As an attempt to minimize energy requirements, a new residence has been constructed in Dallas, Texas (100°F dry bulb; 20°F daily range; W=0.0156). Size the air-conditioning unit (Btu/h) for this residence with the following features:
• No windows
• Inside design conditions 75°F, 60% rh (W = 0.0112)
• One 3 ft by 7 ft, 2 in. thick wood door with storm door on south side, U=0.26 Btu/h ft². °F, one 3 ft by 7 ft, 2 in. thick

Answers

Answer 1

The Air-conditioning unit for this residence should be sized to handle a cooling load of approximately 25,816.5 Btu/h

To size the air-conditioning unit for the residence, we need to calculate the cooling load. The cooling load is the amount of heat that needs to be removed from the indoor space to maintain the desired temperature and humidity conditions.First, we need to calculate the heat gain through the walls and doors of the residence. Let's assume the walls have an insulation value (U-value) of 0.2 Btu/h ft². °F.Walls:Area: Assuming the residence has four walls of equal size, each wall will have an area of (height) × (width) = (8 ft) × (3 ft) = 24 ft².

Heat gain: (Area) × (U-value) × (Temperature difference) = (24 ft²) × (0.2 Btu/h ft². °F) × (100°F - 75°F) = 180 Btu/h.Door:

Area: The door has an area of (height) × (width) = (7 ft) × (3 ft) = 21 ft².Heat gain: (Area) × (U-value) × (Temperature difference) = (21 ft²) × (0.26 Btu/h ft². °F) × (100°F - 75°F) = 136.5 Btu/h.

Next, we calculate the sensible heat load based on the indoor design conditions Sensible heat load:Area of the residence: Assuming a rectangular shape, let's assume a floor area of (length) × (width) = (40 ft) × (30 ft) = 1200 ft².

Sensible heat load: (Area) × (Temperature difference) × (Sensible heat factor) = (1200 ft²) × (100°F - 75°F) × (0.85) = 25500 Btu/h.

The sensible heat factor of 0.85 is an approximation that takes into account factors such as occupancy, appliances, and lighting.Now, we add up the heat gains from the walls, door, and sensible heat load:

Total cooling load = Heat gain from walls + Heat gain from door + Sensible heat load

= 180 Btu/h + 136.5 Btu/h + 25500 Btu/h

= 25816.5 Btu/h

Therefore, the air-conditioning unit for this residence should be sized to handle a cooling load of approximately 25,816.5 Btu/h.

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Answer 2

To size the air-conditioning unit for the new residence in Dallas, we need to calculate the cooling load. The cooling load is the amount of energy required to maintain the inside temperature and humidity at the desired levels.

Since there are no windows in the residence, we don't need to consider their heat gain. However, we need to take into account the heat gain from the door. The total area of the door is 21 square feet (3 x 7), and its U-value is 0.26 Btu/h ft². °F. Therefore, the door contributes to a heat gain of 1.386 Btu/h °F (21 x 0.26).

To calculate the sensible cooling load, we need to use the inside design conditions of 75°F and 60% rh. The moisture content of the air (W) is 0.0112. The sensible cooling load is calculated as follows:

Sensible cooling load (Btu/h) = 1.1 x CFM x (Tin - Tout) + Qdoor
where CFM is the air flow rate, Tin is the inside air temperature, Tout is the outside air temperature, and Qdoor is the heat gain from the door.

Since we don't have information about the air flow rate, we can assume a value of 400 CFM per ton of cooling. Let's assume that we want to maintain the inside temperature at 75°F and the outside temperature is 100°F. The sensible cooling load is calculated as follows:

Sensible cooling load (Btu/h) = 1.1 x 400/12 x (75 - 100) + 1.386
= -1,098 Btu/h + 1.386
= -1,096 Btu/h (negative value indicates a cooling load)

We have a negative sensible cooling load because the inside temperature is higher than the outside temperature. This means that we don't need to provide any sensible cooling, but we need to remove the heat gain from the door. Therefore, the air-conditioning unit needs to have a capacity of at least 1.386 Btu/h to maintain the inside temperature and humidity at the desired levels.

