All categories

2 years ago

Question 1 : Replacement [42 Pts] Consider the following page reference string:

Engineering

2 answers:

mariarad [96]2 years ago

7 0

Answer:

2,970

Explanation:

The string are arranged as follow: a,a,b,b,c,c,d,e,e,g,g,g

There are 12 combinations and with four frames

Then, it can be computed thus: nCr, where n = 12 and r = 4 = n!/r(n - r)!

∴ 12C4 = 12!/4(8)! = 2970

MakcuM [25]2 years ago

4 0

Answer:

Optimal

Time 123456789101112

RS. ecbeagdcegda

F0. eeeeeeeee e e a

F1 c c c c c c c c c c c

F2 b b b g g g g g g g

F3 a a d d d d d d

Page fault? * * * * * * *

Total page fault:7

2. LRU

Time 1 2 3 4 5 6 7 8 9 10 11 12

RS e c b e a g d c e g d a

F0 e e e e e e e c c c c a

F1 c c c c g g g g g g g

F2 b b b b d d d d d d

F3 a a a a e e e e

Page fault? Y Y Y N Y Y Y Y Y N N Y Total page fault:9

3. LRU approximation algorithm: Second chance

Time 1 2 3 4 5 6 7 8 9 10 11 12

RS e c b e a g d c e g d a

F0 0,e 0,e 0,e 1,e 1,e 0,e 0,e 0,e 1,e 1,e 1,e 0,e

F1 0,c 0,c 0,c 0,c 0,g 0,g 0,g 0,g 1,g 1,g 0,g

F20,b0,b0,b0,b0,d0,d0,d0,d1,d0,dF30,a0,a0,a0,c0,c0,c0,c0,a

Page fault? YYYNYYYYNNNY

Total page fault: 8

You might be interested in

A thin, flat plate that is 0.2 m × 0.2 m on a side is oriented parallel to an atmospheric airstream having a velocity of 40 m/s.

Leto [7]

Answer:

The rate of heat transfer from both sides of the plate to the air is 240 W

Explanation:

Given;

area of the flat plate = 0.2 m × 0.2 m = 0.04 m²

velocity of atmospheric air stream, v = 40 m/s

drag force, F = 0.075 N

The rate of heat transfer from both sides of the plate to the air:

q = 2 [h'(A)(Ts -T∞)]

where;

h' is heat transfer coefficient obtained from Chilton-Colburn analogy

$h' = \frac{C_f}{2} \rho u C_p P_r^{-2/3}\\\\\frac{C_f}{2} = \frac{\tau'_s}{2*\rho u^2/2}$

Properties of air at 70°C and 1 atm:

ρ = 1.018 kg/m³, cp = 1009 J/kg.K, Pr = 0.7, v = 20.22 x 10⁻⁶ m²/s

$\frac{C_f}{2} = \frac{(0.075/2)/(0.2)^2}{2*(1.018)(40)^2/2} = 5.756*10^{-4}\\\\Thus,\\h' = 5.756*10^{-4} (1.018*40*1009)*(0.7)^{-2/3}\\\\h' = 30 \ W/m^2.K$

Finally;

q = 2 [ 30(0.04)(120 - 20) ]

q = 240 W

Therefore, the rate of heat transfer from both sides of the plate to the air is 240 W

4 0

2 years ago

Read 2 more answers

Link BD consists of a single bar 36 mm wide and 18 mm thick. Knowing that each pin has a 12-mm diameter, determine the maximum v

MAXImum [283]

Answer:

hello the diagram attached to your question is missing attached below is the missing diagram

answer :

a) 48.11 MPa

b) - 55.55 MPa

Explanation:

First we consider the equilibrium moments about point A

∑ Ma = 0

( Fbd * 300cos30° ) + ( 24sin∅ * 450cos30° ) - ( 24cos∅ * 450sin30° ) = 0

therefore ;<em> Fbd = 36 ( cos ∅tan30° - sin∅ ) kN ----- ( 1 )</em>

A ) when ∅ = 0

Fbd = 20.7846 kN

link BD will be under tension when ∅ = 0, hence we will calculate the loading area using this equation

