Answer:
The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²
Explanation:
Here we have the heat Q given as follows;
Q = 15 × 1075 × (1100 - 
) = 10 × 1075 × (850 - 300) = 5912500 J
∴ 1100 - = 1100/3
= 733.33 K
Where
= Arithmetic mean temperature difference
= Inlet temperature of the gas = 1100 K
= Outlet temperature of the gas = 733.33 K
= Inlet temperature of the air = 300 K
= Outlet temperature of the air = 850 K
Hence, plugging in the values, we have;
Hence, from;
, we have
5912500 = 90 × A × 341.67
Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².
The brakes are being bled on a passenger vehicle with a disc/drum brake system is described in the following
Explanation:
1.Risk: Continued operation at or below Rotor Minimum Thickness can lead to Brake system failure. As the rotor reaches its minimum thickness, the braking distance increases, sometimes up to 4 meters. A brake system is designed to take kinetic energy and transfer it into heat energy.
2.Since the piston needs to be pushed back into the caliper in order to fit over the new pads, I do open the bleeder screw when pushing the piston back in. This does help prevent debris from traveling back through the system and contaminating the ABS sensors
3.There are three methods of bleeding brakes: Vacuum pumping. Pressure pumping. Pump and hold.
4,Brake drag is caused by the brake pads or shoes not releasing completely when the brake pedal is released. ... A worn or corroded master cylinder bore causes excess pedal effort resulting in dragging brakes. Brake Lines and Hoses: There may be pressure trapped in the brake line or hose after the pedal has been released.
Answer: 3/2mg
Explanation:
Express the moment equation about point B
MB = (M K)B
-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α
α = 3g/2L cosθ
express the force equation along n and t axes.
Ft = m (aG)t
mg cosθ – Bt = m [(3g/2L cos) (L/6)]
Bt = ¾ mg cosθ
Fn = m (aG)n
Bn -mgsinθ = m[ω^2 (L/6)]
Bn =1/6 mω^2 L + mgsinθ
Calculate the angular velocity of the rod
ω = √(3g/L sinθ)
when θ = 90°, calculate the values of Bt and Bn
Bt =3/4 mg cos90°
= 0
Bn =1/6m (3g/L)(L) + mg sin (9o°)
= 3/2mg
Hence, the reactive force at A is,
FA = √(02 +(3/2mg)^2
= 3/2 mg
The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg
