Complete Question
The complete question is shown on the first uploaded image.
Answer:
The answer is shown on the second uploaded image
Explanation:
The explanation is also shown on the second uploaded image
Assumptions:
- Steady state.
- Air as working fluid.
- Ideal gas.
- Reversible process.
- Ideal Otto Cycle.
Explanation:
Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):
- Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.
- Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).

- Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
- Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.

- Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
- Exhaust 1-0: the working fluid is vented to the atmosphere.
If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:

where:

Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.

Answer:
See image attached.
Answer:
capacity = 0.555 mAh
capacity = 3600 mAh
Explanation:
given data
battery = 1800 mAh
OCV = 3.9 V
solution
we get here capacity when it is in series
so here Q = 2C
capacity = 2 × ampere × second ...............1
put here value and we get
and 1 Ah = 3600 C
capacity =
capacity = 0.555 mAh
and
when it is in parallel than capacity will be
capacity = Q1 +Q2 ...............2
capacity = 1800 + 1800
capacity = 3600 mAh
Answer:
1561.84 MPa
Explanation:
L=20 cm
d1=0.21 cm
d2=0.25 cm
F=5500 N
a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa
lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16
longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3
(assuming a poisson's ration of 0.3)
ε_l =0.16/0.3 = 0.5333
b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)
σ_true = 1561.84 MPa
ε_true = ln( 1+ε_l)= ln(1+0.5333)
ε_true= 0.4274222
The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.
Answer:
ambages. pitiless or without compassion; cruel; merciless. winding, roundabout paths or ways.
Explanation: