Answer:
The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.
Explanation:
The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"
Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:
Where:
- Safety factor, dimensionless.
- Ultimate shear strength, measured in pascals.
- Maximum allowable shear stress, measured in pascals.
The maximum allowable shear stress is consequently cleared and computed: (
, 
)
Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:
Where:
- Maximum allowable shear stress, measured in pascals.
- Shear force, measured in kilonewtons.
- Cross section area, measured in square meters.
As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:
The minimum allowable cross section area is cleared in the shearing stress equation:
If and 
, the minimum allowable cross section area is:
The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:
The diameter is now cleared and computed:
The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.
Answer: Isentropic process is the process in fluids which have a constant entropy.
Explanation: The isentropic process is considered as the ideal thermodynamical process and has both adiabatic as well as reversible processes in internal form.This process supports no transfer of heat and no transformation of matter .The entropy of the provided mass also remains unchanged or consistent.These processes are usually carried out on material on the efficient device.
Answer:
The engineers disagreed because their jobs were on the line
The ethical factors are:
The reason for the customer dumping the business is yet to be figured out
The need to keep cost to the lowest ebb in order to keep maintain profitability at the expense of employees' welfare
The are several ways of growing customer base which are yet to be exploited.
Explanation:
The engineers disagree because there is no direct connection between the company's loss of the customer and their proposed layoff of the engineers,at least no one strong evidence has been given by the president.
The ethical factors inherent in this case are as follows:
The reason for the customer dumping the business is yet to be figure out
The need to keep cost to the lowest ebb in order to keep maintain profitability.
The are several ways of growing customer base which are yet to be exploited.
There is a need for a fact-finding exercise to establish the main motive behind losing such customer,without which the company can run into more troubles in future,otherwise the company would keep firing its good hands each time a customer dumps it.
Also,the president had resulted into such decision in order to maintain company's margins,where then lies ethics of welfare economics?Welfare economics is about looking beyond margins and looking at issues from a wider perspective of fulfilling the needs of employees in order for them to put in their best performance,at least by granting them job safety.
The company could have also grown business by investing in new technology that sets it apart from competitors instead of just jumping into the conclusions of sacking employees in a business where the company's strength lies in quality of engineers that it has.
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:
V(t) = XLI₀sin(π/2 - ωt)
Explanation:
According to Maxwell's equation which is expressed as;
V(t) = dФ/dt ........(1)
Magnetic flux Ф can also be expressed as;
Ф = LI(t)
Where
L = inductance of the inductor
I = current in Ampere
We can therefore Express Maxwell equation as:
V(t) = dLI(t)/dt ....... (2)
Since the inductance is constant then voltage remains
V(t) = LdI(t)/dt
In an AC circuit, the current is time varying and it is given in the form of
I(t) = I₀sin(ωt)
Substitutes the current I(t) into equation (2)
Then the voltage across inductor will be expressed as
V(t) = Ld(I₀sin(ωt))/dt
V(t) = LI₀ωcos(ωt)
Where cos(ωt) = sin(π/2 - ωt)
Then
V(t) = ωLI₀sin(π/2 - ωt) .....(3)
Because the voltage and current are out of phase with the phase difference of π/2 or 90°
The inductive reactance XL = ωL
Substitute ωL for XL in equation (3)
Therefore, the voltage across inductor is can be expressed as;
V(t) = XLI₀sin(π/2 - ωt)
