Nancy wants to buy a cooking stove that’s electrically insulated and resistant to heat. What material should she choose for the
1 answer:
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Answer:
You need a 120V to 24V commercial transformer (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)
Step by step design:
- Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer. 120 Vrms = 85 V and 24 Vrms = 17V = Vin
- Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
- Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA
Our circuit meet the average voltage (Va) specification:
Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it
Answer:
a) 1
b) 1813.96 MJ/kmol
c) 32.43 MJ/kg , 1980.39 MJ/Kmol
Explanation:
molar mass of ethanol (C2H5OH) = 46 g/mol
molar mass of octane (C8H18) = 114 g/mol
therefore the moles of ethanol and octane
ethanol = 0.85 / 46
octane = 0.15 / 114
a) determine the molar air-fuel ratio and air-fuel ratio by mass
attached below
mass of air / mass of fuel = 12.17 / 1 = 12.17
b ) Determine the lower heating value
LHV of ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol
LHV of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol
LHV ( MJ/kmol) for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol
c) Determine higher heating value ( HHV )
HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol
HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol
HHV in MJ/kg = 0.85 * 29.7 + 0.15 * 47.9 = 32.43 MJ/kg
HHV in MJ /kmol = 0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol
