An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 5.25 kW. Determine (a) the COP of this air conditioner and (b) the rate of heat transfer to the outside air.Answers:(a) 2.38, (b) 1065 kJ/min

Answer: The process of data analysis uses analytical and logical reasoning to gain information from the data. The main purpose of data analysis is to find meaning in data so that the derived knowledge can be used to make informed decisions.

Answer:

i) σ1 = 133.5 MPa

  σ2 = -2427 MPa

ii) 78.89 MPa

Explanation:

Given data:

ε1 = 0.0020 and ε2 = –0.0010

E = 71 GPa

v = 0.35

i) Determine the biaxial stresses  σ1 and σ2 using the relations below

ε1 = σ1 / E - v (σ2 / E)   -----( 1 )

ε2 = σ2 / E - v (σ1 / E)  -------( 2 )

resolving equations 1 and 2

σ1 = E / 1 - v^2 {  ε1 + vε2 } ---- ( 3 )

σ2 = E / 1 - v^2 {  ε2 + vε1 } ----- ( 4 )

input the given data into equation 3 and equation 4

σ1 = 133.5 MPa

σ2 = -2427 MPa

ii) Calculate the value of the maximum shear stress ( Zmax )

Zmax = ( σ1 - σ2 ) / 2

         = 133.5 - ( - 2427 ) / 2

         = 78.89 MPa

The per capita GDP of China in the Calendar year 2021 was found to be around 12,359 U.S. dollars.

GDP termed Gross Domestic Product, has been evaluated with the value producing the economy of the region with the values added with the used products formed to be the less of the economy produced. It has been termed as the measure of the income of a region and not the wealth.

The per capita GDP has been the total income earned by a person in a region during a specified period of time. The calculation has been made by dividing the total gross income of the region by the total population.

China has been the world's most populous country in the East Asian region. It has been found that the per capita GDP of China is low because of its large population. In the calendar year 2021, the per capita GDP of China was 12,359 U.S. dollars.

Learn more about the GDP, here:

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Answer:

B. rotor

Explanation:

The correct answer Is rotor because the others are part of a cars drivetrain

The drive train system exists as a critical element of a vehicle and the transmission exists as an integral part of the drive train. B rotors NOT part of a car's drive train.

A drive train exists not really a single part of your car – it's a set of drive train components that interact with the engine to drive the wheels and different regions of the vehicle to thrust it into motion. These components often contain the transmission, differential, driveshaft, axles, CV joints, and wheels.

Therefore, the correct answer is option B rotors.

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Answer:

i) Cases with Important Inertia

  Large whale ( a ) , Flying duck ( b ) , Large dragonfly ( c )

ii) Cases where viscous effects dominate

  Invertebrate larva ( d ) , bacterium ( e )

iii) Cases where flow is Laminar ( cases where Re is < 2,100 )

    Invertebrate larva ( d ), bacterium ( e )

iv) Cases where flow is turbulent ( case Re is > 2,100 )

    Large whale (a) , Flying duck (b), Large dragonfly ( c ),

Explanation:

Reynolds number is  the the ratio of  Initial forces to viscous forces, hence cases with Large Reynolds number ( > 2100 ) have their inertial forces greater than viscous forces, therefore we can say the Inertia is more important , while cases with smaller Reynolds number ( < 2100 ) have the viscous forces greater than the inertial forces therefore in such case the viscous effect is more important

i) Cases with Important Inertia

  Large whale ( a ) , Flying duck ( b ) , Large dragonfly ( c )

ii) Cases where viscous effects dominate

  Invertebrate larva ( d ) , bacterium ( e )

iii) Cases where flow is Laminar ( cases where Re is < 2,100 )

    Invertebrate larva ( d ), bacterium ( e )

iv) Cases where flow is turbulent ( case Re is > 2,100 )

    Large whale (a) , Flying duck (b), Large dragonfly ( c ),

Answer: the heat flux increases as the velocity boundary layer transitions to laminar to turbulent.

