A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at 2308C by rejecting its waste heat to cooling water that enters the condenser at 188C at a rate of 0.25 kg/s and leaves at 268C. The refrigerant enters the condenser at 1.2 MPa and 658C and leaves at 428C. The inlet state of the compressor is 60 kPa and 2348C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator, and (d) the theoretical maximum refrigeration load for the same power input to the compressor.

Answer: abundant energy sources

Explanation:

A power generating station which is sometimes referred to as the power plant is simply an industrial facility that is used for the generation of power.

Large, centralized power-generating stations are often located near abundant energy sources. It should be noted that most power stations burn fossil fuels like oil, coal, natural gas, in order to be able to generate power and hence, they have to be located where there's abundance of the energy sources.

Answer: In a clean plastic or metal container with a tightly sealed lid

Answer:

Technician B is correct.

Explanation:

Two technicians are discussing the a series-parallel brake lamp circuit that does not illuminate.

The statement made by Technician A saying that an open in one of the parallel branches could be the cause has no ground because an opening in any branches has not to do with the issue since there is no disconnection The statement made by Technician B saying that a shorted (or stuck closed) brake lamp switch wired in series could be the cause is correct.

Answer:

h = 6.35 W/m².k

Explanation:

In order to solve this problem, we will use energy balance, taking the thin hot plate as a system. According to energy balance, the rate of heat transfer to surrounding through convection must be equal to the energy stored in the plate:

Rate of Heat Transfer Through Convection = Energy Stored in Plate

- h A (Ts - T₀) = m C dT/dt

where,

h = convection heat transfer coefficient = ?

A = Surface area of plate through which heat transfer takes place = 2 x 0.3 m x 0.3 m (2 is multiplied for two sides of thin plate) = 0.18 m²

Ts = Surface Temperature of hot thin plate = 225⁰C

T₀ = Ambient Temperature = 25°C

m = mass of plate = 3.75 kg

C = Specific Heat = 2770 J/kg. k

dT/dt = rate of change in plate temperature = - 0.022 K/s

Therefore,

- h (0.18 m²)(225 - 25) k = (3.75 kg)(2770 J/kg.k)(- 0.022 k/s)

h = (- 228.525 W)/(- 36 m².k)

h = 6.35 W/m².k

This question is incomplete, the complete question is;

Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K.

The constant B = 3.56×10¹⁴ (cm⁻³ K^-3/2) and the bandgap voltage E = 1.42eV.

Answer: the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³

Explanation:

Given that;

T = 300k

B = 3.56×10¹⁴ (cm⁻³ K^-3/2)

Eg = 1.42 eV

we know that, the value of Boltzmann constant k = 8.617×10⁻⁵ eV/K

so to find the ni for gallium arsenide;

ni = B×T^(3/2) e^ ( -Eg/2kT)

we substitute

ni = (3.56×10¹⁴)(300^3/2) e^ ( -1.42 / (2× 8.617×10⁻⁵ 300))

ni =  (3.56×10¹⁴)(5196.1524)e^-27.4651

ni = (3.56×10¹⁴)(5196.1524)(1.1805×10⁻¹²)

ni = 2.1837 × 10⁶ cm⁻³

Therefore the value of ni for gallium arsenide (GaAs) is 2.1837 × 10⁶ cm⁻³

Answer:

i)The results are consistent at ; Rows ( 1,2,4,6 ), which simply means that it is fairly consistent and makes sense

ii) Row 5 shows a gain of -99 which is ≈ -100 and there is a possibility of experimental error of (1%)

ii ) The omitted values are : row 3 = 0.99, Row 7 = 5.049

Explanation:

To calculate/determine the missing/omitted values we have to apply this formula:

Vo = ( V1 - V2 ) G

Vo = output = 1

G = gain = - 100

V1 =

we find  V1 along line 3 ( row 3 )

(V1 - V2) G = 1

( V1 - 1 ) * -100 = 1

hence V1 = 0.99

we find V2 along row 7

( V1 - V2 ) G = 1

( 5.10 - V2 ) * -100 = 1

hence V2 = 5.049

Answer:

V = 20.6 m/s

Explanation:

Given that the temperature of the neon molecules = 356 Kelvin

Pressure = 0.9 bar

The mass number of Neon = 21.

