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11111nata11111 [884]
3 years ago
7

If OSHA determines that an employer's response to a non-formal complaint is adequate, what options does the employee filing the

non-formal complaint have?
Engineering
1 answer:
Anuta_ua [19.1K]3 years ago
4 0
Fill it out without telling ur employer as that may cause backlash and have an osha certified employee come check out ur work or job site
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A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 240 kJ of work on the
Sergeu [11.5K]

To solve the problem it is necessary to consider the concepts and formulas related to the change of ideal gas entropy.

By definition the entropy change would be defined as

\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{P_2}{P_1})

Using the Boyle equation we have

\Delta S = C_p ln(\frac{T_2}{T_1})-Rln(\frac{v_1T_2}{v_2T_1})

Where,

C_p = Specific heat at constant pressure

T_1= Initial temperature of gas

T_2= Final temprature of gas

R = Universal gas constant

v_1= Initial specific Volume of gas

v_2= Final specific volume of gas

According to the statement, it is an isothermal process and the tank is therefore rigid

T_1 = T_2, v_2=v_1

The equation would turn out as

\Delta S = C_p ln1-ln1

<em>Therefore the entropy change of the ideal gas is 0</em>

Into the surroundings we have that

\Delta S = \frac{Q}{T}

Where,

Q = Heat Exchange

T = Temperature in the surrounding

Replacing with our values we have that

\Delta S = \frac{230kJ}{(30+273)K}

\Delta S = 0.76 kJ/K

<em>Therefore the increase of entropy into the surroundings is 0.76kJ/K</em>

8 0
3 years ago
A 600 MW coal-fired power plant has an overall thermal efficiency of 38%. It is burning coal that has a heating value of 12,000
velikii [3]

Answer:

See step by step explanations for answer.

Explanation:

600 megawatts =

568 690.272 btu / second

thermal eficiency=work done/Heat supllied

0.38=568690.272/Heat supplied

Heat supplied=1496553.35btu /s

heat emmitted to the atmosphere=heat supplied -work done=(1496553.35-568690.272)=927863.1 btu/s

feed rate=(1496553.35)/12000=124.71 lb/s =10775184.1056 lb/day=5 387.472 ton / day

sulphur content released=(0.03*124.71)/(1.496553)=2.5 lb SO2/million Btu of heat input

so

the degree (%) of sulfur dioxide control needed to meet an emission standard=(2.5/0.15)*100=1666.67 %

the CO2 emission rate=220*(1.496553) =329.241 lb/s =12 903.0802 metric ton / day

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What type of fire extinguisher is used for electrical fires.
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