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Related Questions

... is an actual sequence of interactions (i.e., an instance) describing one specific situation; a ... is a general sequence of interactions (i.e., a class) describing all possible ... associated with a situation. ... are used as examples and for clarifying details with the client. ... are used as complete descriptions to specify a user task or a set of related system features.

Answers

Answer:

ScenarioUse caseScenariosScenariosUse case

Explanation:

A scenario is an actual sequence of interactions (i.e., an instance) describing one specific situation; a use case is a general sequence of interactions (i.e., a class) describing all possible scenarios associated with a situation. Scenarios are used as examples and for clarifying details with the client. Use cases are used as complete descriptions to specify a user task or a set of related system features.

Suppose you are choosing between four different desktop computers: one is an Apple Mac Intosh and the other three are PC-compatible computers that use a Pentium 4, an AMD processor (using the same compiler as the Pentium 4), and a Pentium 5 (which does not yet exist in 2004 but has the same architecture as the Pentium 4 and uses the same compiler). Which of the following statements are true?
a. The fastest computer will be the one with the highest clock rate.
b. Since all PCs use the same Intel-compatible instruction set and execute the same number of instructions for a program, the fastest PC will be the one with the highest clock rate.
c. Since AMD uses different techniques than Intel to execute instructions,they may have different CPIs. But, you can still tell which of the two Pentium-based PCs is fastest by looking at the clock rate.
d. Only by looking at the results of benchmarks for tasks similar to your workload can you get an accurate picture of likely performance.

Answers

Answer:

d.

Explanation:

A storm with a duration of about 24 hours resultsin the following hydrograph at a gaging station on a river. The flow was 52 cubic meters per second (cms) before the rain began. The drainage area above the gaging station is 1,450 square kilometers. Use the observed hydrograph to develop a 24-hour rainfall duration unit hydrograph for this watershed.
Time 0 6 12 18 24 30 36 42 48 54 60 66 72 78 84
(Hours)
Flow 52 52 55 56 97 176 349 450 442 370 328 307 280 259 238
(m3/s)
Time 90 66 102 108 114 120 126 132 138 144 155 156 162 168 176
(Hours)
(m3/s) 214 193 173 145 135 124 114 107 97 86 79 66 62 58 52

Answers

Answer:

attached below

Explanation:

Given data:

Gaged flow = 52 m^3 / sec

Depth covering drainage area = 1450 km^2

Develop a 24-hour rainfall duration unit hydrograph for the watershed using observed hydorgraph

Runoff flow = gaged flow - base flows

                   = 52 - 52 = 0 m^3/sec

For 18 hours time duration

Direct runoff volume ( Vr ) = ∑ ( QΔt1 )

where Δt = 6

           ∑Q = 3666 m^3/sec

hence Vr = Δt * ∑Q  = 79185600 m^3

Next we will convert the Direct runoff volume to its equivalent depth covering the drainage area

= Vr / drainage area depth

= 79185600 / 1450000000 = 5.46 cm

Next we will find the unit hydrograph flows by applying the relation below

[tex]Q_{1.0cm} = Q_{5.46cm} (\frac{1.0cm}{5.46cm} )[/tex]

where 14m^3/sec = [tex]Q_{5.6cm}[/tex]

input value back to the above relation

[tex]Q_{1.0cm} = 2.57 m^3/sec[/tex]

Attached below is The remaining part of the solution  tabulated below

and A graph of the unit hydorgraph for the given watershed

Note :

Base  flow total = 1560

UH total = 671.30

Describe how to contribute to
zero/low carbon work outcomes
within the built environment.

Answers

Answer:

day if you workout without Zero billing that means you're not sweating. Sweating you're not losing anything that means you have zero outcomes

Explanation:

During a shrinkage limit test, a 19.3 cm3 saturated clay sample with a mass of 37 g was placed in a porcelain dish and dried in the oven. The oven-dried sample had a mass of 28 g with a final volume of 16 cm3 . Determine the shrinkage limit and the shrinkage ratio.

