An Otto cycle engine is analyzed using the cold air standard method. Given the conditions at state 1, compression ratio (r), and
pressure ratio (rp). The specific heats for air are Cp = 1.0045 kJ/kg-K and Cv = 0.7175 kJ/kg-K.
T1 = 300K
P1 = 100 kPa
r = 10
rp = 1.65
a) Determine the temperature (K) at state 2.
b) Determine the pressure (kPa) at state 2.
c) Determine the pressure (kPa) at state 3.
d) Determine the temperature (K) at state 3.
e) Determine the temperature (K) at state 4.
f) Determine the pressure (kPa) at state 4.
g) Determine the net work output (kJ/kg/cycle) for the engine.
h) Determine the heat addition (kJ/kg/cycle) for the engine.
i) Determine the efficiency (%) of the engine.
Points will be given to answer showing all the work done to get right answer.
T1 = 300K
P1 = 100 kPa
r = 10
rp = 1.65
a) Determine the temperature (K) at state 2.
b) Determine the pressure (kPa) at state 2.
c) Determine the pressure (kPa) at state 3.
d) Determine the temperature (K) at state 3.
e) Determine the temperature (K) at state 4.
f) Determine the pressure (kPa) at state 4.
g) Determine the net work output (kJ/kg/cycle) for the engine.
h) Determine the heat addition (kJ/kg/cycle) for the engine.
i) Determine the efficiency (%) of the engine.
Points will be given to answer showing all the work done to get right answer.
1 answer:
You might be interested in
Answer:
Explanation:
In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.
We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.
The weigth depends on the size and specific gravity.
W = 1/2 * b * h * L * SG
Then
Teq = 1/2 * b * h * L * SG * b / 2
Teq = 1/4 * b^2 * h * L * SG
The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:
The term sin(30) is because of the slope of the wall
The pressure of water is:
p(y) = SGw * (h - y)
Then:
T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)
T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)
T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)
T1 = 1/3 * SGw * sin(30) * L * h^3
To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)
1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3
In an equilateral triangle h = b * cos(30)
1/4 * b^3 * cos(30) * L * SG > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3
SG > SGw * 4/3* sin(30) * (cos(30))^2
SG > 1/2 * SGw
For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.
This is avergae specific gravity, including holes.
Answer:
The elevation at the high point of the road is 12186.5 in ft.
Explanation:
The automobile weight is 2500 lbf.
The automobile increases its gravitational potential energy in . It means the mobile has increased its elevation.
The initial elevation is of 5183 ft.
The first step is to convert Btu of potential energy to adequate units to work with data previously presented.
British Thermal Unit -
Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:
Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.
is the final elevation and
is the initial elevation.
Replacing W in the Ep equation
Finally, the next step is to replace the variables of the problem.
The elevation at the high point of the road is 12186.5 in ft.