)

m = 0.9994 kg/m₃ × 0.15 m³/s

m= 0.150 kg/s

We calculate the exit temperature of the air from the duct

Te = Ts - (Ts -Ti) exp ( - hAₓ / mCp)

we know that

Ts = 60°C , Ti = 80°C, h = 13.5 W/m².°C , Aₓ = 6.4m², m = 0.150 kg/s , Cp = 1008 J/kg.°C

we substitute

Te = 60 - (60-80) exp(- ((13.5 × 6.4)/(0.15 × 1008))

Te = 71.3°

Now we calculate the rate of heat loss from the duct.

Q = mCp ( Ti -Te )

we substitute again

Q = 0.150 × 1008 × ( 80 - 71.3 )

Q = 1315.44 W

Next we calculate the estimated friction factors by using Haaland equation

1/√f = - 1.8log₁₀ [ 6.9/Re + (E/D)/3.7)¹'¹¹]

we know that E/D = relative roughness = 10⁻³

we substitute

so

1/√f = - 1.8log₁₀ [ (6.9/35765.379) + ( 10⁻³/3.7)¹'¹¹]

1/√f = - 1.8log₁₀ { 0.000192924 + 0.00010947}

1/√f = - 1.8log₁₀ 0.000302324  

√f =   1/6.334

f = (1/6.334)²

f = 0.02492

We calculate the pressure difference between inlet and outlet sections of the duct

ΔPl = fLPVa² / Dh × 2

ΔPl = {0.02492 × 8 × 0.9994 × (3.75)²} / 0.2 × 2

ΔPl = 2.8018 / 0.4

ΔPl = 7.0045 N/m²

Therefore pressure deference is 7.0045 N/m²

Answer:

[tex]w_{out}=319.1\frac{BTU}{lbm}[/tex]

Explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

[tex]h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}[/tex]

Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:

[tex]s_2=s_1[/tex]

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

[tex]x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965[/tex]

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

[tex]h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}[/tex]

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

[tex]h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}[/tex]

Best regards.

Answer:

V2= 1666.6 volts

Explanation:

Given data

primary turns N1= 100 turns

secondary turns N2= 1500 turns

primary voltage V1= 110 volts

secondary voltage V2= ?

We can solve for the output voltage using the turns ration sated below

  Turns Ratio = N1 / N2 = V1 / V2

Substituting our given data into the expression we have

100/1500= 110/V2

Making V2 subject of formula we have

V2= 110/(100/1500)=  1666.6 volts

V2= 1666.6 volts

Hence the secondary voltage is 1666.6 volts

Answer:

The answer is below

Explanation:

1) The synchronous speed of an induction motor is the speed of the magnetic field of the stator. It is given by:

[tex]n_s=\frac{120f_s}{p}\\ Where\ p\ is \ the \ number\ of\ machine\ pole, f_s\ is\ the\ supply \ frequency\\and\ n_s\ is \ the \ synchronous\ speed(speed \ of\ stator\ magnetic \ field)\\Given: f_s=60\ Hz, p=4. Therefore\\\\n_s=\frac{120*60}{4}=1800\ rpm[/tex]

2) The speed of the rotor is the motor speed. The slip is given by:

[tex]Slip=\frac{n_s-n_m}{n_s}. \\ n_m\ is\ the \ motor\ speed(rotor\ speed)\\Slip = 0.05, n_s= 1800\ rpm\\ \\0.05=\frac{1800-n_m}{1800}\\\\ 1800-n_m=90\\\\n_m=1800-90=1710\ rpm[/tex]

3) The frequency of the rotor is given as:

[tex]f_r=slip*f_s\\f_r=0.04*60=2.4\ Hz[/tex]

4) At standstill, the speed of the motor is 0, therefore the slip is 1.

The frequency of the rotor is given as:

[tex]f_r=slip*f_s\\f_r=1*60=60\ Hz[/tex]

Answer: a = change in w =0.00656

b = q = 5.1kj/kg

Explanation:

Find explanation in the attached file

The amount of steam added to the air  a = change in w =0.00656 b = q = 5.1kj/kg

The digital game retail and distribution service Steam is provided by Valve. In order to allow Valve to automatically update its games, it was first released as a software client in September 2003. In late 2005, it was expanded to include the distribution and sale of games from other publishers.

a) We can use the absolute humidity we and wg to determine the amount

of moisture added Aw.

Aww3-W2

Aw= 0.01291 -0.00635

Aw= 0.00656

b) To determine the heat transfer q we will need the enthalpies h and h2.

kJ

kg 9 = 36.2

kJ  kg

31.1

q=5.1

kJ

kg

RESULT

Do = 0.00656

kJ

kg

9 = 5.1

Therefore, The amount of steam added to the air  a = change in w =0.00656 b = q = 5.1kj/kg

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