Answers

Answer:

shrinkae limit = 20.35%

shrinkage ratio = 1.45

Explanation:

1. to get the shrinkage limit we would first calculate the moisture content w.

w = (37-28)/28

= 9/28

= 0.3214

then the formula for shrinkage limit is

[tex][w-\frac{(V-Vd}{wd} ]*100[/tex]

w = 0.3214

V = 19.3

Vd = 16

Wd = 28

when we put these values into the formula:

[tex][0.3214-\frac{(19.3-16)}{28} ]*100\\[/tex]

= 20.35%

2. the shrinkage limit = Wd/V

= 28/19.3

= 1.45

a load of 12tonnes is put along a horizontal plane by a force at 30°to and above the flat. if the coefficient of sliding friction is 0.2 find the frictional force​

Answers

Answer:

20368.917N

Explanation:

Frictional force (F) is the product of the Coefficient of friction and the normal reaction.

F = μN

Coefficient of friction, μ = 0.2

Normal reaction = MgCosθ

Mass, m = 12 tonnes = 12 * 1000 = 12000 kg

N = 12000 * 9.8 * cos30

N = 101844.58

F = 0.2 * 101844.58

F = 20368.917N

The movement of the piston and connecting rod assembly driven by the force of combustion turns the ________.

Answers

Answer:

the answer. for this is Crankshaft

Convert an acceleration of 12m/s² to km/h²​

Answers

Here is your answer , hope you like it

The pressure less than atmospheric pressure is known as:

Suction pressure

Negative gauge pressure

Vacuum pressure

All of the above

Answers

Answer:

answer is option (d) all of the above

Select the correct answer.
Alan is walking on a street and passes by a very busy construction site. He sees engineers hanging one end of various objects to the ceiling and the other end to heavy
weights. They observe the objects and then then take down notes. What are the engineers doing

Answers

Answer: the engineers are performing a tensile test

Explanation:

i took the test and got it right

If the sum of the two numbers is 4 and the sum of their squares minus three times their product is 76,find the number

Answers

Answer:

-2 and 6

Explanation:

Let "x" and "y" be 2 numbers.

The sum of the two numbers is 4. The mathematical expression is:

x + y = 4

y = 4 - x   [1]

The sum of their squares minus three times their product is 76. The mathematical expression is:

x² + y² - 3 x y = 76   [2]

If we substitute [1] in [2], we get:

x² + (4 - x)² - 3 x (4 - x) = 76

x² + 16 - 8 x + x² - 12 x + 3 x² = 76

5 x² - 20 x - 60 = 0

We apply the solving formula for second order equations and we get x₁ = 6 and x₂ = -2.

If we replace these x values in [1], we get:

y₁ = 4 - x₁ = 4 - 6 = -2

y₂ = 4 - x₂ = 4 - (-2) = 6

As a consequence, one of the numbers is 6 and the other is -2.

A cylindrical bar of metal having a diameter of 15.7 mm and a length of 178 mm is deformed elastically in tension with a force of 49100 N. Given that the elastic modulus and Poisson's ratio of the metal are 67.1 GPa and 0.34 respectively, Determine the following:(a) The amount by which this specimen will elongate (in mm) in the direction of the applied stress. The entry field with incorrect answer now contains modified data.(b) The change in diameter of the specimen (in mm). Indicate an increase in diameter with a positive number and a decrease with a negative number.

Answers

Answer:

0.6727

-0.02017

Explanation:

diameter = 15.7

l = 178

E =elastic modulus = 67.1 Gpa

poisson ratio = 0.34

p = force = 49100N

first we calculate the area of the cross section

[tex]A=\frac{\pi }{4} d^{2}[/tex]

[tex]A=\frac{\pi }{4} (15.7)^{2} \\A = \frac{774.683}{4} \\[/tex]

A = 193.6mm²

1. Change in directon of the applied stress

[tex]= \frac{pl}{AE}[/tex]

= 49100*178/193.6*67.1*10³

= [tex]=\frac{8739800}{12990560}[/tex]

δl = 0.6727  mm

2. change in diameter of the specimen

equation for poisson distribution =

m = -(δd/d) / (δl/l)

0.34 = (δd/15.7) / (0.6727/178)

0.34 = (-δd * 178) / 15.7 * 0.6727

0.34 = -178δd / 10.56139

we cross multiply

10.56139*0.34 =-178δd

3.5908726 = -178δd

δd = 3.5908726/-178

δd = -0.02017 mm

the change in dimeter has a negative sign. it decreases

Which of the following choices accurately contrasts a categorical syllogism with a conditional syllogism?


An argument constructed as a categorical syllogism uses deductive reasoning whereas an argument constructed as a conditional syllogism uses inductive reasoning.

A categorical syllogism contains two premise statements and one conclusion whereas a conditional syllogism contains one premise statement and one conclusion.

A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.

An argument constructed as a categorical syllogism is valid whereas an argument constructed as a conditional syllogism is invalid.

Answers

Answer:

The correct option is - A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.

Explanation:

As,

Categorical syllogisms follow an "If A is part of C, then B is part of C" logic.

Conditional syllogisms follow an "If A is true, then B is true" pattern of logic.

So,

The correct option is - A categorical syllogism argues that A and B are both members of C whereas a conditional syllogism argues that if A is true then B is also true.

I don’t know the answer to this question

Answers

Answer:

I dont know the answer either

Explanation:

Answer:

flux

Explanation:

This is silence I couldent find the tab... 30 points plus marked brainliest if corrects!


The most recent evidence supporting the theory of plate tectonics would include
es )
A)
GPS monitoring of plate speeds and movements.
B)
the WWII discovery of paleomagnetic reversals.
Elimi
O
the 1963 mapping of the tectonic plate boundaries.
D
C-14 dating of marine fossils found in the Himalayas.

Answers

Yeah the answer is A

Find the derivative of x ​

Answers

Answer:

this is your answer. if mistake don't mind.

1. What is the productivity rate using cycle time for the following information:
I
Type of Work – Hauling
Average Cycle Time – 35 Minutes
Truck Capacity – 25 Tons
Crew - One Driver
Productivity Factor - 0.85
System Efficiency – 55 Minutes
per
Hour

Answers

ndjdjdhdjshdhdjdjjdjdjdjdhdhhdhdhdhdhdhdhdhdhdhdh

Now, you get a turn to practice writing a short program in Scratch. Try to re-create the program that was shown that turns the sprite in a circle. After you have completed that activity, see if you can make one of the improvements suggested. For example, you can try adding a sound. If you run into problems, think about some of the creative problem-solving techniques that were discussed.

When complete, briefly comment on challenges or breakthroughs you encountered while completing the guided practice activity.


Pls help im giving 100 points for this i have this due in minutes

Answers

Answer:

u need to plan it out

Explanation:

u need to plan it out

Answer:

use the turn 1 degrees option and put a repeat loop on it

Explanation:

u can add sound in ur loop

Which of the following devices is a simple machine?

A.
an engine
B.
a pulley
C.
a motor
D.
a bicycle
E.
a crane

Answers

Answer:

A PULLY

Explanation:

HAD THIS ONE THAT IS THE CORRECT ANWSER

Answer:

The answer is B. a pulley

Explanation:

I hope I answered your question:)

Travel Time Problem: Compute the time of concentration using the Velocity, Sheet Flow Method for Non-Mountainous Orange County and SCS method at a 25 year storm evert.
Location Slope (%) Length (ft) Land Use
1 4.5 1000 Forest light underbrush with herbaceous fair cover.
2 2.5 750 Alluvial Fans (eg. Natural desert landscaping)
3 1.5 500 Open Space with short grasses and good cover
4 0.5 250 Paved Areas (1/4 acre urban lots)

Answers

Answer:

Total time taken = 0.769 hour

Explanation:

using the velocity method

for sheet flow ;

Tt = [tex]\frac{0.007(nl)^{0.8} }{(Pl)^{5}s^{0.4} }[/tex]  

Tt = travel time

n = manning CaH

Pl = 25years

L = how length ( ft )

s = slope

For Location ( 1 )

s = 0.045

L = 1000 ft

n = 0.06 ( from manning's coefficient table )

Tt1 = 0.128 hour

For Location ( 2 )

s = 2.5 %

L= 750

n = 0.13

Tt2 = 0.239 hour

For Location ( 3 )

s = 1.5%

L = 500 ft

n = 0.15

Tt3 = 0.237  hour

For Location (4)

s = 0.5 %

L = 250 ft

n = 0.011

Tt4 = 0.165 hour

hence the Total time taken = Tt1 + Tt2 + Tt3 + Tt4

                                              = 0.128 + 0.239 + 0.237 + 0.165 = 0.769 hour

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [100] direction. If the critical resolved shear stress for this material is 2.00 MPa, calculate the magnitude of applied stress necessary to cause slip to occur on the (111) plane in the direction.

