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3 years ago

If you are a mechanical engineer answer these questions:

Engineering

1 answer:

Natasha_Volkova [10]3 years ago

3 0

Answer:

1. Yes, they are all necessary.

2. Both written and verbal communication skills are of the utmost importance in business, especially in engineering. Communication skills boost you or your teams' performance because they provide clear information and expectations to help manage and deliver excellent work.

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A coal fired power plant geneartes 2.4 lbs. of CO2 per kWh. A lighting system consumes 300,000kWh per year. A corporation is con

Serjik [45]

Answer:

The perceived economic impact of CO2 generated per year by lighting sstem is $8164.67.

Explanation:

The CO2 requirement for the plant is:

Amount of CO2 per year = (2.4 lb / KWh)(300,000 KWh)

Amount of CO2 per year = (720000 lb)(1 ton/ 2204.62 lb)

Amount of CO2 per year = 326.59 ton

The perceived economic impact of CO2 generated per year will then be:

Economic Impact = ($25 / ton)(326.59 ton)

<u>Economic Impact = $8164.67</u>

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3 years ago

What are the well-known effects of electricity

Sever21 [200]

Answer:

Hence, the three effects of electric current are heating effect, magnetic effect and chemical effect.

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3 years ago

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Anyone want to play among us with me the code is MMJSUF

Katarina [22]

Answer:

sure

Explanation:

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3 years ago

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A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas

ch4aika [34]

Answer:

gauge pressure is 133 kPa

Explanation:

given data

initial temperature T1 = 27°C = 300 K

gauge pressure = 300 kPa = 300 × 10³ Pa

atmospheric pressure = 1 atm

final temperature T2 = 77°C = 350 K

to find out

final pressure

solution

we know that gauge pressure is = absolute pressure - atmospheric pressure so

P (gauge ) = 300 × 10³ Pa - 1 × $10^{5}$ Pa

P (gauge ) = 2 × $10^{5}$ Pa

so from idea gas equation

$\frac{P1*V1}{T1} = \frac{P2*V2}{T2}$ ................1

so ${P2} = \frac{P1*T2}{T1}$

${P2} = \frac{2*10^5*350}{300}$

P2 = 2.33 × $10^{5}$ Pa

so gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 2.33 × $10^{5}$ - 1.0 × $10^{5}$

gauge pressure = 1.33 × $10^{5}$ Pa

so gauge pressure is 133 kPa

4 0

3 years ago

Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa, and saturated vapor enters the turbine at

sergeinik [125]

Explanation:

The obtained data from water properties tables are:

Point 1 (condenser exit) @ 8 KPa, saturated fluid

$h_{f} = 173.358 \\h_{fg} = 2402.522$

Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid

$h_{2a} = 489.752\\h_{2b} = 313.2$

Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam

$h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876$

Point 4 (Turbine exit) @ 8 KPa, mixed fluid

$x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119$

Calculate mass flow rates

Part a) @ 18 MPa

mass flow

$\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a} - h_{4a}) - (h_{2a} - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26 - 2241.448938 ) - (489.752 - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}$

Heat transfer rate through boiler

$Q_{in} = mass flow * (h_{3a} - h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W$

Heat transfer rate through condenser

$Q_{out} = mass flow * (h_{4a} - h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W$

Thermal Efficiency

$n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.06485$

Part b) @ 4 MPa

mass flow

$\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b} - h_{4b}) - (h_{2b} - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14 - 2405.54119 ) - (313.12 - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}$

Heat transfer rate through boiler

$Q_{in} = mass flow * (h_{3b} - h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W$

Heat transfer rate through condenser

$Q_{out} = mass flow * (h_{4b} - h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W$

Thermal Efficiency

$n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.038275$

6 0

3 years ago