Check them to see if the cylinder is running rich or lean. This can be determined by looking at the electrodes on the plugs
Hope this helps.
Answer:
The torque torque produced on the turbine from the wind is approximately 955 kN·m
Explanation:
The number of revolution per minute of the turbine = 20 rev/min
The power output of the turbine = 2.0 MW
The power transmitted to a shaft equation is given as follows;

Where;
P = The power transmitted to a turbine shaft = 2.0 MW
N = The number or revolutions per minute = 20 rev/min
T = The torque produced on the turbine by the wind
Therefore;

The torque torque produced on the turbine from the wind = 3/π MN·m ≈ 0.955 MN·m = 955 kN·m.
Answer:
46.3 cm²
216 m
Explanation:
The cross sectional area is the area of the outer circle minus the area of the inner circle.
A = πR² − πr²
R = 300/2 = 150 mm = 15 cm
r = 150 − 5 = 145 mm = 14.5 cm
A = π (15 cm)² − π (14.5 cm)²
A ≈ 46.3 cm²
Volume is area times length:
V = AL
10⁶ cm³ = (46.3 cm²) L
L ≈ 21,580 cm
L ≈ 216 m
Answer:
The work done is 3654.83 J/mol
Explanation:
The process is an isothermal process since the air temperature is kept constant during the release.
W = RT ln(P1/P2)
R is gas constant = 8.314 J/mol.K
T is air temperature = 400 K
P1 is initial pressure of air = 600 kPa
P2 is final temperature of air = 200 kPa
W = 8.314×400 ln(600/200) = 3325.6×ln 3 = 3325.6×1.099 = 3654.83 J/mol