Answer:
Enthalpy at outlet=284.44 KJ
Explanation:
We need to Find enthalpy of outlet.
Lets take the outlet mass m and outlet enthalpy h.
So from mass conservation
m=1+1.5+2 Kg/s
m=4.5 Kg/s
Now from energy conservation
By putting the values
So h=284.44 KJ
Answer:
A) m' = 351.49 kg/s
B) m'= 1036.91 kg/s
Explanation:
We are given;
Pressure Ratio;r_p = 12
Inlet temperature of compressor;T1 = 300 K
Inlet temperature of turbine;T3 = 1000 K
cp = 1.005 kJ/kg·K
k = 1.4
Net power output; W' = 70 MW = 70000 KW
A) Now, the formula for the mass flow rate using the total power output of the compressor and turbine is given as;
m' = W'/[cp(T3(1 - r_p^(-(k - 1)/k)) - T1(r_p^((k - 1)/k))
At, 100% efficiency, plugging in the relevant values, we have;
m' = 70000/(1.005(1000(1 - 12^(-(1.4 - 1)/1.4)) - 300(12^((1.4 - 1)/1.4)))
m' = 70000/199.1508
m' = 351.49 kg/s
B) At 85% efficiency, the formula will now be;
m' = W'/[cp(ηT3(1 - r_p^(-(k - 1)/k)) - (T1/η) (r_p^((k - 1)/k))
Where η is efficiency = 0.85
Thus;
m' = 70000/(1.005(0.85*1000(1 - 12^(-(1.4 - 1)/1.4)) - (300/0.85)(12^((1.4 - 1)/1.4)))
m' = 70000/(1.005*(432.09129 - 364.9189)
m'= 1036.91 kg/s
Answer:
Technician B only is correct
Explanation:
The last stage of gears found between the vehicle transmission system and the wheels is the final drive ratio. The function of the final drive gear assembly is to enable a gear reduction control stage to reduce the rotation per minute and increase the wheel torque, such that the vehicle performance can be adjusted and the final gear ratio can be between 3:1 and 4.5:1 not 1:1
Therefore, technician B only is correct
Answer:
option 3 is the correct answer