Answers

Answer:

Explanation:

From the given information:

The equation for applied stress can be expressed as:

[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda}[/tex]

where;

[tex]\phi[/tex] = angle between the applied stress [100] and [111]

To determine the [tex]\phi[/tex] and [tex]\lambda[/tex] for the system

Using the equation:

[tex]\phi= cos^{-1}\Big [\dfrac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)}}\Big][/tex]

for [100]

[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]

for [111]

[tex]l_1 = 1 , m_1 = 1, n_1 = 1[/tex]

Thus;

[tex]\phi= cos^{-1}\Big [\dfrac{1*1+0*1+0*1}{\sqrt{(1^2+0^2+0^2)(1^2+1^2+1^2)}}\Big][/tex]

[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(3)}}\Big][/tex]

[tex]\phi= 54.74^0[/tex]

To determine  [tex]\lambda[/tex]  for [tex][1 \overline 1 0][/tex]

where;

for [100]

[tex]l_1 = 1, m_1 = 0, n_1 = 0[/tex]

for [tex][1 \overline 1 0][/tex]

[tex]l_1 = 1 , m_1 = -1, n_1 = 0[/tex]

Thus;

[tex]\lambda= cos^{-1}\Big [\dfrac{1*1+0*1+0*0}{\sqrt{(1^2+0^2+0^2)(1^2+(-1)^2+0^2)}}\Big][/tex]

[tex]\phi= cos^{-1}\Big [\dfrac{1}{\sqrt{(2)}}\Big][/tex]

[tex]\phi= 45^0[/tex]

Thus, the magnitude of the applied stress can be computed as:

[tex]\sigma_{app} = \dfrac{\tau_{CRSS}}{cos \phi \ cos \lambda }[/tex]

[tex]\sigma_{app} = \dfrac{2.00}{cos (54.74) \ cos (45) }[/tex]

[tex]\mathbf{\sigma_{app} =4.89 \ MPa}[/tex]

An insulated closed piston–cylinder device initially contains 0.3 m3 of carbon dioxide at 200 kPa
and 27°C. A resistance heater inside the cylinder is turned on and supplied heat to the gas. As a
result, the gas expanded by pushing the piston up, until the volume doubled. During this process,
6
the pressure changed according to = 4, in which the constant 6 has units of kPa.m
a) Find the mass of the hydrogen in the tank in kg.
b) Determine the work done by the gas in kJ.

Answers

I=4.7 A hope it helps

To solve this problem, we can use the ideal gas law and the equation for polytropic process.

What is ideal gas law ?

The ideal gas law is a fundamental law of physics that describes the behavior of an ideal gas. It relates the pressure, volume, temperature, and number of particles of a gas using the following equation:

PV = nRT

a) First, we need to find the mass of the carbon dioxide in the tank. The ideal gas law is:

PV = mRT

where P is the pressure, V is the volume, m is the mass, R is the universal gas constant, and T is the temperature. Rearranging for the mass, we get:

m = PV / RT

Substituting the given values, we have:

m = (200 kPa)(0.3 m3) / [(0.287 kPam3/kgK)(27°C + 273.15)] = 3.87 kg

So the mass of the carbon dioxide in the tank is 3.87 kg.

b) To determine the work done by the gas during the process, we can use the equation for polytropic process:

P1V1^n = P2V2^n

where P1 and V1 are the initial pressure and volume, P2 and V2 are the final pressure and volume, and n is the polytropic index. Substituting the given values, we have:

(200 kPa)(0.3 m3)^n = (4)(0.6 m3)^n

Dividing both sides by (0.3 m3)^n and taking the logarithm of both sides, we get:

log(200) + nlog(0.3) = log(4) + nlog(0.6)

Solving for n, we get:

n = log(4/200) / log(0.6/0.3) ≈ 1.235

Using the polytropic work equation:

W = (P2V2 - P1V1) / (1 - n)

Substituting the given values, we have:

W = [(4 kPa)(0.6 m3) - (200 kPa)(0.3 m3)] / (1 - 1.235) = 233.7 kJ

So the work done by the gas during the process is 233.7 kJ.

