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Answer:
F = 9.11 x 10³ N = 9.11 KN
Explanation:
The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:
ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper
ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)
where,
ΔL = Equivalent Change in Length = ?
ΔT = Change in Temperature = 25°C - 12°C = 13°C
Ls = Length of Steel Segment = 300 mm = 0.3 m
Lb = Length of Brass Segment = 200 mm = 0.2 m
Lc = Length of Copper Segment = 100 mm = 0.1 m
Therefore,
ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)
ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m
ΔL = 123.5 x 10⁻⁶ m ----------------------- equation (1)
Now, we calculate this deflection in terms of an applied force (F):
ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)
ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)
ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)
ΔL = F(13.55 x 10⁻⁹ m/N) --------------------- equation (1)
Comparing equation (1) and equation (2):
123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)
F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)
<u>F = 9.11 x 10³ N = 9.11 KN</u>
Mark brainliest please!
Isothermal work will be less than the adiabatic work for any given compression ratio and set of suction conditions. The ratio of isothermal work to the actual work is the isothermal efficiency. Isothermal paths are not typically used in most industrial compressor calculations.
Compressors
Compressors are used to move gases and vapors in situations where large pressure differences are necessary.
Types of Compressor
Compressors are classified by the way they work: dynamic (centrifugal and axial) or reciprocating. Dynamic compressors use a set of rotating blades to add velocity and pressure to fluid. They operate at high speeds and are driven by steam or gas turbines or electric motors. They tend to be smaller and lighter for a given service than reciprocating machines, and hence have lower costs.
Reciprocating compressors use pistons to push gas to a higher pressure. They are common in natural gas gathering and transmission systems, but are less common in process applications. Reciprocating compressors may be used when very large pressure differences must be achieved; however, since they produce a pulsating flow, they may need to have a receiver vessel to dampen the pulses.
The compression ratio, pout over pin, is a key parameter in understanding compressors and blowers. When the compression ratio is below 4 or so, a blower is usually adequate. Higher ratios require a compressor, or multiple compressor stages, be used.
When the pressure of a gas is increased in an adiabatic system, the temperature of the fluid must rise. Since the temperature change is accompanied by a change in the specific volume, the work necessary to compress a unit of fluid also changes. Consequently, many compressors must be accompanied by cooling to reduce the consequences of the adiabatic temperature rise. The coolant may flow through a jacket which surrounds the housing with liquid coolant. When multiple stage compressors are used, intercooler heat exchangers are often used between the stages.
Dynamic Compressors
Gas enters a centrifugal or axial compressor through a suction nozzle and is directed into the first-stage impeller by a set of guide vanes. The blades push the gas forward and into a diffuser section where the gas velocity is slowed and the kinetic energy transferred from the blades is converted to pressure. In a multistage compressor, the gas encounters another set of guide vanes and the compression step is repeated. If necessary, the gas may pass through a cooling loop between stages.
Compressor Work
To evaluate the work requirements of a compressor, start with the mechanical energy balance. In most compressors, kinetic and potential energy changes are small, so velocity and static head terms may be neglected. As with pumps, friction can be lumped into the work term by using an efficiency. Unlike pumps, the fluid cannot be treated as incompressible, so a differential equation is required:
Compressor Work
Evaluation of the integral requires that the compression path be known - - is it adiabatic, isothermal, or polytropic?
uncooled units -- adiabatic, isentropic compression
complete cooling during compression -- isothermal compression
large compressors or incomplete cooling -- polytropic compression
Before calculating a compressor cycle, gas properties (heat capacity ratio, compressibility, molecular weight, etc.) must be determined for the fluid to be compressed. For mixtures, use an appropriate weighted mean value for the specific heats and molecular weight.
Adiabatic, Isentropic Compression
If there is no heat transfer to or from the gas being compressed, the porocess is adiabatic and isentropic. From thermodynamics and the study of compressible flow, you are supposed to recall that an ideal gas compression path depends on:
Adiabatic Path
This can be rearranged to solve for density in terms of one known pressure and substituted into the work equation, which then can be integrated.
Adiabatic Work
The ratio of the isentropic work to the actual work is called the adiabatic efficiency (or isentropic efficiency). The outlet temperature may be calculated from
Adiabatic Temperature Change
Power is found by multiplying the work by the mass flow rate and adjusting for the units and efficiency.
Isothermal Compression
If heat is removed from the gas during compression, an isothermal compression cycle may be achieved. In this case, the work may be calculated from:
http://facstaff.cbu.edu/rprice/lectures/compress.html
Answer:
All 3 principal stress
1. 56.301mpa
2. 28.07mpa
3. 0mpa
Maximum shear stress = 14.116mpa
Explanation:
di = 75 = 0.075
wall thickness = 0.1 = 0.0001
internal pressure pi = 150 kpa = 150 x 10³
torque t = 100 Nm
finding all values
∂1 = 150x10³x0.075/2x0,0001
= 0.5625 = 56.25mpa
∂2 = 150x10³x75/4x0.1
= 28.12mpa
T = 16x100/(πx75x10³)²
∂1,2 = 1/2[(56.25+28.12) ± √(56.25-28.12)² + 4(1.207)²]
= 1/2[84.37±√791.2969+5.827396]
= 1/2[84.37±28.33]
∂1 = 1/2[84.37+28.33]
= 56.301mpa
∂2 = 1/2[84.37-28.33]
= 28.07mpa
This is a 2 d diagram donut is analyzed in 2 direction.
So ∂3 = 0mpa
∂max = 56.301-28.07/2
= 14.116mpa