Explanation:

The correct statement about the relation between the velocity boundary layer and heat transfer for flow over a flat plate that is uniform in temperature is that the heat flux increases as the velocity boundary layer transitions to laminar to turbulent.

It should be noted that the heat goes in a streamline direction in a laminar flow, thereby the molecules less collide with each other. On the other hand, the direction is zig zag in a turbulent heat flux and this will bring about more collision of the molecules which leads to a rise in the heat flux.

Answer:

a) the expected flow rate is 31.4 ft³/s

b) the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation is -8.4 ft

Explanation:

Given the data in the question;

free surface elevation = 200 ft

total length of pipe required = 1000 ft

diameter = 12 inch

Iron with relative roughness ( k/D ) = 0.0005

H[tex]_{pump[/tex] = 665-0.051Q² [Qinft ]

a) the expected flow rate

given that;

k/D  = 0.0005

k/2R = 0.0005

R/k = 1000

now, we determine the friction factor;

1/√f = 2log₁₀( R/k ) + 1.74

we substitute

1/√f = 2log₁₀( 1000 ) + 1.74

1/√f = 6 + 1.74

1/√f = 7.74

√f = 1/7.74

√f = 0.1291989

f = (0.1291989)²

f = 0.01669

Now, Using Bernoulli theorem between two reservoirs;

(p/ρq)₁ + (v²/2g)₁ + z₁ + H[tex]_p[/tex] = (p/ρq)₂ + (v²/2g)₂ + z₂ + h[tex]_L[/tex]

so

0 + 0 + 0 + 665-0.051Q² = 0 + 0 + 200 + flQ²/2gdA²

665-0.051Q² = 200 + flQ²/2gdA²

665-0.051Q² = 200 +[  ( 0.01669 × 1000 × Q² ) / (2 × 32.2 × (π/4)² × 1⁵ )

665 - 0.051Q² = 200 + [ 16.69Q² / 39.725 ]

665 - 200 - 0.051Q² = 0.420138Q²

665 - 200 = 0.420138Q² + 0.051Q²

465 = 0.471138Q²

Q² = 465 / 0.471138

Q² = 986.97196

Q = √986.97196

Q = 31.4 ft³/s

Therefore, the expected flow rate is 31.4 ft³/s

b) the brake horsepower required to drive the pump (assume an efficiency of 78%).

we know that;

P = ρgH[tex]_p[/tex]Q / η

where; H[tex]_p[/tex] = 665 - 0.051(986.97196) = 614.7

we substitute;

P = ( 62.42 × 614.7 × 31.4 ) / ( 0.78 × 550 )

P = 1204804.6236 / 429

P = 2808.4 bhp

Therefore, the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).

NPSH = ([tex]P_{atom[/tex] / ρg) - h[tex]_s[/tex] - ( P[tex]_v[/tex] / ρg )

we substitute

25  = ( 2116 / 62.42 ) - h[tex]_s[/tex] - ( 30 / 62.42 )

h[tex]_s[/tex] = 8.4 ft

Therefore, the location of pump inlet to avoid cavitation is -8.4 ft

Answer:

Part A

The flow of the bee lava is inertia dominant and laminar

Part B

The flow of a baseball in Mars is inertia dominated and the flow is laminar

Part C

The flow of an elderly on i95 is viscous force dominated and turbulent

Explanation:

The Reynold's number when inertia is dominant is low, and the flow is laminar

When viscous force is dominant, the Reynold's number is high, and we have turbulent flow

The Reynold's number of laminar flow is Re < 2,000

The Reynold's of unstable or intermediate flow is 2,000 < Re < 4,000

The Reynold's number of turbulent flow is Re > 4,000

Therefore, we have;

Part A

The Reynold's number of a bee lava, Re = 0.2, therefore, the flow is inertia and laminar

Part B

The Reynold's number of the ball in Mars is Re = 500, therefore, given that the Reynold's number is less than 2,000, the inertia is dominant, and the flow is laminar flow

Part C

The Reynold's number of the driver on i95 = 1,200,000 which is larger than 4,000, therefore, the flow is viscous force dominated and the flow is turbulent.

