On what single factor does the efficiency of the Otto cycle depend?
1 answer:
Answer:
Compression ratio(r)
Explanation:
Otto cycle:
Otto cycle is an ideal cycle for all working petrol engine.It have four processes in which two are constant volume process and other two are reversible adiabatic or we can say that isentropic processes.All petrol engine works on Otto cycle.
The efficiency of Otto cycle given as follows
Where r is the compression ratio and γ is heat capacity ratio.
So from above we can say that the efficiency of Otto cycle depends onl;y on compression ratio (r).
You might be interested in
Answer:
the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm
Explanation:
Given the data in the question;
yield strength σ = 690 Mpa
plane strain fracture toughness K = 32 MPa-
minimum component thickness for which the condition of plane strain is valid = ?
Now, for plane strain conditions, the minimum thickness required is expressed as;
t ≥ 2.5( K / σ
)²
so we substitute our values into the formula
t ≥ 2.5( 32 / 690 )²
t ≥ 2.5( 0.0463768 )²
t ≥ 2.5 × 0.0021508
t ≥ 0.005377 m or 5.38 mm
Therefore, the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm
Answer:
the angle of twist of B with respect to D is -1.15°
the angle of twist of C with respect to D is 1.15°
Explanation:
The missing diagram that is supposed to be added to this image is attached in the file below.
From the given information:
The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.
For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²
The objective are :
1) To determine the angle of twist of B with respect to D
Considering the Polar moment of Inertia at the shaft
shaft =
where ;
r = 1 in /2
r = 0.5 in
shaft =
shaft = 0.098218
Now; the angle of twist at B with respect to D is calculated by using the expression
where;
are the torques at segments CD and length at segments CD
are the torques at segments BC and length at segments BC
Also ; from the diagram; the following values where obtained:
= 2.5 in
= 0.098218
G = 11 × 10⁶ lb/in²
= -60 lb.ft
= 0 lb.ft
= 5.5 in
− 0.02 rad
To degree; we have
Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction
Thus; the angle of twist of B with respect to D is 1.15°
(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places
For the angle of twist of C with respect to D; we have:
0.02 rad
To degree; we have