Using the formula below

1/2 m v^2 = (3kT)/2

Where

T = temperature = 356 k

K = Bolzmann constant

= 1.38 × 10^-23jk^-1

NA = 6.02214076×10²³ mol⁻¹

Where NA = Avogadro number

Substitute all the parameters Into the formula

1/2 × 21/NA × v^2 = (3 × k × 356)/2

1.744×10^-23V^2 = 7.3692 × 10^-21

Make V^2 the subject of formula

V^2 = (7.37×10^-21)/1.744×10^-23

V^2 = 422.55

V = sqrt( 422.55)

V = 20.55

Therefore, the average (root mean square) velocity (m/s) of Neon molecules at 356 Kelvins and 0.9 bars is 20.55 m/s

Answer:

(a) the maximum tensile stress is 68.2 psi

(b) the maximum normal strain is 2.35 x 10⁻⁶

Explanation:

Given;

modulus of elasticity, E = 29,000 ksi = 29 x 10⁶ psi

(a) the maximum tensile stress

[tex]\tau = \frac{f}{A}[/tex]

f is the maximum force suspended = 20 x 60 = 1200 lb

A is the area of W10 x 60 = 17.6 in²

[tex]\tau = \frac{1200}{17.6} \\\\\tau = 68.2 \ psi[/tex]

(b) the maximum normal strain.

According to Hook's law stress is directional to strain

τ = Eε

[tex]\epsilon = \frac{\tau}{E}\\\\\epsilon = \frac{68.2}{29*10^{6}}\\\\\epsilon = 2.35*10^{-6}[/tex]

Explanation:

1. Weld only in authorized areas. Make sure the area is dry, chemical free, and well ventilated.

2. Inspect the equipment before starting to use it.

3. Keep other people away, unless they are authorized to be there and are wearing the appropriate personal protective equipment.

Answer:

Trash rack

Open channel

Fore bay

Pen stock

Inlet valve

Turbine

Tailrace

Generator

Power house

Answer:

b. satellite surveying

c. aerial surveying

d. optical tooling

e. marketing surveying

f. control surveying

h. statistical surveying

i. telephone surveying

k. alignment surveying

l. mine surveying

m. solar surveying

Explanation:

A Survey is an act of examination of the features of a subject or material under consideration. Land surveying refers to the examination of the natural and man-made features of a piece of land using scientific and mathematical methods.

Land surveying finds application in construction where a survey is made on all the structures found in a constructed property. Topographic surveying deals with examining the natural and man-made feature of a piece of land. As-built survey as the name implies examines the features and location of a building during or recently after construction. These three are examples of land surveys.

Some type of surveying other than land surveying includes satellite, aerial, optical tooling, marketing, control surveying, statistical, telephone, alignment, mine, and solar surveying

A Survey means an examination of a features, subject or material under consideration.

Land surveying refers to the examination of the natural and man-made features of a piece of land using scientific and mathematical methods.

However, some other type of surveying other than land surveying includes satellite, aerial, optical tooling, marketing, control surveying, statistical, telephone, alignment, mine, and solar surveying.

Read more about surveying

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Answer:

Fit test

Explanation:

it's on the automotive technology sight : )

Answer: IMA = 12.042

Explanation:

Given that;

Height from the ground h is 3 ft

Slope of ramp s is 1:12

Horizontal length of the ramp x will be 3 × 12 = 36 ft

Now to get the IMA ( ideal mechanical advantage,)

IMA = length of the ramp over / height of the ramp

IMA = (√( 36² + 3²)) / 3

IMA = (√ 1305 ) / 3

IMA = 36.124 / 3

IMA = 12.042

Answer:

2.35 + j8.34 Ω

Explanation:

Voltage = V[tex]_{L}[/tex] = 240 V rms

supplying power = S[tex]_{s}[/tex] = 8 kVA

power factor = pf[tex]_{s}[/tex] = 0.6

Let P₁ represents one load draws 3kW at unity powder factor

The power angle is:

θ[tex]_{s}[/tex] = cos⁻¹  pf[tex]_{s}[/tex] = cos⁻¹  0.6 = 53.13°

Complex power supplied source is:

S[tex]_{s}[/tex] =  S[tex]_{s}[/tex] < θ[tex]_{s}[/tex] = 8<53.13° kVA

Complex power for first load:

