A O.1m³ rigid tank contains steam initially as 500k pa and 200°C. The steam is now allowed to cool until the temperature drops to 50°C. Determine the amount of heat transfer during the process and the final pressure in the tank​

Answer:

Horizontal component ( Ax) = 0

vertical component ( Ay ) = 1794.87 N

Explanation:

Attached below is the detailed solution and the free body diagram of the question

given data:

density = 7910 kg/m3

thickness of plate = 9 mm = 0.009

determine the horizontal and vertical components of the reaction at the pin A and the tension in the cable BC

Answer:

A) EXIT TEMPERATURE = 14⁰C

b) rate of heat transfer of air = - 13475.78 = - 13.5 kw

Explanation:

Given data :

diameter of duct = 20-cm = 0.2 m

length of duct = 12-m

temperature of air at inlet= 50⁰c

pressure = 1 atm

mean velocity = 7 m/s

average heat transfer coefficient = 85 w/m^2⁰c

water temperature = 5⁰c

surface temperature ( Ts) = 5⁰c

properties of air at 50⁰c and at 1 atm

= 1.092 kg/m^3

Cp = 1007 j/kg⁰c

k = 0.02735 W/m⁰c

Pr = 0.7228

v  = 1.798 * 10^-5 m^2/s

determine the exit temperature of air and the rate of heat transfer

attached below is the detailed solution

Calculate the mass flow rate

= p*Ac*Vmean

= 1.092 * 0.0314 *  7 = 0.24 kg/s

Answer:

Explanation:

MOSFET: The acronym for  "metal oxide semiconductor field-effect transistor"  are mostly used in switching circuits.

There are two classes of MOSFET

1. Depletion mode

2. Enhancement mode

          Generally a  MOSFET is a kind of transistor, it is actually a field effect transistor with tree terminals gate, source and drain terminals, also the MOSFET can be used as an amplifier for the amplification of electronic signals in the electronic circuit/devices

Answer:

Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible

Answer:

forward voltage triggering

temperature triggering

dv/dt triggering

light triggering

gate triggering

Then turning off;

Turn off is accomplished by a "negative voltage" pulse between the gate and cathode terminals

Explanation:

hope it helps

Answer:

A) 13.80 ft/s^2

B) 1.714 Ib

Explanation:

Magnitude of acceleration center G

mass = W / g = 8 / 32.2 = 0.2484 Ib.s^2/ft

calculate the acceleration along x direction

A = ra

r = radius

a = angular acceleration

A = 6 in [tex]\frac{1 ft}{12 in}[/tex] * a

a= 2A

equation of the plane along the x-direction

w sin∅ - F = ma

8* sin40 - F = 0.2484 * a

hence F = 5.1423 - 0.2484 a

next find the moment of inertia along the z axis

I = 1/2 mr^2

  = 1/2 * 0.2484 * (6/12)^2  = 0.03105 Ib.ft.s^2

Applying moment balance equation

F * r = inertia * a

(5.1423 - 0.2484 a)*0.5 = 0.03105 * 2A

2.57115 = 0.1863 A      hence

A = 13.80 ft/s^2  ( acceleration of the cylinder )

B) Calculate the friction force exerted by the ramp on the cylinder

F = 5.1423 - 0.2484 A

  = 5.1423 - 0.2484 ( 13.80 )

  = 1.714 Ib

The magnitudes of the acceleration and the friction force are;

Acceleration = 13.8 ft/s²

Friction Force = 1.714 lb

The image of the solid homogeneous cylinder is missing and so i have attached it.

From the image we see that;

We are given;

      We know that formula for weight is; W = mg

Thus; m = W/g

where g is acceleration due to gravity = 32.2 ft/s²

m = 8/32.2

mass; m = 0.2484 lb.s²/ft

     Now, to get the acceleration along the x-axis, we will use the formula;

a = rα

where α is angular acceleration. Thus;

a = 0.5α

α = 2a   ----- (eq 1)

    Now, resolving forces along the x-direction gives;

W*sinθ - F = ma

Plugging in the relevant values;

8*sin 40 - F = 0.2484a

F = 8*sin 40 - 0.2484a    -----(eq 2)

     Now, moment of inertia of the cylinder along the z-axis is gotten from;

I = ¹/₂mr²

I = ¹/₂ × 0.2484 × 0.5²

I = 0.03105 lb.ft/s²

     Taking equilibrium of moments we have;

F*r = I*α

Thus;

(8*sin 40 - 0.2484a)0.5 = 0.03105α

⇒ 2.57115 - 0.1242a = 0.03105α

⇒ 0.03105α + 0.1242a = 2.57115

From eq 1, α = 2a. Thus;

0.03105(2a) + 0.1242a = 2.57115

0.1863a = 2.57115

a = 2.57115/0.1863

a = 13.8 ft/s²

F = 8*sin 40 - 0.2484a

F = 5.1423 - 0.2484(13.8)

F = 1.714 lb

Read more about cylinder moment of inertia at; https://brainly.com/question/7020147

