Steam is leaving a 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cook
er has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the cross-sectional area of theexit opening is 8mm2. Determine
a)The mass flow rate of the steam and the exit velocity
b)The total and flow energies of the steam per unit mass
c)The rate at which energy leaves the cooker by steam.
a)The mass flow rate of the steam and the exit velocity
b)The total and flow energies of the steam per unit mass
c)The rate at which energy leaves the cooker by steam.
1 answer:
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The velocity of the block when t == 2 s is 60.7 ft./sec.
Equations of Motion.
Here the friction is = 0.2 N
Kinematics.
The velocity of the block as a function of t can be determined by
integrating dv = adt using the initial condition v = 4 ft./s at t = 0.
The displacement as a function of t can be determined by integrating
ds = vdt using
the initial condition s = 0 at t = 0
at t = 2 sec
s = 30 ft.
Thus, at s = 30 ft.,
Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that cause them to move.
Kinematics, as a field of study, is often referred to as the "geometry of motion" and is occasionally seen as a branch of mathematics. A kinematics problem begins by describing the geometry of the system and declaring the initial conditions of any known values of position, velocity and/or acceleration of points within the system.
Then, using arguments from geometry, the position, velocity and acceleration of any unknown parts of the system can be determined. The study of how forces act on bodies falls within kinetics, not kinematics. For further details, see analytical dynamics.
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Draw a diagram to illustrate the problem as shown below.
The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s
Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s
The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
= 21.675 m
Answer: The height is 21.7 m (nearest tenth)
The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s
Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s
The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
= 21.675 m
Answer: The height is 21.7 m (nearest tenth)