Answer:
Disk Capacity = 45056000 bytes
Optimal skew = 26.455 ≈ 27
Maximum Data transfer rate = 262.5 GB per second
Explanation:
given data
heads = 16
cylinders = 400
cylinder zones = 100
each sector contains = 512 bytes
average seek time = 1 msec
disk rotates = 7200 RPM
to find out
disk capacity, optimal track skew, and maximum data transfer rate
solution
first we get total number of sectors that is
total number of sector = number of zones × (number of sectors in different zones
total number of sector = 100 × (160+200+240+280)
total number of sector = 88000
so
Disk Capacity = total number of sectors × size of each sector
Disk Capacity = 88000 × 512
Disk Capacity = 45056000 bytes
and
Rotation time =
Rotation time = 8.33 milli seconds
so Optimal number of sectors in a track = average of ( 160,200,240,280 )
Optimal number of sectors in a track = 220
now New sector is read every i.e = 0.0378 ms
here Optimal skew = seek time ÷ new sector read time
Optimal skew =
Optimal skew = 26.455 ≈ 27
and
here we know that for maximum transfer rate we will select cylinder with maximum number of sectors i.e here 280 sectors
so
capacity of one track with maximum = 280 × 512
capacity of one track with maximum = = 143360 bytes
and Number of rotations in 1 second is =
Number of rotations in 1 second is = 120
so Data transfer rate = Number of heads × Capacity of one track × Number of rotations in one second
Data transfer rate = 16 × 143360 × 120
Data transfer rate = 275251200 bytes per second
Data transfer rate = 262.5 GB per second
Answer:
Given that
Mass flow rate ,m=2.3 kg/s
T₁=450 K
P₁=350 KPa
C₁=3 m/s
T₂=300 K
C₂=460 m/s
Cp=1.011 KJ/kg.k
For ideal gas
P V = m R T
P = ρ RT
ρ₁=2.71 kg/m³
mass flow rate
m= ρ₁A₁C₁
2.3 = 2.71 x A₁ x 3
A₁=0.28 m²
Now from first law for open system
For ideal gas
Δh = CpΔT
by putting the values
Q= - 45.49 KJ/kg
Q =- m x 45.49 KW
Q= - 104.67 KW
Negative sign indicates that heat transfer from air to surrounding
Answer:
nf=1.11 (Goodman criterion)
ny=2.41 (factor of safety for fatigue)
Explanation:
From the table A-20 Deterministic ASTM minimum tensile and yield strengths for HR and CD steels, we have approximately:
Sut=120 kpsi
Sy=66 kpsi
Due Sut<1400 Mpa, the endurance limit is:
The surface condition modification factor is:
The effective diameter is:
The size factor is:
The endurance limit at critical location is:
From Figure A-15-15 chart, the Kf = 2.1
The notch sensitivity is:
The fatigue stress is:
The moment of inertia is:
The maximum stress is:
The mean stress is:
The alternate stress is:
The fatigue factor using Goodman is:
nf=1.11
the factor of safety is:
Answer:
C# codee
using System;
class IntegerFacts
{
static void Main()
{
int[] x = new int[10];
int sum = 0;
int siz = 0, high = 0, low = 0, avg = 0;
siz = FillArray(x);
Statistics(x, ref high, ref low, ref sum, ref avg, siz);
Console.Write("The highest value is " + high);
Console.Write("\nThe lowest value is " + low);
Console.Write("\nThe sum of the values is " + sum);
Console.Write("\nThe average is " + avg);
}
static void Statistics (int [] b, ref int h, ref int l, ref int s, ref int a, int size)
{
if (size == 0)
h = l = s = a = 0;
else
{
int i =0;
l = 999;
h = 0;
s = 0;
for (; i < size;++i)
{
if(b[i] > h)
h = b[i];
if(b[i] < l)
l = b[i];
s += b[i];
}
a = s / size;
}
}
static int FillArray (int[] a)
{
int i = 0, count = 0;
int intTemp =0 ;
string temp;
for(i = 0 ; i < 10 ; i++)
{
Console.Write("Enter element number"+(i+1) +" : ");
temp = Console.ReadLine();
if (int.TryParse(temp, out intTemp)) {
if (intTemp != 999)
{
a[i] = intTemp;
count++;
}
else
break;
}
else
{
Console.Write("\n\nOops.. You entered a wrong number. Try again \n\n");
i--;
}
}
return count;
}
}
Explanation:
The output of the above program is given in the attached file.