A = ( b - d ) t

b = 12 mm

d = 36 mm

t = 18

therefore loading area ( A ) = 432 mm^2

determine the maximum value of average normal stress in link BD using the relation below

бbd = $\frac{Fbd}{A}$ = 20.7846 kN / 432 mm^2 = 48.11 MPa

b) when ∅ = 90°

Fbd = -36 kN

the negativity indicate that the loading direction is in contrast to the assumed direction of loading

There is compression in link BD

next we have to calculate the loading area using this equation ;

A = b * t

b = 36mm

t = 18mm

hence loading area = 36 * 18 = 648 mm^2

determine the maximum value of average normal stress in link BD using the relation below

бbd = $\frac{Fbd}{A}$ = -36 kN / 648mm^2 = -55.55 MPa

4 0

1 year ago

A 0.2-m^3 rigid tank equipped with a pressure regulator contains steam at 2MPa and 320C. The steam in the tank is now heated. Th

prohojiy [21]

Answer:

Q=486.49 KJ/kg

Explanation:

Given that

V= 0.2 m³

At initial condition

P= 2 MPa

T=320 °C

Final condition

P= 2 MPa

T=540°C

From steam table

At P= 2 MPa and T=320 °C

h₁=3070.15 KJ/kg

At P= 2 MPa and T=540°C

h₂=3556.64 KJ/kg

So the heat transfer ,Q=h₂ - h₁

Q= 3556.64 - 3070.15 KJ/kg

Q=486.49 KJ/kg

7 0

2 years ago

Water (cp = 4180 J/kg·°C) enters the 2.5 cm internal diameter tube of a double-pipe counter-flow heat exchanger at 17°C at a rat

aksik [14]

Answer:

Length = 129.55m, 129.55m

Explanation:

Given:

cp of water = 4180 J/kg·°C

Diameter, D = 2.5 cm

Temperature of water in = 17°C

Temperature of water out = 80°C

mass rate of water =1.8 kg/s.

Steam condensing at 120°C

Temperature at saturation = 120°C

hfg of steam at 120°C = 2203 kJ/kg

overall heat transfer coefficient of the heat exchanger = 700 W/m2 ·°C

U = 700 W/m2 ·°C

Since Temperature of steam is at saturation,

temperature of steam going in = temperature of steam out = 120°C

Energy balance:

Heat gained by water = Heat loss by steam

Let specific capacity of steam = 2010kJ/Kg .°C

Find attached the full solution to the question.

3 0

2 years ago

An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160kPa. Determine (a) the temperature, (b) the quality, (c) th

makvit [3.9K]

Answer:

temperature -15.6 C, quality x=0.646, enthalpy h=667.20 KJ, volume of vapor phase Vg= 79.8 L

Explanation:

property table for R-134a

https://www.ohio.edu/mechanical/thermo/property_tables/R134a/R134a_PresSat.html

at 160 KPa , temperature = -15.66 C

quality x=mass of vapour/ total mass of liq-vap mixture

alternaternately: x=(v-vf)/(vg-vf)

v=total volume i.e. volume of container"80L" 80L=0.08 cubic meter

vf=vol of liquid phase vg=vol of vapor phase vf, vg values at 160Kpa

x=(0.08-0.0007437)/(0.1235-0.0007437)=0.646

enthalpy

h=hf+xhfg hf, hfg values at 160Kpa

h=hf+xhfg=31.2+0.646(209.9)=166.80 KJ/Kg

for 4Kg R-134a h=m(166.80 KJ/Kg )=667.20 KJ

volume of vapor phase

vg at 160Kpa=0.1235 cubic meter for quality=1.

in this case quality=0.646 , so it will occupy 64.6% space of the vapor phase at quality=1.

vol. of vapor phase=0.646*0.1235=0.0798 cubic meter=79.8 L

7 0

2 years ago