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Drupady [299]
3 years ago
6

Explain the purpose of the checkpoint mechanism. How often should checkpoints be performed?How does the frequency of checkpoints

affect:
System performance when no failure occurs?
The time it takes to recover from a system crash?
The time it takes to recover from a disk crash?
Engineering
1 answer:
Olegator [25]3 years ago
3 0

Explanation:

Answer:

1) Check point mechanism is a system, point or mechanism within the database management system that allows for data, information, transactions and general logs passed through the system be retrieved and stored into a separate storage system and talking about how often the check point should be performed, it is advised that checkpoint should be done as often as possible because the more often the checkpoints is done, the lower the probability that obsolete data will be updated or retrieved during the checkpoint process.

How does the frequency of check point affects system performance when no failure occurs?

Answer: When failure does not occur, the process of checkpoint mechanism have only succeeded in accruing unnecessary cost of performing the process which at this time would have been necessary.

How does the frequency of check point affects the time it takes to recover from a system crash ?

Answer: It is important to note that the purpose of a check point mechanism in the first place is for us to be able to recover our data in the database before a time when the database became corrupt. So if we constantly run the process of check point mechanism, it will drastically reduce the time it takes to recover from a system crash because we would have previously detected obsolete data and updated the system before the crash.

How does the frequency of check point affects the time it takes to recover from a disk crash?

Answer: This can also be related to the same time of recovery as that of the system crash.  We can substitute the answer for time to recover from a system crash for that of a disc crash also. The only difference here is that it applies to a disc.

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Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l
sergeinik [125]

The local convection heat transfer coefficient at 1 m from the leading edge is  0.44 \frac{W}{m^{2} \times K} ,  the average convection heat transfer coefficient over the entire plate is  0.293 \frac{W}{m^{2} \times K}and the total heat flux transfer to the plate is 61.6 KJ.

Explanation:

It is case of heat and mass transfer in which due to temperature difference between gas  and surface. Further temperature  boundary layer will developed on flat plate in longitudinal direction.  

Hot carbon dioxide exhaust gas

physical properties

r= 1.05 \frac{kg}{m^{3}}

c_p = 1.02 \frac{kJ}{Kg \times K}

m= 231 \times 10^{7}  \frac{N \times s }{m^2}

υ = 21.8 \times 10^{6}  \frac{m^2}{s}

k = 32.5 \times 10^{3} \frac{W}{m \times K}

\alpha = 30.1 \times 10^{6} \frac{m^{2}}{s}

Pr = 0.725

Apart from these other data arr given below,

v= 3 \frac{m}{s}  \\ p= 1 atm \\ L_c = 1.5m \\T_g= 220 C \\ T_s = 80 C

To find the local convection heat transfer coefficient at 1 m from the leading edge, we use correlation used for laminar flow over flat plate,

Nu = \frac{ h \times L }{k}  = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })

where h= Average heat transfer coefficient

           L= Length of a plate

           k= Thermal Conductivity of carbon dioxide

           Re = Reynold's Number

           Pr  = Prandtle Number

(a) Convection heat transfer coefficient at 1 m from the leading edge

    is referred as local convection heat transfer coefficient.

   

   To find convection heat transfer coefficient at 1 m from leading edge,

  Nu = \frac{ h_local \times L }{k}  = 0.332 \times (Re^{\frac{1}{2} }) \times (Pr^{\frac{1}{3} })

  Here, first we have to find Re and Pr,

   Re = \frac{r \times v \times L}{m}

   Re = \frac{1.0594 \times 3 \times 1}{231 \times 10^{7}}

   Re = 20.63 \times  10^{-10}

   Pr number is take from physical property data and Pr is 0.725.

   Putting value of Re and Pr in main equation,

   we get

   Nu = \frac{ h_local \times 1 }{32.5 \times 10^{3}}  = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })

    h_local   = 32.5 \times 10^{3} \times  0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })

    h_local   =  0.44 \frac{W}{m^{2} \times K}

(b)  To find average convection heat transfer coefficient,

      it can be find out as case (a), only difference is that instead of L=1 m,        L=1.5 m would come,  

   Therefore,

    Nu = \frac{ h \times 1.5 }{32.5 \times 10^{3}}  = 0.332 \times ( (20.63 \times 10^{-10})^{\frac{1}{2} }) \times (0.725^{\frac{1}{3} })

    Finally,

      h  = \frac{0.44}{1.5}

      h  = 0.293 \frac{W}{m^{2} \times K}

(C) Total heat flux transfer to the plate is found out by,

     Q = h \times (T_g - T_s)

     Q = 0.293 \times (220-80) \\ Q= 0.293 \times 140  \\ Q= 61.6 KJ

     

     

   

   

     

   

     

   

   

 

   

   

   

   

8 0
2 years ago
In multi-grade oil what is W means?
irga5000 [103]

Answer:

winter viscosity grades

Explanation:

The “W”/winter viscosity grades describe the oil's viscosity under cold temperature engine starting conditions. There's a Low Temperature Cranking Viscosity which sets a viscosity requirement at various low temperatures to ensure that the oil isn't too thick so that the starter motor can't crank the engine over.

4 0
3 years ago
Why is data driven analytics of interest to companies?
Svetlanka [38]

Answer:

Advantages of data analysis

Ability to make faster and more informed business decisions, backed by facts. Helps companies identify performance issues that require action. ... can be seen visually, allowing for faster and better decisions.

7 0
2 years ago
Π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...
leva [86]

Answer: Pi= 4 - 4/3 + 4/5 - 4/7 + 4/9 ...

Explanation:

Is the same as the example,

If Π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...

Then

(Π/4 )*4= 4*(1 - 1/3 + 1/5 - 1/7 + 1/9 ...)

Π =4 - 4/3 + 4/5 - 4/7 + 4/9 ...

The way to write this is

Sum(from n=0 to n=inf) of (-1)^n 4/(2n+1)

(photo)

6 0
3 years ago
Compute L, T, M, LC, and R and stations of the BC and EC for the circular curve with the given data of: I (delta) = 22°15′00" an
Mars2501 [29]

Answer:

L = 475.718

T = 240.89 ft

M = 23.0195

LC = 472.728

R = 1225 ft

Explanation:

See the attached file for the calculation.

8 0
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