Answer:
If the heat engine operates for one hour:
a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.
b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.
In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.
Explanation:
The Carnot efficiency is obtained as:
Where 
is the atmospheric temperature and 
is the maximum burn temperature.
For the case (B), the efficiency we will use is:
The work done by the engine can be calculated as:
where Hv is the heat value.
If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.
If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.
Answer:
Both Technician A and B are correct
Explanation:
Due to a contamination of the clutch disc friction surface, the clutch chatters
The contamination that causes clutch chattering includes
1) Worn or broken motor mounts
2) Bell housing bolts becoming loose
3) Clutch link damage
4) Warped flywheel, due to overheating, the flywheel can become warped such that the non uniform surface interfaces the clutch resulting in clutch chattering
5) Engine or transmission oil contaminating the disc.
Parallel Resistor Equation
If the two resistances or impedances in parallel are equal and of the same value, then the total or equivalent resistance, RT is equal to half the value of one resistor. That is equal to R/2 and for three equal resistors in parallel, R/3, etc.
Answer:
Rate of heat transfer is 0.56592 kg/hour
Explanation:
Q = kA(T2 - T1)/t
Q is rate of heat transfer in Watts or Joules per second
k is thermal conductivity of the styrofoam = 0.035 W/(mK)
A is area of the cubical picnic chest = 6L^2 = 6(0.5)^2 = 6×0.25 = 1.5 m^2
T1 is initial temperature of ice = 0 °C = 0+273 = 273 K
T2 is temperature of the styrofoam = 25 °C = 25+273 = 298 K
t is thickness of styrofoam = 0.025 m
Q = 0.035×1.5(298-273)/0.025 = 1.3125/0.025 = 52.5 W = 52.5 J/s
Mass flow rate = rate of heat transfer ÷ latent heat of melting of ice = 52.5 J/s ÷ 3.34×10^ 5 J/kg = 1.572×10^-4 kg/s = 1.572×10^-4 kg/s × 3600 s/1 hr = 0.56592 kg/hr
