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Scorpion4ik [409]
3 years ago
7

A motor driven water pump operates with an inlet pressure of 96 kPa (absolute) and mass flow rate of 120 kg/min. The motor consu

mes 0.8 kW of electrical power, determine the maximum water pressure at the discharge of the pump.
Engineering
1 answer:
NeX [460]3 years ago
8 0

Answer:

The maximum water pressure at the discharge of the pump (exit) = 496 kPa

Explanation:

The equation expressing the relationship of the power input of a pump can be computed as:

E _{pump,u} = \dfrac{m(P_2-P_1)}{\rho}

where;

m = mass flow rate = 120 kg/min

the pressure at the inlet P_1 = 96 kPa

the pressure at the exit P_2 = ???

the pressure \rho = 1000 kg/m³

∴

0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}

0.8 \times 10^{3}\times 1000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}

800000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}

\dfrac{800000}{2} = P_2-96000

400000 = P₂ - 96000

400000 +  96000 = P₂

P₂ = 496000 Pa

P₂ = 496 kPa

Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa

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Answer:

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Explanation:

Since, there are two coils, they can be used independently or in series or parallel. The power is given as:

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I = V/R

therefore,

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Hence, the resistance (R) and (P) are inversely proportional. Therefore, the maximum value of resistance will give minimum power, that is, 300 W. And the maximum resistance will be in series arrangement, as in series the total resistance gets higher than, any individual resistance.

Therefore,

Rmax = V²/Pmin = R1 + R2

R1 + R2 = (220 V)²/300 W

R1 + R2 = 161.333 Ω    ______ en (1)

Similarly, the minimum resistance will give maximum power. And the minimum resistance will occur in parallel combination. Because equivalent resistance of parallel combination is less than any individual resistance.

Therefore,

(R1 R2)/(R1 + R2) = (220 V)²/2000 W

using eqn (1), we get:

(R1 R2) / 161.333 Ω = 24.2 Ω

R1 R2 = 3904.266 Ω²

R1 = 3904.266 Ω²/R2  _____ eqn (2)

Using this value of R1 in eqn (1), we get:

3904.266/R2 +R2 = 161.333

(R2)² - 161.333 R2 +3904.266 = 0

Solving this quadratic eqn we get two values of R2 as:

R2 = 131.7 Ω     OR     R2 = 29.64 Ω

when ,we substitute these values in eqn (1) to find R1, we get get the same two values as R2, alternatively. This means that the two coils have these resistance, and the order does not matter.

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Answer:

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For this problem, you may not look at any other code or pseudo-code (even if it is on the internet), other than what is on our w
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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,

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

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

current = parent

(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

else if A[parent] < A[current]

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:

8 0
4 years ago
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