Question 2 Given the following gas Volume (%) Component Methane 58.7 Ethane 16.5 9.9 Propane Butanes 5.0 Pentanes 3.5 Hydrogen Suflide 6.4 100 Total I Calculate: 1. Average molecular weight 1. Density at 320F and 14.7 psia ni. Density at OoC and 760 mm mercury iv. Specific gravity and gross and net heating value​

Answer:

[tex]E_f=400<-17.4volts[/tex]

Explanation:

From the question we are told that:

Load [tex]V=480[/tex]

Poles [tex]p=6[/tex]

Power [tex]P=150hp[/tex]

3-Phase

Load:

[tex]Tl=0.05*\omega_s^2Nm[/tex]

Motor:

[tex]Ef=400V\\\\X_d=1ohm[/tex]

Generally the equation for Synchronous speed is mathematically given by

[tex]N_s=\frac{120F}{p}=\frac{120*60}{6}[/tex]

[tex]N_s=1200rpm\\\\N_s=125.66 rads/sec[/tex]

Therefore

[tex]Tl=0.05*\omega_s2Nm[/tex]

With

[tex]\omega=N_s[/tex]

We have

[tex]Tl=0.05*(125.66)^2Nm[/tex]

[tex]T_l=789.52 Nm[/tex]

Therefore

Load Power

[tex]P_l=T_l*\omega_s\\\\P_l=789.52*125.66[/tex]

[tex]P_l=9922watts[/tex]

Generally the equation for Load Power is mathematically given by

[tex]P_l=\frac{\sqrt{3}*E_f.V_t}{x_d}*sin\theta\\\\9922=\frac{\sqrt{3}*480*400}{1}*sin\theta[/tex]

[tex]\theta=17.4 \textdegre3[/tex]

Therefore

Voltage

[tex]E_f=400<-17.4volts[/tex]

Answer:

A-personal income in a year this is the answer

Explanation:

We assume the T flip-flop changes state on the rising edge of the clock input.

The first stage is connected to the clock. The second stage clock is connected to the inverse of the Q output of the first stage, so that when the first stage Q makes a 1 to 0 transition, the second stage changes state.

Answer:

you have to report to the federal aviation administration upon request

Explanation:

An unmanned aircraft can also be a drone it is an aircraft that does not have passengers or a human pilot.

Such an aircraft could be fully autonomous but most times they have a human pilot who controls it remotely.

Now you have to report this deviation to the federal aviation administration. This body is responsible for the enforcement of regulations that have to do with the manufacture, operation and maintenance of aircrafts. The body makes sure there is an efficient and safe system for a navigation and they also control air traffic.

Answer:

The governments receiving aid were generally experienced in industrial development. ... During the 1950s, little attention was given to differences in the Third World's conditions and needs, until these appeared to create obstacles to achieving high levels of industrial output

I hope it is helpful

Answer:

B

Explanation:

Anxiety is very common especially nowadays but it's especially common in psychological disorders

Answer:

The main sections included in the bill of quantities are Form of Tender, Information, Requirements, Pricing schedule, Provisional sums, and Day works.

Answer:

a. Composition of Vector.

Explanation:

When a bird flies in the air, it stretches its wings into air and this movement helps it move in a certain direction. This is an example of composition of vector. Air strikes the wings in opposite direction and bird wing movement helps it move against the wind.

Answer:

I think it would be C but im not sure sorry if its wrong

Explanation:

Network and Switching Subsystem (NSS)

Base-Station Subsystem (BSS)

Mobile station (MS)

Operation and Support Subsystem (OSS)

Answer:

  4.5 m/s = 16.2 km/h

Explanation:

The speed is the ratio of distance to time:

  speed = (180 m)/(40 s) = 4.5 m/s

Answer:

I didn't understand your question or is it a fun fact

Answer:

  5.9 watts

Explanation:

The secondary voltage is the primary voltage multiplied by the turns ratio:

  (120 V)(41) = 4920 V

The power is the product of voltage and current:

  (4920 V)(1.2·10^-3 A) = (4.92)(1.2) W = 5.904 W

The power consumed is about 5.9 watts.

