Answer:
Thermal efficiency of the Otto Cycle is higher than Diesel Engine becuase of their larger area under the PV diagram.
Reason for increasing use of diesel engine instead of Otto Cycle is that compression ratio of Diesel Engine is much higher than Otto Cycle so at same compression ratio, Diesel engine is more efficient making it more feasible to use at most operating points & that is why in practical approach, Diesel engines are more in use instead of Otto Cycle.
Explanation:
The thermal efficiency of the Otto cycle is higher than diesel engine becuse of the fact that area under the PV diagram is larger because the Otto cycle has heat addition at constant volume rather than constant pressure which increases the amount of work engine can produce during expansion by increasing mean effective pressure.
whereas in the diesel engine heat addition is at constant pressure, being analogous to the heat being added while the piston is moving down in the expansion stroke. Hence part of the potential heat energy is not fully made in use in the expansion stroke of a diesel engine.
The reason for increasing use of diesel engine instead of Otto cycle has to deal with the fact that in practical world, compression ratios for diesel engine is so much higher than the otto cycle. So at same compression ratio, deisel engine is more efficient at most oprating points making it more feasible to use even at low loads particularly since they operate both unthrotlled and lean burn. Diesel engines are also more efficient at near or equal to full load since they still run lean while Otto engines run rich. That is why. when we talk about real world phenomenon, diesel engines are more in use than Otto Cycle.
Density of cylinder = 11.4g/cm3
Volume of cylinder = h r2
but height (h) = 5.50cm
radius(r) = d/2 = 2.50/2 = 1.25cm
Hence Volume = 5.50 × (1.25)2
= 8.6cm3
Density = Mass/Volume
Hence Mass of the cylinder =
Density × Volume
Mass = 11.4 × 8.6 = 94.04g.
Answer:
969.68N
Explanation:
d₁=0.04 m A₁=
d₂=0.02 m A₂=
Q=1.2m³/min Q=1.2/60=0.02m³/s
using continuity equation
Q₁=A₁v₁
v₁=Q₁/A₁=0.02/0.00125=16m/s
Q₂=A₂v₂
v₂=Q₂/A₂=0.02/0.00031=64.5m/s
Force on the nozzle=F_{outlet}-F_{inlet}
= 1289.68-320
=969.68N
Answer:
The efficiency of a DC generator is maximum when those losses proportional to the square of the load current are equal to the constant losses of the DC generator. This relation applies equally well to all rotating machines, regardless of the type of machine.
Explanation: