Air at 1.00 atm is passed through a horizontal 1.5-in. Schedule 40 (steam-jacketed) steel pipe at a velocity of 2.10 ft/s and an inlet and outlet temperature of 25.0 °C and 95.5 °C, respectively. The pipe wall temperature is maintained at 121.1 °C. (HINT: You may neglect the viscosity correction in your calculations.) a) Assuming only heat transfer by forced convection, how long must

Answer:

Bourdon Tube gauge

Explanation:

The most popular type of pressure gauge in several countries is the Bourdon pressure tube gauge, that is used to determine medium and high loads. Bourdon tube  will measure pressures ranging between 600 mbar - 4,000 bar. While the inner pressure is greater than the exterior pressure, the tube pushes forward, and vise versa.

Answer:

A = 1.41 m²

Explanation:

The ara can be found out by using Stefan Boltzman Law:

[tex]P = \sigma AT^4[/tex]

where,

A = Area = ?

P = Radiation Power = 1200 W

σ = Stefan Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴

T = surface temperature = 350 k

Therefore:

[tex]1200\ W = (5.6703\ x\ 10^{-8}W/m^2.k^4)(A)(350\ k)^4\\\\A = \frac{1200\ W}{850.9\ W/m^2}[/tex]

A = 1.41 m²

Answer:fall arrest harness

Explanation:cuz it’s just right

Answer:

B is correct

Explanation:

Answer:

attached below

Explanation:

Applying the state transition Formula

state - next = ( state *2 + in ) % 5

how this works : remainder of previous cycle is doubled to enable the calculation of the new remainder.  

Input of current cycle is represented as either 1 or 0

since the dividing number = 5 . possible remainders = 1,2,3,4,0

each remainder is represented as :

S0 = zero remainder , S1 = 1 remainder , S2 = 2 remainder, S3 = 3 remainder,

S4 = 4 remainder

Answer:

  entering drivers

Explanation:

Drivers entering a traffic circle must yield the right-of-way to those already in the traffic circle. (Among other things, this rule helps avoid gridlock.)

Answer:

Answer:

  12.44Ω

Explanation:

For a question such as this, the assumption must be that the resistance is inversely proportional to the square of the diameter. The resistance will be ...

  20Ω × (1.262/1.6)² ≈ 12.44Ω

Answer:

[tex]ohms[/tex]

Explanation:

hope it helps

Answer:

i) 25.04 W/m^2 .k

ii) 23.82 minutes = 1429.2 secs

Explanation:

Given data:

Diameter of steel ball = 15 cm

uniform temperature = 350°C

p = 8055 kg/m^3

Cp = 480 J/kg.k

surface temp of ball drops to  250°C

average surface temperature = ( 350 + 250 ) / 2 = 300°C

i) Determine the average convection heat transfer coefficient during the cooling process

Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table  " properties of air "  contained in your textbook

average convection heat transfer coefficient = 25.04 W/m^2 .k

ii) Determine how long this process has taken

Time taken by the process = 23.82 minutes = 1429.2 seconds

Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs

attached below is the detailed solution of the given question

Answer:

Ans: To see if the results are similar

Answer:

- the force of the rope on the tugboat is 14.87 kN

- Force acting on the tugboat is 24.79 kN

Explanation:

 Given the data in the question;

tugboat increases its speed uniformly to 25 km/h

v₁ = 25 km/h = (25 × 1000) / ( 1 × 60min × 60sec )

= 25000m / 3600s

=  6.94 m/s

Now, lets determine the force on the rope using the following relation;

T = [tex]m_b[/tex]v₁ / t₁

[tex]m_b[/tex] is mass of barge( 75 Megagram = 75 × 10³ Kilogram  ), time t₁ is 35 s and v₁ is 6.94 m/s

so we substitute

T = [(75 × 10³) × 6.94 ]  / 35

T = 520500 / 35

T = 14871.43 N

T = 14871.43 / 1000

T = 14.87 kN

Therefore, the force of the rope on the tugboat is 14.87 kN

Now, to determine F acting on the tugboat;

[tex]Ft_1[/tex] = ( [tex]m_t[/tex] + [tex]m_b[/tex] )[tex]v_1[/tex]

we solve for F

F = ( [tex]m_t[/tex] + [tex]m_b[/tex] )[tex]v_1[/tex] / [tex]t_1[/tex]

where [tex]m_t[/tex] is mass of tugboat (50 Megagram = 50 × 10³ Kilogram  )

so we substitute

F = [( (50 × 10³) + (75 × 10³) )6.94] / 35

F = [ 125000 × 6.94 ) / 35

F = 867500 / 35

F = 24785.7 N

F = 24785.7 / 1000

F = 24.79 kN

Therefore, Force acting on the tugboat is 24.79 kN

Answer:

I’m sorry I don’t understand this is there more steps?‍♀️

Explanation:

Answer:

skate-themed video games

library research

interviews with skaters

documentaries on skater culture

(I got a 90% on the test so I dont  know if this was the one that gave me the 90% but it said it was right)

Answer:

Product Teardown 28 pieces (1) Plastic packaging: protect and display product for purchase. (4) Exterior screws: hold case halves together. (1) Right case half: acts as part of a handle and contains the rest of the parts. (1) Left case half: acts as part of a handle and contains the rest of the parts.

Explanation:

A product teardown process is an orderly way to know about a particular product and identify its parts, system functionality to recognize modeling improvement and identify cost reduction opportunities. Unlike the traditional costing method, tear down analysis collects information to determine product quality and price desired by the consumers.

Answer:

?

Explanation:

Answer:

3

Explanation:

it is compulsory to have a bachelor's degree

Answer: True

Explanation:

Answer:

a) the output voltage is 50 V

b)

- the average inductor current is 10 A

- the maximum inductor current is 13 A

- the maximum inductor current is 7 A

c) the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components is 4 A

Explanation:

Given the data in the question;

a) the output voltage

V₀ = V[tex]_s[/tex]/( 1 - D )

given that; V[tex]_s[/tex] = 20 V, D = 0.6

we substitute

V₀ = 20 / ( 1 - 0.6 )

V₀ = 20 / 0.4

V₀ = 50 V

Therefore, the output voltage is 50 V

b)

- the average inductor current

[tex]I_L[/tex] = V[tex]_s[/tex] / ( 1 - D )²R

given that R = 12.5 Ω, V[tex]_s[/tex] = 20 V, D = 0.6

we substitute

[tex]I_L[/tex] = 20 / (( 1 - 0.6 )² × 12.5)

[tex]I_L[/tex] = 20 / (( 0.4)² × 12.5)

[tex]I_L[/tex] = 20 / ( 0.16 × 12.5 )

[tex]I_L[/tex] = 20 / 2

[tex]I_L[/tex] = 10 A

Therefore, the average inductor current is 10 A

- the maximum inductor current

[tex]I_{Lmax[/tex] = [V[tex]_s[/tex] / ( 1 - D )²R] + [ V

given that, R = 12.5 Ω, V[tex]_s[/tex] = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

[tex]I_{Lmax[/tex] = [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]

[tex]I_{Lmax[/tex] = [20 / 2 ] + [ 60 / 20 ]    

[tex]I_{Lmax[/tex] = 10 + 3

[tex]I_{Lmax[/tex] = 13 A

Therefore, the maximum inductor current is 13 A

- The minimum inductor current

[tex]I_{Lmax[/tex] = [V[tex]_s[/tex] / ( 1 - D )²R] - [ V

given that, R = 12.5 Ω, V[tex]_s[/tex] = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz

we substitute

[tex]I_{Lmin[/tex] = [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]

[tex]I_{Lmin[/tex] = [20 / 2 ] -[ 60 / 20 ]    

[tex]I_{Lmin[/tex] = 10 - 3

[tex]I_{Lmin[/tex]  = 7 A

Therefore, the maximum inductor current is 7 A

c)  the output voltage ripple

ΔV₀/V₀ = D/RCf

given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz

we substitute

ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )

ΔV₀/V₀ = 0.6 / 100

ΔV₀/V₀ = 0.006 or 0.6%V₀

Therefore, the output voltage ripple is 0.006 or 0.6%V₀

d) the average current in the diode under ideal components;

under ideal components; diode current = output current

hence the diode current will be;

[tex]I_D[/tex] = V₀/R

as V₀ = 50 V and R = 12.5 Ω

we substitute

[tex]I_D[/tex] = 50 / 12.5

[tex]I_D[/tex] = 4 A

Therefore, the average current in the diode under ideal components is 4 A

Answer:

A) Not possible, B) Posible, C) Possible, D) Not possible.

Explanation:

The maximum theoretical efficiency for any thermal engine is defined by Carnot's cycle, whose energy efficiency ([tex]\eta[/tex]), no unit, is expressed below:

[tex]\eta = 1-\frac{T_{L}}{T_{H}}[/tex] (1)

Where:

[tex]T_{L}[/tex] - Cold reservoir temperature, in Kelvin.

[tex]T_{H}[/tex] - Hot reservoir temperature, in Kelvin.

If we know that [tex]T_{L} = 250\,K[/tex] and [tex]T_{H} = 750\,K[/tex], then the maximum theoretical efficiency for the thermal engine is:

[tex]\eta = 1-\frac{T_{L}}{T_{H}}[/tex]

[tex]\eta = 0.667[/tex]

For real thermal engines, the following inequation is observed:

[tex]0 \le \eta_{r} \le \eta[/tex] (2)

Where [tex]\eta_{r}[/tex] is the efficiency of the real heat engine, no unit.

