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:
(a)Volume in liters=5.3 liters.
(b)Volume in liters/minute=31.8 liters/minute.
Explanation:
Given:
Diameter of cylinder ,D=150 mm
Stroke,L=300 mm
Time ,t=10 sec
we know that swept volume of cylinder
So
(a) Volume in liters =5.3 liters ( 1=1000 liters)
(b) When we divide swept volume by time(in minute) we will get liters/minute.
We know that 1 minute=60 sec
⇒10 sec= minute
So volume displace in liters/minute=31.8 liters/minute.
Answer:
patience
Explanation:
you can wait it out,
you can wait to solve,
you can wait for another opinion,
you can wait for an answer,
you can wait to get help,
or you can wait till it's not an issue.
Answer:
The efficiency of this ideal and reversible engine is 85 percent.
The efficiency of the Carnot cycle represents the efficiency of a thermal machine with no irreversibilities, hence, it is impossible for any real engine operating between the two reservoirs cannot be more efficient than this engine.
Explanation:
Let assume that the temperature of the atmosphere is 300 K. From Thermodynamics we know that the efficiency of the Carnot's cycle (), dimensionless, is:
(1)
Where:
- Temperature of the kerosene combustor (hot reservoir), measured in kelvins.
- Temperature of the atmosphere (cold reservoir), measured in kelvins.
If we know that and , then the efficieny of this ideal and reversible engine is:
The efficiency of this ideal and reversible engine is 85 percent.
The efficiency of the Carnot cycle represents the efficiency of a thermal machine with no irreversibilities, hence, it is impossible for any real engine operating between the two reservoirs cannot be more efficient than this engine.