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3 years ago

As an engineer which types of ethical issues or problem you can face in industrial environment.

Engineering

1 answer:

Anarel [89]3 years ago

8 0

Explanation:

Answer ⬇️

Social and ethical issues in engineering, ethical principles of engineering, professional code of ethics, some specific social problems in engineering practice: privacy and data protection, corruption, user orientation, digital divide, human rights, access to basic services.

➡️dhruv73143(⌐■-■)

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A 230 V shunt motor has a nominal armature current of 60 A. If the armature resistance is 0.15 ohm, calculate the following: a.

hjlf

Answer:

a) $E_{b} = 221 V$

b) P = 13,800 W

c) $P_{mech} = 13260 W$

di) $I_{initial} = 1533.33 A$

dii) $R_{start} = 1.85 ohms$

Explanation:

Voltage, Vt = 230 V

Armature current, $I_{a} = 60 A$

Armature Resistance, $R_{a} = 0.15 ohms$

a) The back emf is calculated as follows:

$E_{b} = V_{t} - I_{a} R_{a} \\E_{b} = 230 - (60 * 0.15)\\E_{b} = 230 - 9\\E_{b} = 221 V$

b) The power supplied to the armature (W)

$P =V_{t} I_{a}$

P = 230 * 60

P = 13,800 W

c) Mechanical power developed by the motor

$P_{mech}$ = Power supplied to the armature - Power lost in the armature

Power lost in the armature, $P_{a} = I_{a} ^{2} R_{a}$

$P_{a} = 60^{2} *0.15\\P_{a} = 540 W$

$P_{mech} = 13800 - 540\\P_{mech} = 13260 W$

d)( i)The initial starting current if the motor is directly connected across the 230 V line

At starting, there is no back emf, $E_{b} = 0$

$V_{t} = I_{initial} R_{a}$

$230 = I_{initial} * 0.15\\ I_{initial} = 230/0.15\\ I_{initial} = 1533.33 A$

ii) Value of the starting resistor needed to limit the initial current to 115 A

$V_{t} = I_{initial} (R_{a} + R_{start})$

$230= 115 (0.15 + R_{start})\\230/115 = 0.15 + R_{start}\\2 - 0.15 = R_{start}\\ R_{start} = 1.85 ohms$

5 0

4 years ago

A water contains 50.40 mg/L as CaCO3 of carbon dioxide, 190.00 mg/L as CaCO3 of Ca2 and 55.00 mg/L as CaCO3 of Mg2 . All of the

Galina-37 [17]

Answer:

Total sludge = 123426kg/d

Explanation:

The reaction is given as;

H2Co3 + Ca(OH)2 ⇆ CaCo3 + 2H20

1 1 1 2 moles

Calculating the concentration of C02, we have

Concentration of C02 = concentration of CaCo3/Molecular weight of Caco3

= 50.4/100.09

= 0.5035mol/L

Sludge of Co2 = Conc. of Co2 * Q * MW of CaCo3 *10^-6

= 0.5035 * 253.6 *10^6 * 100.09 * 10^-6

= 12780kg/d

From the equation Ca2+ + 2HCo3- + Ca(OH)2 ⇄ 2CaCo3 + 2H2O

1 mole of calcium yields 2 moles of CaCo3

Therefore, Concentration of Ca2+ = Conc. of CaCo3/Mw of CaCO3

= 190-30/100.09

=1.599mol/L

Calculating sludge of calcium:

Sludge of Ca = 2 * Conc. of ca * Q * mw of CaCO3 * 10^-6

= 2 * 1.599 *253.6*10^6* 100.09 * 10^-6

= 811742kg/d

From the equation,

Mg2+ +2HCO3- + Ca(OH)2 ⇄ MgCO3 + 2CaCO3 + 2H2O

1 mole of mg yields 2 moles CaCO3 and 1 mole of Mg(OH)2

Concentration of Mg2+ = Conc, of CaCO3 /Mw of CaCo3

= 55- 10/100.09

= 0.4496mol/L

Sludge of Mg = 2 * Conc. of Mg * Q * mw of CaCO3 * 10^-6 +* Conc. of Mg * Q * mw of Mg(OH)2 * 10^-6

= 2 * 0.4496 * 253.5*10^6 * 100.09 * 10^-6 + 0.4996* 253.5*10^6 58.3 * 10^-6

= 29472kg/d

Total Sludge = Sludge of CO2 + Sludge of Ca + Sludge of Mg

12780+ 81174 + 29472

= 123426kg/d

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3 years ago

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One reason the shuttle turns on its back after liftoff is to give the pilot a view of the horizon. Why might this be useful?

nordsb [41]

To look at your surroundings and see if there’s any tall mountains because they can effect air currents

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3 years ago

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An existing building is suffering from cracks in the exterior walls. The investigating engineer wants to ensure that the foundat

jek_recluse [69]

Answer:

$18 ft^{2}$

Explanation:

Soil bearing pressure= $\frac {Load}{Area}$

Since we're given pressure of 2500 psf and load of 45000 pounds

The area= $\frac {45000}{2500}=18$

Therefore, the smallest area of safe footings should not be less than $18 ft^{2}$

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3 years ago

Determine the reactor volume (assume a CSTR activated sludge aerobic reactor at steady state) required to treat 5 MGD of domesti

12345 [234]

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

4 0

3 years ago