This question is incomplete, the missing diagram is uploaded along this answer below;

Answer:

from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

Explanation:

Given the data in the question;

We write the expression for the local Nusselt number for Laminar flow over the flat plate;

Nu = [tex]C[/tex][tex](Re_x)^{0.5[/tex] [tex](Pr)^{1/3[/tex]

Nu = [tex]C(\frac{Vx}{v})^{0.5}[/tex] [tex](Pr)^{1/3[/tex]

[tex]\frac{h_xx}{k}[/tex] = [tex]C(\frac{V}{v})^{0.5}[/tex]  [tex](Pr)^{1/3[/tex]  [tex](x)^{0.5[/tex]

[tex]h_x[/tex] = [tex]\frac{1}{x^{1/2}}[/tex]

Next we write down the expression for the local heat flux from the plate with  uniform surface temperature;

q = [tex]h_xA([/tex] T[tex]_s[/tex] - T∞ )

q ∝ [tex]h_x[/tex]

∴

q ∝  [tex]\frac{1}{x^{1/2}}[/tex]

The local heat flux decreases with the position as it is inversely proportional to the square root of the position from the leading edge and it will not be zero at the end of the plate.

Therefore, from the diagram, the curve that most closely describes the variation of the local heat transfer coefficient with position along the plate is Option D

Explanation:

6/12 = 0.5ft

length = 16

width = 14

The volume of cinder concrete = 0.5 * 16 *14 = 112

the resultant force that is caused by the dead load

density of cinder concrete * volume

density is assumed to be 108

dead load = 108 * 112 = 12096 lb

resultant force caused by the live load

liveload = 125lb/ft2

= 125 * 14 * 16

= 28000 lb

Solution :

Given data:

Diameter of the copper cylindrical rod = 16 mm

Yield strength  = 250 MPa

Calculating the percent cold work

[tex]$\text{Percentage Cold Work} = \frac{\pi\left(\frac{d_0}{2}\right)^2-\pi\left(\frac{d_d}{2}\right)^2}{\pi\left(\frac{d_0}{2}}\right)^2} \times 100$[/tex]

                                  [tex]$ = \frac{\pi\left(\frac{16}{2}\right)^2-\pi\left(\frac{11.3}{2}\right)^2}{\pi\left(\frac{16}{2}}\right)^2} \times 100$[/tex]

                                  = 50% CW

Therefore, at [tex]50\% \ CW[/tex], the yield strength of copper will be of the order of 330 MPa.

The ductility will be 4% elongation (EL).

Rather than performing drawing in single operation, we draw some of the fraction of total deformation, and then the anneal them to recrystallize and also finally w do cold work on the material  for the second time to achieve its final diameter, ductility and yield strength.

[tex]21\% \ CW[/tex]is required for a yield strength of [tex]250 \ MPa[/tex]. Similarly, a maximum of [tex]23\% \ CW[/tex] is required for[tex]12\% \ EL[/tex].

The average of the two values is [tex]22\% \ CW[/tex]. To achieve both the [tex]\text{specified yield strength and ductility,}[/tex] the copper should be deformed to[tex]22\% \ CW[/tex]. The[tex]\text{ final diameter}[/tex] after the first drawing and the initial diameter for the second drawing is [tex]$d_0'$[/tex] , then

[tex]$22\% \ CW = \frac{\pi\left(\frac{d_0'}{2}\right)^2-\pi\left(\frac{11.3}{2}\right)^2}{\pi\left(\frac{d_0'}{2}}\right)^2} \times 100$[/tex]

[tex]$d_0'=\frac{11.3}{\sqrt{1-\frac{22\% \ CW}{100}}}$[/tex]

[tex]$d_0'=12.8 \ mm$[/tex]

Answer:

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Answer:

The correct answer will be "2".