Answer:

special type

Explanation:

As per the classification of milling cutters. This cutter can handle deep and long open slots in a more comfortable manner, which increase the productivity.

Answer:

manager and supervisor

Explanation:

Answer:

[tex]Q_2 = 32[/tex] mL/s

Explanation:

Given :

The flow is incompressible viscous flow.

The initial flow rate, [tex]Q_1[/tex] = 1 mL/s

Initial diameter, [tex]D_1= D_0[/tex]

Initial length, [tex]L_1=L_0[/tex]

The initial pressure difference to maintain the flow, [tex]P_1=P_0[/tex]

We know for a viscous flow,

[tex]$\Delta P = \frac{32 \mu V L}{D^2}$[/tex]

[tex]$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$[/tex]

[tex]$Q \propto \Delta P \times D^4$[/tex]

[tex]$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{1}{32}$[/tex]

∴ [tex]Q_2 = 32[/tex] mL/s

The flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0 is; Q2 = 32 mL/s

We are given;

Initial flow rate; Q1 = 1 mL/s

Initial uniform diameter; D0

Initial Length; L0

Initial Pressure difference; P0

Relationship between pressure, flow rate and diameter for vicious flow is given by;

Q1/Q2 = (P1/P2) × (D1/D2)⁴

Where;

Q1 is initial flow rate

Q2 is final flow rate

P1 is initial pressure difference

P2 is final pressure difference

D1 is initial diameter

D2 is final diameter

We are told that the pressure difference was increased to 2P0 and the diameter was increased to 2D0. Thus;

P2 = 2P0

D2 = 2D0

Thus;

1/Q2 = (P0/2P0) × (D0/2D0)⁴

>> 1/Q2 = ½ × (½)⁴

1/Q2 = 1/32

Q2 = 32 mL/s

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Answer:

Explanation:

In the development of a torsion formula for a circular shaft, the following assumptions are made: Material of the shaft is homogeneous throughout the length of the shaft. Shaft is straight and of uniform circular cross section over its length. Torsion is constant along the length of the shaft.

Answer:

  B. 10,000 ohms

Explanation:

|8000 +j6000| = √(8000² +6000²) = 1000√(64 +36) = 1000√100 = 10,000

The impedance is 10,000 ohms.

The correct responses are;

(i) Weight percent of nitrogen: 58.6%

Weight percent of oxygen: 14.65%

Weight percent of carbon dioxide: 29.59%

(ii) Volume percent of nitrogen: 64.38%

Weight percent of oxygen: 14.1%

Weight percent of carbon dioxide: 21.53%

(iii) Partial pressure of nitrogen: 19.314 psia

Partial pressure of oxygen: 4.23 psia

Partial pressure of carbon dioxide: 6.459 psia

(iv) Partial pressure of nitrogen: 19.314 psia

Partial pressure of oxygen: 4.23 psia

Partial pressure of carbon dioxide: 6.459 psia

(v) The average molecular weight is approximately 32.02 g/mole

(vi) At 20 psia, 60 °F, the density is 0.11275 lb/ft.³

At 14.7 psia, 60 °F, the density is 0.0844 lb/ft.³

At 14.7 psia, 32 °F, the density is 0.0892 lb/ft.³

(vii) The specific gravity of the mixture 0.169

Reasons:

The volume of the tank = 40 ft.³ = 1.132675 m³

Content of the tank = Carbon dioxide and air

The pressure inside the tank = 30 psia = 206843 Pa

The temperature of the tank = 70 °F ≈ 294.2611 K

Mass of carbon dioxide placed in the tank = 2 lb.

Percent of nitrogen in the tank = 80%

Percent of oxygen in the tank = 20%

(i) 2 lb ≈ 907.1847 g

Molar mass of carbon dioxide = 44.01 g/mol

Number of moles of carbon dioxide = [tex]\displaystyle \mathbf{ \frac{907.1847 \, g}{44.01 \, g/mol}} \approx 20.613 \, moles[/tex]

Assuming the gas is an ideal gas, we have;

[tex]\displaystyle n = \mathbf{\frac{206843\times 1.132674}{8.314 \times 294.2611} }\approx 95.7588[/tex]

The number of moles of nitrogen and oxygen = 95.7588 - 20.613 = 75.1458

Let x represent the mass of air in the mixture, we have;

[tex]\displaystyle \mathbf{ \frac{0.8 \cdot x}{28.014} + \frac{0.2 \cdot x}{32}}= 75.1458[/tex]

Solving gives;

x ≈ 2158.92 grams

Mass of the mixture = 2158.92 g + 907.1847 g ≈ 3066.1047 g

[tex]\displaystyle Weight \ percent \ of \ nitrogen= \frac{0.8 \times 2158.92}{3066.105} \times 100 \approx \underline{58.6 \%}[/tex]

[tex]\displaystyle Weight \ percent \ of \ oxygen = \frac{0.8 \times 2158.92}{3066.105} \times 100 \approx \underline{14.65\%}[/tex]

[tex]\displaystyle Weight \ percent \ of \ carbon \ dioxide = \frac{907.184}{3066.105} \times 100 \approx \underline{29.59\%}[/tex]

ii.

[tex]\displaystyle Number \ of \ moles \ nitrogen= \frac{0.8 \times 2158.92 \, g}{28.014 \, g/mol} \approx 61.65\, moles[/tex]

[tex]\displaystyle Number \ of \ moles \ oxygen= \frac{0.2 \times 2158.92 \, g}{32 \, g/mol} \approx 13.5\, moles[/tex]

Number of moles of carbon dioxide = 20.613 moles

Sum = 61.65 moles + 13.5 moles + 20.613 moles ≈ 95.763 moles

In an ideal gas, the volume is equal to the mole fraction

[tex]\displaystyle Volume \ percent \ of \ nitrogen= \frac{61.65}{95.763} \times 100 \approx \underline{64.38 \%}[/tex]

[tex]\displaystyle Volume \ percent \ of \ oxygen = \frac{13.5}{95.763} \times 100 \approx \underline{14.1\%}[/tex]

[tex]\displaystyle Volume \ percent \ of \ carbon \ dioxide = \frac{20.613 }{95.763} \times 100 \approx \underline{21.53\%}[/tex]

(iv) The partial pressure of a gas in a mixture, [tex]P_A = \mathbf{X_A \cdot P_{total}}[/tex]

Partial pressure of nitrogen, [tex]P_{N_2}[/tex] = 0.6438 × 30 psia ≈ 19.314 psia

Partial pressure of oxygen, [tex]P_{O_2}[/tex] = 0.141 × 30 psia ≈ 4.23 psia

Partial pressure of carbon dioxide, [tex]P_{CO_2}[/tex] = 0.2153 × 30 psia ≈ 6.459 psia

(v) The average molecular weight is given as follows;

[tex]\displaystyle Average \ molecular \ weight = \frac{3066.105\, g}{95.7588 \, moles} = 32.02 \, g/mole[/tex]

(vi) At 20 psia 70 °F, we have;

Converting to SI units, we have;

[tex]\displaystyle n = \frac{137895.1456\times 1.132674}{8.314 \times 294.2611} \approx 63.84[/tex]

The number of moles, n ≈ 63.84 moles

The mass = 63.84 moles × 32.02 g/mol  ≈ 2044.16 grams ≈ 4.51 lb

[tex]\displaystyle Density = \frac{4.51\ lb}{40 \ ft.^3} \approx \underline{0.11275\, lb/ft.^3}[/tex]

When the pressure is 14.7 psia = 101352.93 Pa, and 60 °F

[tex]\displaystyle n = \frac{101352.9\times 1.132674}{8.314 \times 288.7056} \approx 42.8247[/tex]

The mass = 47.8247 moles × 32.02 g/mol  ≈ 1531.35 grams = 3.376 lb.

[tex]\displaystyle Density = \frac{3.376 \ lb}{40 \ ft.^3} \approx \underline{ 0.0844\, lb/ft.^3}[/tex]

When the pressure is 14.7 psia = 101352.93 and 32 °F

[tex]\displaystyle n = \frac{101352.9\times 1.132674}{8.314 \times 273.15} \approx 50.5482[/tex]

The mass = 50.5482 moles × 32.02 g/mol  ≈ 1618.55 grams = 3.568 lb.

