Given :
Power, P = 20 kW
Speed, N = 430 rpm
Allowable shear stress, τ = 65 MPa
Torque in the shaft is given by :
[tex]$P=\frac{2 \pi NT}{60}$[/tex]
[tex]$T=\frac{60 \times 20 \times 10^3}{2 \pi \times 430}$[/tex]
T = 444.37 N.m
Diameter of the solid shaft is
[tex]$d=\sqrt[3]{\frac{16 T}{\pi \tau}}[/tex]
[tex]$d=\sqrt[3]{\frac{16 \times 444.37}{3.14 \times 65}}[/tex]
[tex]$d=\sqrt[3]{34.83} $[/tex]
d = 3.265 m
d = 326.5 mm
Internal diameter of the hollow shaft is :
[tex]$\frac{T}{\frac{\pi}{32} \left( d_0^4 - d_i^4 \right)}=\frac{\tau}{d_0/2}$[/tex]
[tex]$\frac{444.37}{\frac{3.14}{32} \left( 0.036^4 - d_i^4 \right)}=\frac{65 \times 10^6}{0.036/2}$[/tex]
[tex]$\frac{444.37}{0.09 \left( 1.6 \times 10^{-6} - d_i^4 \right)}=\frac{65 \times 10^6}{0.018}$[/tex]
[tex]$\frac{7.99}{ \left( 1.6 \times 10^{-6} - d_i^4 \right)}=5850000$[/tex]
[tex]$1.3\times 10^{-6} = 1.6 \times 10^{-6} - d_i^4 \right)}$[/tex]
[tex]$d_i^4=300000$[/tex]
[tex]$d_i = 23.40$[/tex] mm
Percentage savings in the weight is given by :
Percentage saving = [tex]$\frac{W_{solid}-W_{hollow}}{W_{solid}}\times100$[/tex]
[tex]$=\frac{V_{solid}-V_{hollow}}{V_{solid}}\times100$[/tex]
[tex]$=\frac{d^2 - (d_0^2 - d_i^2)}{d^2} \times 100$[/tex]
[tex]$=\frac{(326.5)^2 - (0.036^2 - (32.40)^2)}{(326.5)^2} \times 100$[/tex]
[tex]$=\frac{106602 - \left(1.29 \times 10^{-3} - 1049.76 \right)}{106602} \times 100$[/tex]
[tex]$=\frac{106602 - 1049 }{106602} \times 100$[/tex]
[tex]$=\frac{105553 }{106602} \times 100$[/tex]
= 99.01 %
8- Concentration polarization occurs on the surface of the.......
a- cathode.
b- anode.
C- both
d-ption 4
Explanation:
Concentration overpotential, ηc,
I hope it helps you
On a ship the price gift is 24 euros .What is the difference in the price on a day when the exchange rate is £1=2378
Answer:
The answer is "[tex]\$29.7072[/tex]".
Explanation:
Please find the complete question in the attachment file.
As the original prices are 24 euros, [tex]\$ 1.2378[/tex] must be multiplied by 24.
[tex]\to 1.2378 \times 24 = \$29.7072[/tex]
Therefore the new price value will be [tex]\$ 29.7072[/tex]
Technician A says that a continuously variable transmission is an automatic transmission that does not shift gears. Technician B says that a continuously variable transmission has a maximum of six distinct gear ranges. Which technician is correct?
Answer:
Only Technician A
Explanation:
Continuously variable transmission (CVT) does not shift gears because it simply doesn't have gears. It instead uses 2 pulleys which change constantly and continuously in size, and they are linked by a belt. Continuously variable transmission (CVT) is automatic transmission as mentioned in the question above, largely due to the power the belt continuously transmits, making the vehicle's engine work in the optimum power range.
While Technician B saying, that a continuously variable transmission has a maximum of six distinct gear ranges, which is not correct because continuously variable transmission (CVT) works with an unlimited number of gear ratios within a fixed range.
The input sin(20) is sampled at 20 ms intervals by using impulse train sampling: i. Construct the input and sampled signal spectra.
Solution :
Let [tex]$x(t) = \frac{\sin (20 \pi t)}{\pi t}$[/tex]
[tex]$T_s = 20$[/tex] ms, so [tex]$f_s=\frac{1}{T_s}[/tex]
[tex]$=\frac{1}{20}$[/tex]
= 0.05 kHz
[tex]$f_s=50 $[/tex] Hz , ws = [tex]$2 \pi f_s = 100 \pi$[/tex] rad/s
We know that,
FT → [tex]$\frac{\sin (20 \pi \omega)}{\pi \omega}$[/tex]
The sampled signal is :
[tex]$XS(\omega) = \frac{1}{T_s} \sum_{k=- \infty}^{\infty}X (\omega-k\omega S)[/tex]
So, [tex]$XS(\omega) = \frac{1}{20 \times 10^{-3}} \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]
[tex]$XS(\omega) = 50 \sum_{k=- \infty}^{\infty}X (\omega-100 k \pi)[/tex]
You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are currently reviewing the scenario involving the failure of a nitrogen regulator that provides inert padding to the vapor space of the reactor. Your calculations show that the maximum discharge rate of nitrogen through the existing relief system of the vessel is 0.5 kgls, However, your calculations also show that the flow of nitrogen through the l-in supply pipe will be much greater than this. Thus under the current configuration a failure of the nitrogen regulator will result in an over pressuring of the reactor. One way to solve the problem is to install an orifice plate in the nitrogen line, thus limiting the flow to the maximum of 0.5 kg/s. Determine the orifice diameter (in cm) required to achieve this flow. Assume a nitrogen source supply pressure of 15 bar absolute. The ambient temperature is 25°C and the ambient pressure is 1 atm. 3.
Answer:
[tex]D=0.016m[/tex]
Explanation:
From the question we are told that:
Discharge Rate [tex]F_r=0.5kgls[/tex]
Pressure [tex]P=15Kpa[/tex]
Temperature [tex]T=25=>298K[/tex]
Ambient pressure is 1 atm.
Generally the equation for Density is mathematically given by
[tex]\rho=\frac{PM}{RT}[/tex]
[tex]\rho=\frac{15*10^5*28.0134*10^{-3}}{8.314*298}[/tex]
[tex]\rho=16.958kg/m^2[/tex]
Generally the equation for Flow rate is mathematically given by
[tex]F_r=\mu A\sqrt{Q \rho P(\frac{2}{Q+1})^{\frac{Q+1}{Q-1}}}[/tex]
Where
[tex]Q=Heat coefficient\ ratio\ of\ Nitrogen[/tex]
[tex]Q=1.4[/tex]
[tex]\mu= Discharge\ coefficient[/tex]
[tex]\mu=0.68[/tex]
Therefore
[tex]0.5=0.68 A\sqrt{1.4 16.958 15*10^{5}(\frac{2}{1.4+1})^{\frac{1.4+1}{1.4-1}}}[/tex]
[tex]A=2.129*10^{-4}[/tex]
Where
[tex]A=\frac{\pi}{4}D^2[/tex]
[tex]\frac{\pi}{4}D^2=2.129*10^{-4}[/tex]
[tex]D=0.016m[/tex]
The number-average molecular weight of a poly (styrene-butadiene) alternating copolymer is 1,350,000 g/mol. What is the average number of styrene and butadiene repeat units per molecule.
a) 6,806
b) 6,944
c) 4,801
d) 8,544
if a person is injured at the hospital during a natural disaster a correct action to take is
Increasing following distance to
when encountering other motorists who follow too closely
is an example of appropriate implementation of the IPDE defensive driving strategy for the
maintenance of an appropriate Safety Cushion.
Two-seconds
Enree-seconds
Four-seconds
Twenty-seconds
Answer:
Increasing following distance to Four-seconds when encountering other motorists who follow too closely is an example of appropriate implementation of the IPDE defensive driving strategy for the maintenance of an appropriate Safety Cushion.