Answer:

nosenose

Explanation:

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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: Tl = - 13.3°C

the lowest outdoor temperature is - 13.3°C

Explanation:

Given that;

Temperature of Th = 21°C = 21 + 273 = 294 K

the rate at which heat lost is Qh = 5400 kJ/h°C

the power input to heat pump Wnet = 6 kw

The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined by;

COPhp = Th/(Th - Tl)

COPhp = Qh/Wnet

Qh/Wnet = Th/(Th -Tl)

the amount of heat loss is expressed as

Qh = 5400/3600(294 - Tl)

the temperature of sink

( 5400/3600(294 - Tl)) / 6 = 294 / ( 294 - Tl)

now solving the equation

Tl = 259.7 - 273

Tl = - 13.3°C

so the lowest outdoor temperature is - 13.3°C

Answer:

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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:

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while n<1:

c=(n!)/((n-r)!(r!))

end

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Answer:

current is measured in Ampere (A)

Answer:

Ampere (A)

Explanation:

The ampere is defined so the elementary charge e is 1.602 176 634 × 10−19 C or A•s.

Hope this helps <3

Answer:

A) The relative humidity : 0.818 (81.8%), Temperature at C = 14.4⁰c

B) The rate of energy destruction = 0.0477 kw  

Explanation:

Given data :

at point 1 : m1 = 3 kg/min , T1 = 30⁰c, p1 = 1 bar,  ∅ = 0.50 ( 50%)

at point 2 : T2 = 5⁰c,  P2 = 1 bar, m2 = 5 kg/min

at point 3 : p3 = 1 bar

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.

solution from c hegg. hope it helps. see photos explanation

Answer:

  11.94 V

Explanation:

Generally the regulated voltage drops as load increases. When the voltage has dropped by 0.5%, it will be 60 mV less than the nominal value:

  12.0 V - 0.06 V = 11.94 V . . . . full load voltage

Answer:

  leakage

Explanation:

That current is "leakage current."

Answer:

a) the minimum time (minutes) for the aluminium casting to solidify is 2.86 min

b) the minimum time (minutes) for the grey iron casting to solidify is 2.13 min. Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

Explanation:

Given that; diameter of Disk = 500 mm, thickness t = 20, mold constant Cm = 2.0 sec/mm²

first we find the volume and Area;

Volume V = πD²t / 4

Volume V = π × (500)² × 20 / 4 = 3,926,991 mm³

Area A = 2πD²/ 4 + πDt

Area A = {[π × (500)²] / 2} +{ π × (500) × (20)}

Area A = 392,699.08 + 31,415.93

Area A = 424,115 mm²

a)

Chvorinov’s rule

T(aluminium) = Cm (V/A) ²

T(aluminium) =  2.0 × (3,926,991 / 424,115) ²

T(aluminium) = 171.5 s = 2.86 min

∴ the minimum time (minutes) for the casting to solidify is 2.86 min

b)

For cast iron

Cm (mold constant = 1.488 sec/mm²)

Chvorinov’s rule

T(iron) = Cm (V/A) ²

T(iron) = 1.488 × (3,926,991 / 424,115) ²

T(iron) = 127.5720s = 2.13 min

Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

Explanation:

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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: 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

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

[tex]$J=\frac{\pi}{32}d^4$[/tex]

[tex]$J=\frac{\pi}{32}\times (46)^4$[/tex]

J = 207.6 [tex]mm^4[/tex]

So the shear stress at point  A is :

[tex]$\tau_A =\frac{Tc_A}{J}$[/tex]

[tex]$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$[/tex]

[tex]$\tau_A = 4913.29 \ MPa$[/tex]

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

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:

b. inversely proportional to radius of curvature

Explanation:

In curvilinear motions, the normal acceleration which is also called the centripetal acceleration is always directed towards the center of the circular path of motion. This acceleration has a magnitude that is directly proportional to the square of the speed of the body undergoing the motion and inversely proportional to the radius of the curvature of the motion path. The centripetal or normal acceleration a,  can be given by;

a = [tex]\frac{v^2}{r}[/tex]

Where;

v = speed of the body

r = radius of curvature.

Answer:

hello your question has some missing part attached below is the complete question

answer : steady state temperature = 419.713k ≈ 218.7⁰c

              Time required to reach a junction ≈ 5 secs

Explanation:

The detailed solution of the given problem is attached below but the solution to the subsequent problem from which the question you asked is referenced to( problem 1 ), is not attached because it was not part of the question you asked

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: y' = - x'

Explanation:

Let f(x) = 2x + y

then f'(x) = 2 + y'

Let f(y) = x - 2y

then f'(y) = x' - 2

Given:  f'(x) + f'(y) = 0

         2 + y' + x' - 2 = 0

                y' + x'= 0

                 y' = -x'

This can also be written as:       [tex]\dfrac{dy}{dx}=-\dfrac{d}{dx}[/tex]

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.

To learn more about the scale model, refer to the below link:

https://brainly.com/question/14341149

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