S₁ = P₁ = 3kVA

Since the power angle of first load is  θ₁ = 0°

According to principle of conservation of AC power, the power of second load is:

S₂ =  S[tex]_{s}[/tex] - S₁

    = 8<53.13° - 3

    = 6.65<74.29° kVA

Since the second load is a Y connected load the phase voltage:

V[tex]_{p}[/tex] =  V[tex]_{L}[/tex] / [tex]\sqrt{3}[/tex]

    = 240/1.732051

    = 138.564

    = 138.56 V

Complex power of second load:

S₂ = 3 V[tex]_{p}[/tex]² / Z[tex]_{p}[/tex]

impedance per phase of the second load:

Z[tex]_{p}[/tex] =  3 V[tex]_{p}[/tex]² / S₂

   = 3 (138.56)² /  6.65<74.29°

   = 3(19198.8736) / 6.65<74.29°

   = 57596.6208 / 6.65<74.29°

Z[tex]_{p}[/tex] = 2.35 + j8.34Ω

Answer:

The following are the answer to this question:

Explanation:

In point a, Calculating the are of flow:

[tex]\bold{Area =B \times D_f}[/tex]

         [tex]=6\times 5\\\\=30 \ ft^2[/tex]

In point b, Calculating the wetter perimeter.

[tex]\bold{P_w =B+2\times D_f}[/tex]

      [tex]= 6 +2\times (5)\\\\= 6 +10 \\\\=16 \ ft[/tex]

In point c, Calculating the hydraulic radius:

[tex]\bold{R=\frac{A}{P_w}}[/tex]

   [tex]=\frac{30}{16}\\\\= 1.875 \ ft[/tex]

In point d, Calculating the value of Reynolds's number.

[tex]\bold{Re =\frac{4VR}{v}}[/tex]

     [tex]=\frac{4V \times 1.875}{1 \times 10^{-5} \frac{ft^2}{s}}\\\\[/tex]

     [tex]=750,000 V[/tex]

Calculating the velocity:

[tex]V= \sqrt{\frac{8gRS}{f}}[/tex]

   [tex]= \sqrt{\frac{8\times 32.2 \times 1.875 \times 0.02}{f}}\\\\=\frac{3.108}{\sqrt{f}}\\\\[/tex]

[tex]\sqrt{f}=\frac{3.108}{V}\\\\[/tex]

calculating the Cole-brook-White value:

[tex]\frac{1}{\sqrt{f}}= -2 \log (\frac{K}{12 R} +\frac{2.51}{R_e \sqrt{f}})\\\\ \frac{1}{\frac{3.108}{V}}= -2 \log (\frac{2 \times 10^{-2}}{12 \times 1.875} +\frac{2.51}{750,000V\sqrt{f}})\\[/tex]

[tex]\frac{V}{3.108} =-2\log(8.88 \times 10^{-5} + \frac{3.346 \times 10^{-6}}{750,000(3.108)})[/tex]

After calculating the value of V it will give:

[tex]V= 25.18 \ \frac{ft}{s^2}\\[/tex]

In point a, Calculating the value of Froude:

[tex]F= \frac{V}{\sqrt{gD}}[/tex]

[tex]= \frac{V}{\sqrt{g\frac{A}{\text{Width flow}}}}\\[/tex]

[tex]= \frac{25.18}{\sqrt{32.2\frac{30}{6}}}\\\\= \frac{25.18}{\sqrt{32.2 \times 5}}\\\\= \frac{25.18}{\sqrt{161}}\\\\= \frac{25.18}{12.68}\\\\= 1.98[/tex]

The flow is supercritical because the amount of Froude is greater than 1.  

Calculating the channel flow rate.

[tex]Q= AV[/tex]

   [tex]=30x 25.18\\\\= 755.4 \ \frac{ft^3}{s}\\[/tex]

Answer:

599.7 m

approximately 600 m

Explanation:

initial speed of the rocket = 0

net acceleration upwards = 7 m/s^2

the engine cuts out 10 sec after take off

maximum height reached = ?

we neglect air resistance

To get the velocity of the rocket at the point where the engine cuts off, we use the equation

v = u + at

where

v is the velocity at this point where the engine stops = ?