Explanation:

Given:

Heat generated through wall,

= 15000 kJ/h

= [tex]\frac{15000}{3600}[/tex]

= [tex]4.17 \ KW[/tex]

Heat generated by bulbs,

= [tex]10(100) \ W[/tex]

= [tex]1000[/tex]

= [tex]1 \ KW[/tex]

Heat generated by people,

= [tex]40(360) \ KJ/h[/tex]

= [tex]\frac{14400}{3600}[/tex]

= [tex]4 \ KW[/tex]

Heat removed by 1 AC,

= 5 KW

Now,

The total increase in heat will be:

= [tex]4+1+4.17[/tex]

= [tex]9.17 \ KW[/tex]

hence,

The required number of AC will be:

⇒ [tex]5n=9.17[/tex]

⇒   [tex]n=\frac{9.17}{5}[/tex]

⇒       [tex]=1.83[/tex]

0r,

⇒       [tex]2[/tex]

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

Derive the unit hydrograph using the inverse procedure

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh *  duration

        = 29700 * 6 hours ( 216000 secs )

        = 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000  / 185 mi^2

                    = 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh /  volume of runoff in equivalent depth

                                = 50 ft^3 / 1.49 in  =  33.56 ft^3/sec.in

Answer:

Following are the application to the given question:

Explanation:

To,

          The Chief Operating Officer,

           Maidan Hospitals, USA

02.10.2020

Subject: HIV/AIDS disease spreading control

Respected Mr Dolman,

The media as well as the individuals living in the area have indeed been notified again that the spread of Hiv / Aids must be regulated and managed immediately. A proper environment must be created artificially to heal this sickness and help stop it from spreading.

Its fight for the sickness ought to be a forward end to bring it through to the high levels instead of being fearful of disease. The disease also is seen in the hospital as it has grown via infusion so that the nursing staff must keep a check mostly on the issue frequently.

This position taken through distributing petitions in the community proposes that a small number of clinics must be established, as this could help to stop the disease from becoming spread. They have also seen that hospital nurses eat, sleep, washing dishes, etc. in such a single place.

To stop this same nursing staff that lives in the health facilities shortly, I suggest that perhaps it should be broken into combinations of 4 individuals and relocated to newly built hospitals, in which they can stay individually and retain hygiene, preserve the atmosphere, help educate, etc. Those who should be accountable. Each clinic and physicians visit every day to examine their clients.

Many patients might well be healed without great faith using these specific HIV/AIDS measures because all norms and laws are observed and mainly social distance is maintained of each patient. Its clinics, as well as the environment, must be sanitized but kept from outside so that the community members recover the confidence and trust of the health departments beforehand. More importantly, a regulatory body should be established that can monitor the operations properly, handle cash, etc. to run smoothly.

Here are a few of my suggestions and promise that it could be successful if it is kept & rigorously implemented.

Thank you and regards,

Joy Roy.

(PR. Director)

Answer:

Explanation:

From the given information;

There is no change or any difference in velocity in between the inlet and the outlet.

Therefore by using Bernoulli's equation, we have:

[tex]\dfrac{V_1^2}{2g}+ \dfrac{P_1}{\gamma}+ z_1 + Epump= \dfrac{V_2^2}{2g}+ \dfrac{P_2}{\gamma}+ z_2+ H_L[/tex]

By dividing like terms on both sides, the equation is reduced to:

[tex]z_1 + E_{pump} = z_2+H_L \\ \\ E_{pump} =(z_2-z_1)+H_L[/tex]

where;

[tex]\Delta z = 400[/tex]

[tex]\Delta z = z_2-z_1[/tex]

[tex]\text{total head loss}= 408.5[/tex]

[tex]E_{pump} =(400)+408.5[/tex]

[tex]E_{pump} = 808.5 \ ft[/tex]

The required power input can be determined by using the formula:

