Answer:
a) 1253 kJ
b) 714 kJ
c) 946 C
Explanation:
The thermal efficiency is given by this equation
η = L/Q1
Where
η: thermal efficiency
L: useful work
Q1: heat taken from the heat source
Rearranging:
Q1 = L/η
Replacing
Q1 = 539 / 0.43 = 1253 kJ
The first law of thermodynamics states that:
Q = L + ΔU
For a machine working in cycles ΔU is zero between homologous parts of the cycle.
Also we must remember that we count heat entering the system as positiv and heat leaving as negative.
We split the heat on the part that enters and the part that leaves.
Q1 + Q2 = L + 0
Q2 = L - Q1
Q2 = 539 - 1253 = -714 kJ
TO calculate a temperature for the heat sink we must consider this cycle as a Carnot cycle. Then we can use the thermal efficiency equation for the Carnot cycle, this one uses temperatures:
η = 1 - T2/T1
T2/T1 = 1 - η
T2 = (1 - η) * T1
The temperatures must be given in absolute scale (1453 C = 1180 K)
T2 = (1 - 0.43) * 1180 = 673 K
673 K = 946 C
A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.
Explanation:
From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).
To find the resistance of 260 ft (79.25 m) of size 4 AWG,
R= K * L/ A
K = 0.0214 ohm mm²/m
L = 79.25 m
A = 21.2 mm²
R = 0.0214 * 
= 0.0214 * 3.738
= 0.0792 ohm.
Thus the resistance of uncoated copper wire is 0.0792 ohm
Answer:
Diesel cycle:
All diesel engine works on diesel cycle.It have four processes .These four processes are as follows
1-2.Reversible adiabatic compression
2-3.Heat addition at constant pressure
3-4.Reversible adiabatic expansion
4-1.Heat addition at constant volume
When air inters in the piston cylinder after that it compresses and gets heated due to compression after that heat addition take place at constant pressure after that power is produces when piston moves to bottom dead center.
From the diagram of P-v And T-s we can understand so easily.
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:
Tire rotation is the least likely cause of tire wear. So, the option D is correct.
Explanation:
Step1
Under-inflation is the process of tire failure under low pressure. This contributes the wear on tire.
Step2
On breaking, kinetic energy changes to heat energy because of rubbing of tire. So, rubbing action increases the wear on the tire.
Step3
Acceleration on the vehicle increases the rubbing action as well as the wear and tear on the tire. So, acceleration is an also a major cause of tire wear.
Step4
Tire rotation has least amount of wear and tear due to no rubbing action. It has less amount surface contact with the surface in rotation.
Thus, tire rotation is the least likely cause of tire wear. So, the option D is correct.