u is the initial velocity of the rocket from rest = 0 m/s

a is the net acceleration upwards = 7 m/s^2

t is the time the engine runs = 10 s

substituting, we have

v = 0 + (7 x 10)

v = 70 m/s

to get the distance from the ground to this point, we use the equation

[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as

where

v is the final velocity at the the height where the engine is cut out = 70 m/s

u is the initial speed at the ground = 0 m/s

a is the net acceleration on the rocket = 7 m/s^2

s is the distance from the ground to this point

substituting, we have

[tex]70^{2}[/tex] = [tex]0^{2}[/tex] + 2(7 x s)

4900 = 14s

s = 4900/14 = 350 m

After this point when the engine cuts out, the rocket experiences an acceleration proportional to the acceleration due to gravity 9.81 m/s^2 downwards, and slows down gradually before coming to a stop at the maximum height.

To get this height, we use the equation

[tex]v^{2}[/tex] = [tex]u^{2}[/tex] - 2gs   (the negative sign is due to the downward direction of the acceleration g)

where

v is the final velocity at the maximum height = 0 m/s (it comes to a stop)

u is the speed at the instance that the engine is cut out = 70 m/s

g is the acceleration due to gravity = 9.81 m/s^2

s is the distance from this point to the maximum height

substituting values, we have

[tex]0^{2}[/tex] = [tex]70^{2}[/tex] - 2(9.81 x s)

0 = 4900 - 19.62s

4900 = 19.62s

s = 4900/19.62 = 249.7 m

The maximum height that will be reached = 350 m + 249.7 m = 599.7 m

approximately 600 m

Answer:

Portable appliance testing (PAT) is the term used to describe the examination of electrical appliances and equipment to ensure they are safe to use. Most electrical safety defects can be found by visual examination but some types of defect can only be found by testing.

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Answer: Central Processing Unit

Answer:

[tex]\frac{ML}{T^2}=\frac{ML}{T^2}[/tex]

Hence it is proved that Stokes-Oseen formula is dimensionally homogenous.

Explanation:

For equation to be dimensionally homogeneous both side of the equation must have same dimensions.

For given Equation:

F= Force, μ= viscosity, D = Diameter, V = velocity, ρ= Density

Dimensions:

[tex]F=\frac{ML}{T^2}[/tex]

[tex]\mu=\frac{M}{LT}[/tex]

[tex]D=L\\\\V=\frac{L}{T}\\ \\\rho=\frac{M}{L^3}[/tex]

Constants= 1

Now According to equation:

[tex]\frac{ML}{T^2}=[\frac{M}{LT}][L] [\frac{L}{T}] + [\frac{M}{L^3}][\frac{L^2}{T^2}][L^2][/tex]

Simplifying above equation, we will get:

[tex]\frac{ML}{T^2}=2*\frac{ML}{T^2}[/tex]

Ignore "2" as it is constant with no dimensions. Now:

[tex]\frac{ML}{T^2}=\frac{ML}{T^2}[/tex]

Hence it is proved that Stokes-Oseen formula  is  dimensionally homogenous.

Answer:

A) 48

B) Pitch diameters : pinion = 42.164 mm, Gear = 101.19 mm

C) standard addendum : pinion = 46.3804, Gear = 105.406

    standard dedendum : pinion = 37.265 mm, Gear = 96.312 mm

D) 71.672 mm

E) 94.989 N , 101.0858 N,  34.573 N

Explanation:

Given Data :

∅ = 20⁰ , Tp = 20 ( tooth pinion ),

diameter pitch = 12, Np = 1776 rpm ,

Ng = 740 rpm,

attached below is the detailed solution of the given problems

Answer:

number of mole is 31342.36 moles

mass is 125.369 kg

Explanation:

Diameter of the spherical balloon d = 9 m

radius r = d/2 = 9/2 = 4.5 m

The volume pf the sphere balloon ca be calculated from

V = [tex]\frac{4}{3} \pi r^3[/tex]

V = [tex]\frac{4}{3}* 3.142* 4.5^3[/tex] = 381.75 m^3

Temperature of the gas T = 20 °C = 20 + 273 = 293 K

Pressure of the helium gas = 200 kPa = 200 x 10^3 Pa

number of moles n = ?