Explanation:
Maintaining the required safety cushion by utilizing the IPDE defensive driving strategy to manage the nine to fifteen space driving zones involves continuous scanning. Therefore, motorists should be able to identify objects and hazards in the driving scene, line of sight, and path of travel. They should predict points of driving conflicts. They should determine appropriate and safe driving actions to take, when, and where. Finally, action is required to ensure that conflicts are avoided.
hỗ trợ mình với được không các bạn
Answer:
Explanation:
Be bop
Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? When would you recommend auaching fins both inside and outside the tubes?
Answer:
Fins should be attached outside the tube Fins can be attached on both sides when convection coefficient of air inside the tube is equal to the convection coefficient of atmospheric air outside the tubeExplanation:
The main function of the fins that are to be added is to ensure the speedy transfer of heat from the Hot air.
The fins should be attached outside the tube because the convection coefficient of air is higher inside the tube than the convection coefficient of the outside air ( atmospheric air ), BUT
When convection coefficient of air inside the tube is equal to the atmospheric air outside the tube, it is recommended that the fins can be added on both sides of the tube ( i.e. in and outside the tube )
17- The cathodic polarization is ..... *
O
a- activation.
O
b- concentration
O c- both.
A 2-stage dcv that has an internal pilot does not work well (if at all) on
Answer:
i really font onow why tbh eot you
deep invasion lead to???
Answer:
2
Explanation:
usjshshshjeoeieieiiekeiekeiekekekekekkske
Answer:
2
Explanation:
High separation between deep...........
If the constant is added to every observation of data then arithmatic mean obtained is
Answer:
Explanation:
Increased by the constant. Take a very simple case.
4 + 5 + 6 = 15
The mean is 5 (obtained by dividing the total (15) by the number of terms (3).
Now add a constant say 6
4 + 6 = 10
5 + 6 = 11
6 + 6 = 12
Total = 33/3 = 11
So the mean 5 is increased by the constant 6.
Now do the same thing more symbolically.
4 + c
5 + c
6 + c
Total = 15 + 3c
Divide by 3 you get 5 + c
If you want a more formal proof involving n terms, leave a note.
A binary system of species 1 and 2 consists of vapor and liquid phases in equilibrium
at temperature T. The overall mole fraction of species 1 in the system is z1 = 0.65. At
temperature T, lnγ1 = 0.67 x2
2; lnγ2 = 0.67 x1
2; P1
sat = 32.27 kPa; and P2
sat = 73.14 kPa.
Assuming the validity of Eq. (13.19),
Final PDF to printer
13.10. Problems 511
smi96529_ch13_450-523.indd 511 01/06/17 03:27 PM
(a) Over what range of pressures can this system exist as two phases at the given T and z1?
(b) For a liquid-phase mole fraction x1 = 0.75, what is the pressure P and what molar
fraction of the system is vapor?
(c) Show whether or not the system exhibits an azeotrope
Different metabolic control systems have different characteristic time scales for a control response to be achieved. Match the time scale with the control system.
a. Covalent modification
b. Allosteric control
c. Gene expression
1. Seconds to minutes
2. Milliseconds
3. Hours
Answer:
a. Covalent modification = Seconds to minutes
b. Allosteric control = Milliseconds
c. Gene expression = Hours
Explanation:
Covalent modifications refer to the addition and/or removal of chemical groups by the action of particular enzymes such as methylases, acetylases, phosphorylases, phosphatases, etc. For example, histones are chromatin-associated proteins covalently modified by enzymes that add methyl groups (histone methylation), acetyl groups (histone acetylation), phosphate groups (histone phosphorylation), etc. Moreover, allosteric control, also known as allosteric regulation, is a type of regulation of the enzyme activity by binding an effector molecule (allosteric modulator) at a different site than the enzyme's active site, thereby triggering a conformational change on the enzyme upon binding of an effector. Finally, gene expression encompasses the cellular processes by which genetic information flows from genes to proteins (i.e., transcription >> translation). In metabolic pathways, enzymes that are able to catalyze irreversible reactions represent sites of control (for example, during glycolysis, pyruvate kinase is an enzyme that catalyzes an irreversible reaction, thereby serving as a control site). In turn, enzymatic activity is modulated by covalent modifications or reversible binding of allosteric effectors. Finally, metabolic pathways are also modulated by gene regulatory mechanisms that control the transcription of specific enzymes required for such pathways. During these processes, the times required for allosteric regulation, covalent modification (e.g., phosphorylation) and transcriptional control can be counted in milliseconds, seconds, and hours, respectively.
What is protection scheme?
Answer:
The objective of a protection scheme is to keep the power system stable by isolating only the components that are under fault, whilst leaving as much of the network as possible still in operation.
Explanation:
The devices that are used to protect the power systems from faults are called protection devices.
By using order of magnitude analysis, the continuity and Navier-Stokes equations can be simplified to the Prandtl boundary-layer equations. For steady, incompressible, and two-dimensional flow, neglecting gravity, the result is delta u/ delta x + delta v/ delta y= 0; u delta u/ delta x +v delta u/ delta y= -1/p(delta u/ delta x)+ v delta^2 u/ delta y^2 Use L and V0 as characteristic length and velocity, respectively. Non-dimensionalize these equations and identify the similarity parameters that result.
Answer: Attached below is the well written question and solution
answer:
i) Attached below
ii) similar parameter = [tex]\frac{V}{VoL } = 1 / Re[/tex]
Explanation:
Using ; L as characteristic length and Vo as reference velocity
i) Nondimensionalize the equations
ii) Identifying similarity parameters
the similar parameters are = [tex]\frac{V}{VoL } = 1 / Re[/tex]
Attached below is the detailed solution
You are responsible for notifying the DMV within 5 days of the sale using a Notice of Release of Liability form if you sell or transfer a vehicle to someone else.
Answer:
True
Explanation:
After you have sold or have made the transfer of ownership of a motor vehicle to another person, you are required to fill a notice of transfer and release of liability. You do this as a notification to let DMV be aware of a change of ownership of this vehicle. And it also serves to protect the previous owner from liabilities such as parking violations or traffic violations. This notification should be done within 5 Days of the transfer of ownership.
Thank you.
State the factor that influence the frequency of the induced emf of an alternating quantity
Explanation:
conductor, flux, and movement of conductor in magnetic field are some of the factors that induce emf.
Which statement describes the relay between minerals and rocks ?
Answer:
•○●□■hey hi!■□●○•Explanation:
Minerals and rocks are the same. Aggregates of minerals form rocks. Minerals determine the texture of a rock. Most rocks are made of a single mineral type.
☆♡hope this helps♡☆A technician wants to implement a dual factor authentication system that will enable the organization to authorize access to sensitive systems on a need-to-know basis. What should be implemented during the authorization stage?
Answer: Biometrics
Explanation:
Dual factor authentication refers to an electronic authentication method whereby a user will only be granted an access to an application or a website after the user has successfully been able to present two pieces of evidence which then grants access to the application or website.
Since the technician wants to implement a dual factor authentication system, the biometrics should be implemented during the authorization stage.
Biometrics refers to the body measurements and the calculations that are related to the characteristics of humans. Biometric authentication is used as a form of identification.
I want to explain what 2000 feet looks like to young children so that they can imagine it in class
Answer:
maybe take a really common toy kids play with or often see, find the average height for the toy and do the math to see how many of those toys stacked ontop of eachother would make up 2000 feet. For example (this isn't accurate btw just an idea of what it would sound like but) "Have you ever seen a barbie doll? well if you stack 400 barbie dolls ontop of their head it would be equal to 2000 feet."
Explanation:
sometimes taking common or beloved objects children have into your examples makes them have a better image of how small or how big something is.
can you guys please introduce yourself
Answer: why?
Explanation:
You toss two coins. If you get heads with the first coin, you stop. If you get tails, you toss it again. The second coin is tossed regardless. What is the ratio of heads to tails?
Answer:
1:1
Explanation:
A coin has either a head or tail.
Thus, probability of head = ½
Probability of tail = ½
The first coin is tossed twice and probability of head or tail on both tosses is still ½. Thus, they will have a ratio 1:1.