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Which option identifies what is missing in the categorical syllogistic argument below? Quadrupeds have four legs so horses are quadrupeds.
Conclusion
Minor premise
Inference
Major premise

Answers

It is a inference because there isn’t any facts stated

If the hypotenuse of a right triangle is 12 and an acute angle is 37 degrees find leg a and leg b lengths

Answers

scrity añao devid codicie

An air heater may be fabricated by coiling Nichrome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter D=1 mm, electrical resistivity rhoe=10−6Ω⋅m, thermal conductivity k=25W/m⋅K, and emissivity ε=0.20. The heater is designed to deliver air at a temperature of T[infinity]=50∘C under flow conditions that provide a convection coefficient of h=250W/m2⋅K for the wire. The temperature of the housing that encloses the wire and through which the air flows is Tsur=50∘C. If the maximum allowable temperature of the wire is Tmax=1200∘C, what is the maximum allowable electric current I? If the maximum available voltage is ΔE=110V, what is the corresponding length L of wire that may be used in the heater and the power rating of the heater? Hint: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this assumption.

Answers

Solution :

Assuming that the wire has an uniform temperature, the equivalent convective heat transfer coefficient is given as :

[tex]$h_T= \epsilon \sigma (T_s+T_{surr})(T_s^2 +T^2_{surr})$[/tex]

[tex]$h_T= 0.20 \times 5.67 \times 10^{-8} (1473+323)(1473^2 +323^2)$[/tex]

[tex]$h_T=46.3 \ W/m^2 .K$[/tex]

The total heat transfer coefficient will be :

[tex]$h_T=(250+46.3) \ W/m^2 .K$[/tex]

    [tex]$=296.3 \ W/m^2 .K$[/tex]

Now calculating the maximum volumetric heat generation :

[tex]$\dot {q}_{max}=\frac{2h_t}{r_0}(T_s-T_{\infty})$[/tex]

[tex]$\dot {q}_{max}=\frac{2\times 296.3}{0.0005}(1200-50)$[/tex]

        [tex]$= 1.362 \times 10^{9} \ W/m^3$[/tex]

The heat generation inside the wire is given as :

[tex]$\dot{q} = \frac{I^2R}{V}$[/tex]

Here, R is the resistance of the wire

         V is the volume of the wire

∴ [tex]$\dot{q} = \frac{I^2\left( \rho \times \frac{L}{A} \right)}{A \times L}$[/tex]

     [tex]$=\frac{I^2 \rho}{\left(\frac{\pi}{4}D^2 \right)}$[/tex]

where, ρ is the resistivity.

[tex]$I_{max}= \left(\frac{\dot{q}_{max}}{\rho} \right)^{1/2} \times \frac{\pi}{4}D^2$[/tex]

[tex]$I_{max}= \left(\frac{1.36 \times 10^9}{10^{-6}} \right)^{1/2} \times \frac{3.14}{4}(1 \times 10^{-3})^2$[/tex]

        = 28.96 A

Now considering the relation for the current flow through the finite potential difference.

[tex]$E=I_{max} \times R$[/tex]

[tex]$E=I_{max} \times \rho \times \frac{L}{A}$[/tex]

[tex]$L=\frac{AE}{I_{max} \ \rho}$[/tex]

[tex]$L=\frac{\frac{\pi}{4} \times (1 \times 10^{-3})^2 \times 110}{28.96 \times 10^{-6}}$[/tex]

   = 2.983 m

Now calculating the power rating of the heater:

[tex]$P= E \times I_{max}$[/tex]

[tex]$P= 110 \times 28.96}$[/tex]

   = 3185.6 W

   = 3.1856 kW

Where does Burj Khalifa located? and how many meters?​

Answers

Answer:

Burj Khalifa is located in dubai UAE at over 828m

Explanation:

828 metres

Answer:

The Burji Khalifa, known as the Burj Dubai prior to its inauguration in 2010, is a skyscraper in Dubai, United Arab Emirates. With a total hight of 829.8 m and a roof hight of 828 m, the Burji Khalifa has been the tallest structure and building in the world since its topping out in 2009.

What current works best when the operator
encounters magnetic arc blow?