There are two possible criteria to determine if a given heat engine is real:

Efficiency

[tex]\eta_{r} = 1 - \frac{Q_{L}}{Q_{H}}[/tex] (3)

Where:

[tex]Q_{L}[/tex] - Heat rejected to the cold reservoir, in kilojoules.

[tex]Q_{H}[/tex] - Heat received from the hot reservoir, in kilojoules.

Power output

[tex]W = Q_{H}-Q_{L}[/tex] (4)

Where [tex]W[/tex] is the power output, in kilojoules.

Now we proceed to verify each case:

A) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 600\,kJ[/tex], [tex]W_{m} = 400\,kJ[/tex]

[tex]\eta_{r} = 0.333[/tex]

[tex]0 \le \eta_{r} \le \eta[/tex]

[tex]W = 300\,kJ[/tex]

[tex]W \ne W_{m}[/tex]

This engine is not possible.

B) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 500\,kJ[/tex], [tex]W_{m} = 400\,kJ[/tex]

[tex]\eta_{r} = 0.444[/tex]

[tex]0 \le \eta_{r} \le \eta[/tex]

[tex]W = 400\,kJ[/tex]

[tex]W = W_{m}[/tex]

The engine is possible.

C) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 300\,kJ[/tex], [tex]W_{m} = 600\,kJ[/tex]

[tex]\eta_{r} = 0.667[/tex]

[tex]0 \le \eta_{r} \le \eta[/tex]

[tex]W = 600\,kJ[/tex]

[tex]W = W_{m}[/tex]

The engine is possible.

D) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{L} = 100\,kJ[/tex], [tex]W_{m} = 800\,kJ[/tex]

[tex]\eta_{r} = 0.889[/tex]

[tex]\eta_{r} > \eta[/tex]

[tex]W = 800\,kJ[/tex]

[tex]W = W_{m}[/tex]

The engine is possible.

Answer:

They do not cause phase-distortion of an input signal. ( option 2 )

Explanation:

Linear-phase filters are desirable in many applications simply because  they do not cause phase-distortion of an input signal.  

The signals inputted into a Linear-phase filter ( FIR type ) are delayed equally as frequency is delayed in the pass band but they are are distorted.

The main special function of the Linear-phase filter is to reduce signal distortion.

Answer:

Hello the frequency response is missing attached below is the missing image

answer : Elliptic Filter

Explanation:

The filter with such frequency response that is an analog band-pass filter as well is called an Elliptic filter.

And this is because an Elliptic filter posses a constant ripple in the stop band and a unity gain is seen in the pass band

Answer:

Power= Heat energy\ time

heat energy= mc∆T

specific heat of water = 4180

1.5*4180*(60-25)

=219450

time = 10× 60= 600 secs

power = 219450/600

Power= 365.75 Watt

=0.37 KW

Explanation:

the actual has to be bigger because the heater might be required to handle more ...also to accommodate extra energy.

Answer:

please give me brainlist and follow

Explanation:

Mild steel can be converted into high carbons steel by which of the following heat treatment process? Explanation: Case hardening, also referred as carburizing increases carbon content of steel, thus, imparting hardness to steel.

Answer:

the required expression is [tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]

Explanation:

Given the data in the question;

Q = -0.2[ 1 - [tex]e^(-0.05t)[/tex] ]

ω = 100 rad/s

Torque T = 18 N-m

Electric power input = 2.0 kW

now, form the first law of thermodynamics;

dE/dt = dQ/dt + dw/dt = Q' + w'

dE/dt = Q' + w'  ------ let this be equation 1

w' is the net power on the system

w' = [tex]w_{elect[/tex] - [tex]w_{shaft[/tex]

[tex]w_{shaft[/tex] = T × ω

we substitute

[tex]w_{shaft[/tex]= 18 × 100

[tex]w_{shaft[/tex] = 1800 W

[tex]w_{shaft[/tex] = 1.8 kW  

so

w' = [tex]w_{elect[/tex] - [tex]w_{shaft[/tex]

w' = 2.0 kW - 1.8 kW

w' = 0.2 kW

hence, from equation 1, dE/dt = Q' + w'

we substitute

dE/dt = -0.2[ 1 - [tex]e^{(-0.05t)[/tex] ] + 0.2

dE/dt = -0.2 + 0.2[tex]e^{(-0.05t)[/tex] ] + 0.2

dE/dt = 0.2[tex]e^{(-0.05t)[/tex]

Now, the change in total energy, increment E, as a function of time;

ΔE = [tex]\int\limits^t_0}\frac{dE}{dt} . dt[/tex]

ΔE = [tex]\int\limits^t_0} 0.2e^{(-0.05t)} dt[/tex]

ΔE = [tex]\int\limits^t_0} \frac{0.2}{-0.05} [e^{(-0.05t)}]^t_0[/tex]

[tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]

Therefore, the required expression is [tex]E_2 - E_1[/tex] = 4[ 1 - [tex]e^{(-0.05t)}[/tex] ]

Answer:

Solving

Explanation: