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zvonat [6]
3 years ago
14

Water at 20◦C is pumped through 1000 ft of 0.425 ft diameter pipe at a volumetric flowrate of 1 ft3/s through a cast iron pipe t

hat connects to connects two reservoirs. The elevation difference between the two reserviors is 120 ft. Find the pumping power delivered to the water. The minor losses only include a wide-open globe valve with KL = 10. Use a density of 1.94 slug/ft3 and a dynamic viscosity of 2.32 × 10−5 lbf·s/ft2 .
Engineering
1 answer:
vichka [17]3 years ago
8 0

Answer:

7582.9 ft.Ibf/s

Explanation:

Given

L=1000ft,d=0.425ft,Q=1ft^3/s,z2-z1=120ft,Kl=10,d=1.94slug/ft^3, vicosity u= 2.32*10-5ibf.s/ft2

Reynold Re= Density*diameter*velocity/ viscosity

But Q=AV

V= 4/3.142*0.425=2.99ft/s

Re= 1.94*0.425*2.99/2.32*10-5)=106455.3

Friction factor=1/√f=-1.8log[((e/d)/3.7)^1.11+6.9/Re] is very neglible hence equals 0

Pump head Hp= z2-z1+v^2/2g[FL/f+KL]

Hp=120+2.99^2/2*32.2(0+10)=121.4ft

Pump power = density*g*Q*hp

1.94*32.2*121.4=7582.9 ft.Ibf/s

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Answer:

(a)

(i) pseudo code :-

current = i

// assuming parent of root is -1

while A[parent] < A[current] && parent != -1 do,

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swap(A[current],A[parent])

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(ii) In heap we create a complete binary tree which has height of log(n). In shift up we will take maximum steps equal to the height of tree so number of comparison will be in term of O(log(n))

(b)

(i) There are two cases while comparing with grandparent. If grandparent is less than current node then surely parent node also will be less than current node so swap current node with parent and then swap parent node with grandparent.

If above condition is not true then we will check for parent node and if it is less than current node then swap these.

pseudo code :-

current = i

// assuming parent of root is -1

parent is parent node of current node

while A[parent] < A[current] && parent != -1 do,

if A[grandparent] < A[current] // if current element is bigger than parent then shift it up

swap(A[current],A[parent])

swap(A[grandparent],A[parent])

current = grandparent

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swap(A[parent],A[current])

current = parent

(ii) Here we are skipping the one level so max we can make our comparison half from last approach, that would be (height/2)

so order would be log(n)/2

(iii) C++ code :-

#include<bits/stdc++.h>

using namespace std;

// function to return index of parent node

int parent(int i)

{

if(i == 0)

return -1;

return (i-1)/2;

}

// function to return index of grandparent node

int grandparent(int i)

{

int p = parent(i);

if(p == -1)

return -1;

else

return parent(p);

}

void shift_up(int A[], int n, int ind)

{

int curr = ind-1; // because array is 0-indexed

while(parent(curr) != -1 && A[parent(curr)] < A[curr])

{

int g = grandparent(curr);

int p = parent(curr);

if(g != -1 && A[g] < A[curr])

{

swap(A[curr],A[p]);

swap(A[p],A[g]);

curr = g;

}

else if(A[p] < A[curr])

{

swap(A[p],A[curr]);

curr = p;

}

}

}

int main()

{

int n;

cout<<"enter the number of elements :-\n";

cin>>n;

int A[n];

cout<<"enter the elements of array :-\n";

for(int i=0;i<n;i++)

cin>>A[i];

int ind;

cout<<"enter the index value :- \n";

cin>>ind;

shift_up(A,n,ind);

cout<<"array after shift up :-\n";

for(int i=0;i<n;i++)

cout<<A[i]<<" ";

cout<<endl;

}

Explanation:

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Formula for <u>distance per hole</u>: 0.5 + A + 1.75

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Total time required to drill 100 holes (t):

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7 0
3 years ago
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