[tex]\displaystyle Density = \frac{3.568 \ lb}{40 \ ft.^3} \approx \underline{ 0.0892 \, lb/ft.^3}[/tex]

(vii) [tex]\displaystyle The \ specific \ gravity \ of \ the \ mixture \ = \frac{\frac{3066.1047\, g}{40 \, ft.^3} }{1.00} = \frac{\frac{6.7596 \, lbs}{40 \, ft.^3} }{1.00} \approx \underline{0.169}[/tex]

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Answer:

Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A​

Explanation:

Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A​

Answer:

Voltage Regulator

Technician A is correct.

Explanation:

Technician B is not correct.  The voltage regulator is not installed between the output terminal of the alternator and the positive terminal of the battery as claimed by Technician B.  Technician A's opinion that the voltage regulator controls the strength of the rotor's magnetic field is correct.  The computer can also be used to control the output of the alternator by controlling the field current.

Answer:

  v = 8.95 rad / s,    a = 22.3 rad / s²

Explanation:

This is an exercise in kinematics, where we must use the definitions of velocity and acceleration

          v = dr / dt

we perform the derivative

          v = 0+ 2 (-sin 5t) 5

          v = -10 sin 5t

we calculate for t = 0.85 t,

remember angles are in radians

           v = -10 sin (5 0.85)

           v = 8.95 rad / s

acceleration is defined by

           a = dv / dt

we perform the derivatives

          a = -10 (cos 5t) 5

          a = - 50 cos 5t

we calculate for t = 0.85 s

          a = -50 are (5 0.85)

          a = 22.3 rad / s²

Answer:

1.0MG

Explanation:

to solve this problem we use this formula

S₀-S/t = ksx --- (1)

the values have been given as

concentration = S₀ = 250mg

effluent concentration = S= 10mg

value of K = 0.04L/day

x = 3000 mg

when we put these values into this equation,

250-10/t = 0.04x10x3000

240/t = 1200

we cross multiply from this stage

240 = 1200t

t = 240/1200

t = 0.2

remember the question says that 5MGD is required to be treated

so the volume would be

v = 0.2x5

= 1.0 MG

Answer:

a. impossible

b. possible and reversible

c. possible and irreversible

Explanation:

a. 1000kw

Qh - Wnet

we have

QH = 1500

wnet = 1000

1500 - 1000

= 500kw

σcycle = [tex]-[\frac{QH}{TH} -\frac{QC}{TC} ][/tex]

Qh = 1500

Th = 1000

Tc = 500

Qc = 500

[tex]-[\frac{1500}{1000} -\frac{500}{500} ][/tex]

solving this using LCM

= -0.5

the cycle is impossible since -0.5<0

b. 750Kw

Qc = 1500 - 750

=750Kw

Qh = 1500

Th = 1000

Tc = 500

Qc = 750

σ-cycle

[tex]-[\frac{1500}{1000} -\frac{750}{500} ]\\= 1.5 -1.5\\= 0[/tex]

This cycle is possible and it is also reversible

c. 0 kw

Qc = 1500-0

= 1500

Qh = 1500

Th = 1000

Tc = 500

Qc = 1500

σ- cycle

[tex]-[\frac{1500}{1000} -\frac{1500}{500} ]\\-(1.5-3)\\-(-1.5)\\= 1.5[/tex]

1.5>0

so this cycle is possible and irreversible

Answer:

80⁰C

Explanation:

80°C.

Normal flat plate collectors can deliver heat at temperatures up to 80°C. Deficiency rates for normal flat plate collectors can be classified as visual losses, which produce with cumulative angles of the incident sunshine, and thermal losses, which upsurge fast with the working temperature intensities

The three most common software development methods are the Waterfall Approach, the Incremental Approach, and the SPIRAL Approach. These methods depend on the team size and specific goals.

Software development is the sequential procedure that involves the division of the work into smaller and parallel stages in order to improve software design and product management.

The software development methods depend on both the team size and specific objectives.

The most common methodologies for software development include:

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