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.
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.
1 answer:
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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a) The stress in the bar when F is 32,000 is approximately 7,100 psi
b) The load P that can be supported by the bar if the axial stress must not exceed is approximately 110,000 lb
Explanation:
The question topic relates to stresses in structures;
The given parameters of the steel bar are;
The width of the steel bar, W = 4.0 in.
The thickness of the steel bar, t = 1.125 in.
The formula for stress in a bar is given as follows;
The cross sectional area of bar, A = W × t = 4.0 in. × 1.125 in. = 4.5 in.²
∴ The cross sectional area of bar, A = 4.5 in.²
a) The stress in the bar for F = 32,000 lb, is given as follows;
The stress in the bar when F is 32,000 is σ = 7,111. psi ≈ 7,100 psi
b) The load P that can be supported by the bar if the axial stress must not exceed, σ = 25,000 psi is given as follows;
Therefore;
P = σ × A = 25,000 psi × 4.5 in² = 112,500 lb
For the axial stress of 25,000 psi not to be exceeded, the maximum load that can be supported by the bar, P = 112,500 lb ≈ 110,000 lb.