Using

PV = nRT

where

P is the pressure of the gas

V is the volume of the gas

n is the mole number of the gas

R is the gas constant = 8.314 m^3⋅Pa⋅K^−1⋅mol^−1

T is the temperature of the gas (must be converted to kelvin K)

substituting values, we have

200 x 10^3 x 381.75 = n x 8.314 x 293

number of moles n  = 76350000/2436 = 31342.36 moles

We recall that n = m/MM

or m = n x MM

where

n is the number of moles

m is the mass of the gas

MM is the molar mass of the gas

For helium, the molar mass = 4 g/mol

substituting values, we have

m = 31342.36 x 4

m = 125369.44 g

m = 125.369 kg

The mole number and the mass of the helium in the balloon are 250.801 kilomoles and 1003.706 kilograms, respectively.

Let suppose that helium contained in the spherical balloon behaves ideally, the mole number ([tex]n[/tex]), in kilomoles, is determined by the following expression:

[tex]n = \frac{P\cdot V}{R_{u}\cdot T}[/tex] (1)

Where:

The volume and mass of helium ([tex]m[/tex]), in kilograms, is described by these two formulas:

[tex]V = \frac{4\pi}{3}\cdot R^{3}[/tex] (2)

[tex]m = n\cdot M[/tex] (3)

Where:

If we know that [tex]P = 200\,kPa[/tex], [tex]R = 9\,m[/tex], [tex]R_{u} = 8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}[/tex], [tex]T = 293.15\,K[/tex] and [tex]M = 4.002\,\frac{kg}{kmol}[/tex], then the mole number and the mass of the helium in the balloon are:

[tex]V = \frac{4\pi}{3}\cdot (9\,m)^{3}[/tex]

[tex]V \approx 3053.628\,m^{3}[/tex]

[tex]n = \frac{(200\,kPa)\cdot (3053.628\,m^{3})}{\left(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)\cdot(293.15\,K)}[/tex]

[tex]n = 250.580\,kmol[/tex]

[tex]m = (250.801\,kmol)\cdot \left(4.002\,\frac{kg}{kmol} \right)[/tex]

[tex]m = 1003.706\,kg[/tex]

The mole number and the mass of the helium in the balloon are 250.801 kilomoles and 1003.706 kilograms, respectively.

We kindly invite to check this question on ideal gases: https://brainly.com/question/16211117

Answer: (B) 100

Explanation:

Given that;

Pstatic = 20 psig , hz = 160ft, hf = 20ft

Now total head will be;

T.h = hz + hf

T.h= 160 + 20

T.h = 180ft

Minimum pressure = Psatic + egh

we know that specific weight of water is 62.4 (lb/ft3)

so

P.min = (20 bf/in² ) + (62.4 b/ft³ × 180 fr

P.min = (20 bf/in² ) + ( 62.4 × 180 × 1 ft²/144 in²)

P.min = 20 + 78

P.min = 98 lbf/in²

Therefore the minimum pressure rating (psi) of the piping system is most nearly B) 100

Answer:

Set high and clear expectations for quality work

Don't attempt to cram all your studying into one session

Explanation:

Participate in Class

eat a well balanced diet

Find perfect place to Study.

HavE a NIce dAY ;}

Answer:

CMRR(Common Mode Rejection Ratio) is the ratio of differential gain and the common mode gain.

Explanation:

Answer:

The answer would be ISP address

Explanation:

An Internet Protocol address is a numerical label assigned to each device connected to a computer network that uses the Internet Protocol for communication. An IP address serves two main functions host or network interface identification and location addressing which can help you a lot.

The type of address the computer use to find something on network is ISP address.

ISP is simply Internet service provider. They are companies which provides internet services to people. They charge based on the quantity of their service used at a particular time. ISP companies also provide softwares for its users.

An Internet Protocol address refers to a numerical label assigned to each device connected to a computer network that uses the Internet Protocol for communication.

Therefore, the correct answer is option D

Learn more about IP address:

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Answer: true

Explanation:

Answer:

Number of cylinders of each gas

Explanation:

The Department of Transportation regulations Title 49 of the United States Code of Federal Regulations specifies that the number of cylinders of each gas being transported should be recorded on the shipping paper for gases classified under hazard class 2.2 of which one of them are the Non Flammable compressed gases.

The definition of gases in division 2.2 encompasses compressed gas, liquefied gas, compressed gas in solution, asphyxiant gas and oxidizing gas. They must meet the absolute pressure of 280kPa or greater at 20 degree Celsius.