The second coin is still tossed regardless and thus continuously and so should have same ratio of 1:1.
Therefore, the ratio of the both of them must, also be 1:1.
Plexiglas, polymethylmethacrylate (PMMA), is often used as display windows and cases in art galleries and museums. If its flame imparts 32 kW/m^2 to the surface when it burns estimate the energy release rate of a 3 m^2 square sheet (one side). Assume its vaporization temperature is 350°C. Calculate with units, provide work.
This question is incomplete, the complete question is;
Plexiglas, polymethylmethacrylate (PMMA), is often used as display windows and cases in art galleries and museums. If its flame imparts 32 kW/m^2 to the surface when it burns estimate the energy release rate of a 3 m^2 square sheet (one side). Assume its vaporization temperature is 350°C. Calculate with units, provide work.
a ⇒ [tex]q_{rad[/tex] = σT⁴, where σ = 5.67 × 10⁻¹¹ kW/m².K⁴
b ⇒ [tex]q"[/tex] = [tex]q_{flame[/tex] - [tex]q_{rad[/tex]
c ⇒ [tex]m"[/tex] = [tex]q"[/tex]/L, where L = 1.6 kJ/g
d ⇒ Q = [tex]m"[/tex] × A × ΔHc, where ΔHc is 24.9 kJ/g
Answer:
a) [tex]q_{rad[/tex] = 8.54 kW/m²
b) [tex]q"[/tex] = 23.46 kW/m²
c) [tex]m"[/tex] = 14.6625 g/m²
d) Q = 1095.2888 kJ
Explanation:
Given the data in the question;
[tex]q_{flame[/tex] = 32 kW/m²
Area; A = 3m²
vaporization temperature; T = 350°C = ( 350 + 273 )K = 623 K
Now,
a) ⇒ [tex]q_{rad[/tex] = σT⁴, where σ = 5.67 × 10¹¹ kW/m².K⁴
we substitute
[tex]q_{rad[/tex] = ( 5.67 × 10⁻¹¹ kW/m².K⁴ ) × ( 623 K)⁴
[tex]q_{rad[/tex] = ( 5.67 × 10⁻¹¹ kW/m².K⁴ ) × 150644120641 K⁴
[tex]q_{rad[/tex] = 8.54 kW/m²
b) ⇒ [tex]q"[/tex] = [tex]q_{flame[/tex] - [tex]q_{rad[/tex]
we substitute
[tex]q"[/tex] = 32 kW/m² - 8.54 kW/m²
[tex]q"[/tex] = 23.46 kW/m²
c) ⇒ [tex]m"[/tex] = [tex]q"[/tex]/L, where L = 1.6 kJ/g
we substitute
[tex]m"[/tex] = 23.46 / 1.6
[tex]m"[/tex] = 14.6625 g/m²
d) ⇒ Q = [tex]m"[/tex] × A × ΔHc, where ΔHc is 24.9 kJ/g
we substitute
Q = 14.6625 × 3 × 24.9
Q = 1095.2888 kJ
In a major human artery with an internal diameter of 5mm, the flow of blood, averaged over the cardiac cycle is 5cm3·s−1. The artery bifurcates (splits) into two identical blood vessels that are each 3mm in diameter. What are the average velocity and the mass flow rate upstream and downstream of the bifurcation? The density of blood is 1.06g·cm−3
9514 1404 393
Answer:
see attached
Explanation:
Assuming flow is uniform across the cross section of the artery, the mass flow rate is the product of the volumetric flow rate and the density.
(5 cm³/s)(1.06 g/cm³) = 5.3 g/s
If we assume the blood splits evenly at the bifurcation, then the downstream mass flow rate in each artery is half that:
(5.3 g/s)/2 = 2.65 g/s
__
The average velocity will be the ratio of volumetric flow rate to area. Upstream, that is ...
(5 cm³/s)/(π(0.25 cm)²) ≈ 25.5 cm/s
Downstream, we have half the volumetric flow and a smaller area.