•DCEP

•ACEN

•CC

•AC

Answers

Answer:

AC

Explanation:

One situation when alternating current would work better than direct current is if the operator is encountering magnetic arc blow.

Current works best when the operator  encounters magnetic arc blow is AC

Magnetic arc blow is simply defined as the arc deflection due to the warping of the magnetic field that is produced by electric arc current.

This is caused as a result of the following;

- if the material being welded has residual magnetism at an intolerable level

- When the weld root is being made, and the welding current is direct current which indicates constant direction and maintains constant polarity (either positive or negative).

Since it is caused by DC(Direct Current) which means constant polarity , it means the opposite will be better which is AC(alternating current) because it means that electricity direction will be switching to and fro and as such the polarity will also be revered in response to this back and forth switch manner.

Thus, Current works best when the operator  encounters magnetic arc blow is AC

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Two sites are being considered for wind power generation. On the first site, the wind blows steadily at 7 m/s for 3000 hours per year. On the second site, the wind blows steadily at 10 m/s for 2000 hours per year. The density of air on the both sites is 1.25 kg/m3 . Assuming the wind power generation is negligible during other times.Calculate the maximum power of wind on each site per unit area, in kW/m2 .

Answers

Solution :

Given :

[tex]$V_1 = 7 \ m/s$[/tex]

Operation time, [tex]$T_1$[/tex] = 3000 hours per year

[tex]$V_2 = 10 \ m/s$[/tex]

Operation time, [tex]$T_2$[/tex] = 2000 hours per year

The density, ρ = [tex]$1.25 \ kg/m^3$[/tex]

The wind blows steadily. So, the K.E. = [tex]$(0.5 \dot{m} V^2)$[/tex]

                                                             [tex]$= \dot{m} \times 0.5 V^2$[/tex]

The power generation is the time rate of the kinetic energy which can be calculated as follows:

Power = [tex]$\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$[/tex]

Regarding that [tex]$\dot m \propto V$[/tex]. Then,

Power [tex]$ \propto V^3$[/tex] → Power = constant x [tex]$V^3$[/tex]

Since, [tex]$\rho_a$[/tex] is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, [tex]$P_1= \text{const.} \times V_1^3$[/tex]

            [tex]$P_1 = \text{const.} \times 343 \ W$[/tex]

For the second site,

Power, [tex]$P_2 = \text{const.} \times V_2^3 \ W$[/tex]

           [tex]$P_2 = \text{const.} \times 1000 \ W$[/tex]

Tech A says that LED brake lights illuminate faster than incandescent bulbs. Tech B says that LED brake lights have
more visibility and last longer. Who is correct?

Answers

Answer:

Both

Explanation:

A sign structure on the NJ Turnpike is to be designed to resist a wind force that produces a moment of 25 k-ft in one direction. The axial load is 30 kips. Soil conditions consist of a normally consolidated clay layer with following properties; su=800 psf, andγsat= 110 pcf. Design for a FOS of 3. Assume frost depth to be 3ft below grade

Answers

Solution :

Finding the cohesion of the soil(c) using the relation:

[tex]$c = \frac{q_u}{2}$[/tex]

Here, [tex]$q_u$[/tex] is the unconfined compression strength of the soil;

[tex]$c = \frac{800}{2}$[/tex]

   = 400 psf

∴ The cohesion value is greater than 0

So the use of the angle of internal friction is 0

Referring to the table relation between bearing capacity factors and angle of internal friction.

For the angle of inter friction [tex]$0^\circ$[/tex]

    [tex]$N_c = 5.14$[/tex]

   [tex]$N_q = 1.0$[/tex]

   [tex]$N_r = 0$[/tex]

Therefore,

[tex]$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$[/tex]

     =  2386 psf

∴ Allowable bearing capacity [tex]$q_{a} = \frac{Q_{allow}}{A}$[/tex]

                                                     [tex]$=\frac{30}{B^2}$[/tex]

∴ [tex]$q_a = \frac{q_{ult}}{F.O.S}$[/tex]

  [tex]$\frac{30}{B^2} = \frac{2386}{3}$[/tex]

∴ B = 0.2 ft

Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft

                                                                                 [tex]$=0.04 \ ft^2$[/tex]

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