Answer:
The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.
Explanation:
From Fluid Mechanics, we remember that absolute pressure (
), measured in pounds per square inch, is the sum of the atmospheric pressure and the working pressure (gauge pressure). That is:
(1)
Where:
- Atmospheric pressure, measured in pounds per square inch.
- Working pressured of the boiler (gauge pressure), measured in pounds per square inch.
If we suppose that
and
, then the absolute pressure is:


The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.
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:
Outside temperature =88.03°C
Explanation:
Conductivity of air-soil from standard table
K=0.60 W/m-k
To find temperature we need to balance energy
Heat generation=Heat dissipation
Now find the value
We know that for sphere

Given that q=500 W
so

By solving that equation we get
=88.03°C
So outside temperature =88.03°C
Answer:
maximum stress is 2872.28 MPa
Explanation:
given data
radius of curvature = 3 ×
mm
crack length = 5.5 ×
mm
tensile stress = 150 MPa
to find out
maximum stress
solution
we know that maximum stress formula that is express as
......................1
here σo is applied stress and a is half of internal crack and t is radius of curvature of tip of internal crack
so put here all value in equation 1 we get
σm = 2872.28 MPa
so maximum stress is 2872.28 MPa