(2.5 cm³/s)/(π(0.15 cm)²) ≈ 35.4 cm/s
A power cycle receives QH by heat transfer from a hot reservoir at TH = 1200 K and rejects energy QC by heat transfer to a cold reservoir at TC = 400 K. For each of the following cases, determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
a.QH = 900 kJ, Wcycle= 450 kJ
b. QH = 900 kJ, Qc = 300 kJ
c. Weycle = 600 kJ, Qc= 400 kJ
d. η = 75%
Answer:
a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.
Explanation:
Maximum theoretical efficiency for a power cycle ([tex]\eta_{r}[/tex]), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:
[tex]\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\%[/tex] (1)
Where:
[tex]T_{C}[/tex] - Temperature of the cold reservoir, in Kelvin.
[tex]T_{H}[/tex] - Temperature of the hot reservoir, in Kelvin.
The maximum theoretical efficiency associated with this power cycle is: ([tex]T_{C} = 400\,K[/tex], [tex]T_{H} = 1200\,K[/tex])
[tex]\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%[/tex]
[tex]\eta_{r} = 66.667\,\%[/tex]
In exchange, real efficiency for a power cycle ([tex]\eta[/tex]), no unit, is defined by this expression:
[tex]\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\%[/tex] (2)
Where:
[tex]Q_{C}[/tex] - Heat released to cold reservoir, in kilojoules.
[tex]Q_{H}[/tex] - Heat gained from hot reservoir, in kilojoules.
[tex]W_{C}[/tex] - Power generated within power cycle, in kilojoules.
A power cycle operates irreversibly for [tex]\eta < \eta_{r}[/tex], reversibily for [tex]\eta = \eta_{r}[/tex] and it is impossible for [tex]\eta > \eta_{r}[/tex].
Now we proceed to solve for each case:
a) [tex]Q_{H} = 900\,kJ[/tex], [tex]W_{C} = 450\,kJ[/tex]
[tex]\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 50\,\%[/tex]
Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.
b) [tex]Q_{H} = 900\,kJ[/tex], [tex]Q_{C} = 300\,kJ[/tex]
[tex]\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 66.667\,\%[/tex]
Since [tex]\eta = \eta_{r}[/tex], the power cycle operates reversibly.
c) [tex]W_{C} = 600\,kJ[/tex], [tex]Q_{C} = 400\,kJ[/tex]
[tex]\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%[/tex]
[tex]\eta = 60\,\%[/tex]
Since [tex]\eta < \eta_{r}[/tex], the power cycle operates irreversibly.
d) Since [tex]\eta > \eta_{r}[/tex], the power cycle is impossible.
Race cars at the Indianapolis Speedway average speeds of 185 mi/h. After determining the altitude of Indianapolis, find the Mach number of these cars and estimate whether compressibility might affect their aerodynamics.
Answer:
- the Mach number is 0.24.
- Compressibility becomes effective when Mach number is greater than 0.3, the Mach number of the race cars is less than 0.3, hence, compressibility will not affect their aerodynamics.
Explanation:
Given the data in the question;
Average speed V = 185 miles per hour = ( 185 /2.237 ) m/s = 82.7 m/s
From Almanac, we can find that Indianapolis is at 220 m altitude.
So from table, at that altitude, the standard speed of sound will be 339.4 m/s .
Mach number of the race car will be;
Mach Number = Velocity / sound speed
we substitute
Mach Number = ( 82.7 m/s ) / ( 339.4 m/s )
Mach Number = 0.24
Therefore the Mach number is 0.24.
We know that, compressibility becomes effective when the Mach number is greater than 0.3.
Since the Mach number of the race cars is less than 0.3, compressibility will not affect their aerodynamics.
Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the ____________ of the pipe.
Answer:
Inlet
Explanation:
Consider a laminar forced flow inside a pipe with constant wall temperature, the heat flux will have a higher value near the INLET of the pipe.
This is because the friction factor is experienced at the highest level when a laminar forced flow is at the tube inlet where the thickness of the boundary layer is zero. Also, the friction factor decreases step by step at a lower rate to the fully augmented value.
