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andrezito [222]
3 years ago
14

Diesel-fueled generators are frequently used as backup electrical power sources for homes and hospitals. Consider a Diesel power

ed generator with an efficiency of 39 percent for an engine speed of idle to about 1,800 rpm (revolutions per minute). If Diesel fuel has a chemical formula of C12H23.
a. Determine the chemical reaction for 1 kmol of Diesel fuel burning with the stoichiometric amount of air.
b. For each kg of Diesel fuel burned, how much CO2 is generated, in kg?
c. Find the higher heating value (HHV) for C12H23 at 25°C, 1 atm.
d. Calculate the amount of Diesel fuel, in kg and in gallons, required to produce a power output of 18 kW to a home for a period of 8 h.
e. Comment on your results.

Engineering
1 answer:
Sergeu [11.5K]3 years ago
3 0

Answer:

a. C12H23 + 84.5 moles of air —-> 12CO2(g)+ 11.5H2O(g)

b. 3.2kg of CO2 per 1kg of C12H23

c. HVV of C12H23 is -1724.5 KCal/mol

d. Total weight required is 30.742kg

e. The amount of CO2 produced per kg of C12H23 is too much. CO2 is harmful to the environment and should be produced in weights as low as possible

Explanation:

Please check attachment for complete solution and step by step explanation

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What is the definition of a duty cycle?
ira [324]

Answer:

D=\frac{PW}{T}*100

Explanation:

In electrical terms, is the ratio of time in which a load or circuit is ON compared to the time in which the load or circuit is OFF.

The duty cycle or power cycle, is expressed as a percentage of the activation time. For example, a 70% duty cycle is a signal that 70% of the time is activated and the other 30% disabled. Its equation can be expressed as:

D=\frac{PW}{T}*100

Where:

D=Duty\hspace{3}Cycle

PW=Pulse\hspace{3}Active\hspace{3}Time

T=Period\hspace{3}of\hspace{3}the\hspace{3}Signal

Here is a picture that will help you understand these concepts.

5 0
3 years ago
A 2 m3 insulated rigid tank contains 3 kg of nitrogen at 90 kPa. Now work is done on the system until the pressure in the tank r
Nutka1998 [239]

Answer: \Delta S = 1.47kJ/K

Explanation: <u>Entropy</u> is the measure of a system's molecular disorder, i.e, the unuseful work a system does.

The nitrogen gas in the insulated tank can be described as an ideal gas, so it can be used the related formulas.

For the entropy, the ratio of initial and final temperatures is needed and as volume is constant, we use:

\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}

\frac{P_{2}}{P_{1}} =\frac{T_{2}}{T_{1}}

\frac{T_{2}}{T_{1}} =\frac{175}{90}

\frac{T_{2}}{T_{1}} =1.94

<u>Specific</u> <u>Heat</u> is the quantity of heat required to increase the temperature 1 degree of a unit mass of a substance. Specific heat of nitrogen at constant volume is c_{v}= 0.743kJ/kg.K

The change in entropy is calculated by

\Delta S= m[c_{v}ln(\frac{T_{2}}{T_{1}})-Rln(\frac{V_{2}}{V_{1}} )]

For the nitrogen insulated in a rigid tank:

\Delta S= m[c_{v}ln(\frac{T_{2}}{T_{1}})]

Substituing:

\Delta S= 3[0.743ln(1.94)]

\Delta S= 1.47

The entropy change of nitrogen in an insulated rigid tank is 1.47kJ/K

6 0
3 years ago
Consider a thin-walled cylindrical tube having a radius of 65 mm that is to be used to transport pressurized gas. (a) (10 points
AlladinOne [14]

Answer:

The minimum required thickness for a steel pressure vessel (t) is 2.275 mm

Explanation:

Minimum required thickness is the thickness of a material without corrosion allowance for each component  based on the appropriate design that consider pressure, mechanical and structural loading.

Given that:

radius (r) = 65 mm = 65 × 10⁻³ m

Factor of safety (N) = 3.5

Inside presssure (P_{in}) = 11 MPa

Outside pressure (P_{out}) = 1 MPa

Yield strength (\sigma_y) = 1000 MPa

Therefore:

\sigma_y =\frac{\sigma_y}{N}, Substituting values,

\sigma_y =\frac{\sigma_y}{N}=\frac{1000}{3.5}=285.714 MPa

The minimum required thickness for a steel pressure vessel (t) is given by the equation:

t=\frac{r.(P_{in}-P_{out})}{\sigma_y}. Substituting values

t=\frac{r.(P_{in}-P_{out})}{\sigma_y}=\frac{65*10^{-3} *(11-1)10^{6} }{285.714*10^{6} } =2.275*10^{-3} =2.275 mm

The minimum required thickness for a steel pressure vessel (t) is 2.275 mm

7 0
3 years ago
I need help!!!!!!!!!
katovenus [111]

Answer:

buy a new one

Explanation:

go on newegg and you will most likely find one there

8 0
3 years ago
It has been estimated that 139.2x10^6 m^2 of rainforest is destroyed each day. assume that the initial area of tropical rainfore
Dmitry [639]

Answer:

A. 6.96 x 10^-6 /day

B. 22.466 x 10^12 m^2

C. 9.1125 x 10^14 kg of CO2

Explanation:

A. Rate of rainforest destruction = 139.2 x 10^6 m^2/day

Initial area of the rainforest = 20 x 10^12 m^2

Therefore to calculate exponential rate in 1/day,

Rate of rainforest destruction/ initial area of rainforest

= 139.2 x 10^6/20 x 10^12

= 6.96 x 10^-6 /day

B. Rainforest left in 2015 using the rate in A.

2015 - 1975 = 40 years

(40 * 365 )days + 10 days (leap years)

= 14610 days

Area of rainforest in 1975 = 24.5 x 10^12m^2

Rate of rainforest destruction = 139.2 x 10^6 m^2/day

Area of rainforest in 2015 = 14610 * 139.2 x 10^6

= 2.034 x 10^12 m^2

Area left = area of rainforest in 1975 - area of rainforest destroyed in 40years

= 24.5 x 10^12 - 2.034 x 10^12

= 22.466 x 10^12 m^2

C. How much CO2 will be removed in 2025

Recall: Photosynthesis is the process of plants taking in CO2 and water to give glucose and O2.

So CO2 removed is the same as rainforest removed so we use the rate of rainforest removed in a day

Area of rainforest in 1975 = 24.5 x 10^12 m^2

Area of rainforest removed in 2025 = 18262 days * 139.2 x 10^6

= 2.54 x 10^12 m^2

Area of rainforest removed between 1975 - 2025 = 24.5 x 10^12 - 2.54 x 10^12

= 21.958 x 10^12 mC2 of rainforest removed

CO2 = 0.83kg/m^2.year

CO2 removed between 1975 - 2025 = 0.83 * 21.958 x 10^12 * 50 years

= 9.1125 x 10^14 kg of CO2 was removed between 1975 - 2025

6 0
3 years ago
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