[tex]P= \dfrac{\gamma_wQH_{pump}}{\eta}[/tex]

Assuming the missing pump efficiency = 70% and the flow rate Q= 1.34

Then:

[tex]P= \dfrac{62.40\times 1.34 \times 808.5}{0.7}[/tex]

[tex]P = \dfrac{96576.48 \ ft.lb/s}{550\dfrac{ ft*lb/s}{hp}}[/tex]

P = 175.594 hp

Answer:

lalalalapumpe

Explanation:

Answer:

[tex]GM<0[/tex]

So the bouy does not float with its axis vertical

Explanation:

From the question we are told that:

Diameter [tex]d=2m[/tex]

Length [tex]l=2.5m[/tex]

Weight [tex]W=22kN[/tex]

Specific weight of sea water [tex]\mu= 10.25kN/m^3[/tex]

Generally the equation for weight of cylinder is mathematically given by

Weight of cylinder = buoyancy Force

[tex]W=(pwg)Vd[/tex]

Where

[tex]V_d=\pi/4(d)^2y[/tex]

Therefore

[tex]22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m[/tex]

Therefore

Center of Bouyance B

[tex]B=\frac{y}{2}=0.26m\\\\B=0.75[/tex]

Center of Gravity

[tex]G=\frac{I.B}{2}=2.6m[/tex]

Generally the equation for\BM is mathematically given by

[tex]BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\[/tex]

Therefore

[tex]BG=2.6-0.476\\\\BG=0.64m[/tex]

Therefore

[tex]GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\[/tex]

Therefore

[tex]GM<0[/tex]

So the bouy does not float with its axis vertical

Answer:

Haha I'm a great guy but my friend has been in a day of the day and a lot to be able and I'm happy holi and a lot to the world of the day and day to

Answer:

the SI base unit of electrical current.

Answer:

An ampere is the SI base unit of electrical current

Answer:

2.68

Explanation:

Percentage by Mass of each Aggregate :

Pa = 65% ; Pb = 35%;

Bulk Specific gravity of each aggregate :

Ga = 2.45 ; Gb = 3.25

Gsb = (Pa + Pb) / (Pa/Ga + Pb/Gb)

Gsb = (65 + 35) / (65/2.45 + 35/3.25)

Gsb = (65 + 35) / 37.299843

Gsb = 100 / 37.299843

Gsb = 2.68

Answer:

The correct solution is "20 volt".

Explanation:

Given that:

Current,

I = 10 mA

or,

 = [tex]10\times 10^{-3} \ A[/tex]

Resistance,

R = 2 K ohm

or,

  = [tex]1\times 10^3 \ ohm[/tex]

Now,

The voltage will be:

⇒ [tex]V=IR[/tex]

By putting the values, we get

        [tex]=10\times 10^{-3}\times 2\times 10^{3}[/tex]

        [tex]= 20 \ volt[/tex]

Explanation:

Light travels in straight lines

Once light has been produced, it will keep travelling in a straight line until it hits something else. Shadows are evidence of light travelling in straight lines. An object blocks light so that it can't reach the surface where we see the shadow.

Answer:

The two factors are Air and Object

Answer:

The head added by the pump is approximately 51.8 meters

Explanation:

The given parameters of the water are;

The initial diameter of the pump ≈ The final diameter of the pump

The inlet side pressure, p₁ = 30 kPa

The intended outlet side pressure, p₂ = 500 kPa

The power of the pump, P = ρ·g·Q·H

Where, the pressure added by the pump, Δp = p₂ - p₁ = ρ·g·H

ρ = The density of the water ≈ 997 kg/m³

g = The acceleration due to gravity ≈ 9.81 m/s²

H = The head added by the pump

Therefore, we have;

500 kPa - 30 kPa = 997 kg/m³ × 9.81 m/s² × H

H = (500 kPa - 30 kPa)/(997 kg/m³ × 9.1 m/s²) ≈ 51.8 m

The head added by the pump, H ≈ 51.8 meters.