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navik [9.2K]
3 years ago
13

Hot engine oil with heat capacity rate of 4440 w/k (product of mass flow rate and specific heat) and an inlet temperature of 150

°c flows through a double pipe heat exchanger. the double pipe heat exchanger is constructed of a 1.5-m-long copper pipe (k = 250 w/m·k) with an inner tube of inside diameter 2 cm and outside tube diameter of 2.25 cm. the inner diameter of the outer tube of the double pipe heat exchanger is 6 cm. oil flowing at a rate of 2 kg/s through inner tube exits the heat exchanger at a temperature of 50°c. the cold fluid, i. e., water enters the heat exchanger at 20°c and exits at 70°c. assuming the fouling factor on the oil side and water side to be 0.00015 m2 ·k/w and 0.0001 m2 ·k/w, respectively, determine the overall heat transfer coefficient on inner and outer surface of the copper tube.
Engineering
1 answer:
Over [174]3 years ago
4 0

Explanation:

fluid, i. e., water enters the heat exchanger at 20°c and exits at 70°c. assuming the fouling factor on the oil side and water side to be 0.00015 m2 ·k/w and 0.0001 m2 ·k/w, respectively, determine the overall heat transfer coefficient on inner and outer surface of the copper tube.xgjicbb .follow me

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A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang
AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

i)

We know that angular speed given as

\omega =\dfrac{d\theta}{dt}

We know that for one revolution

θ=2π

Given that time t= 2 hr

So

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

V=\sqrt{\dfrac{GM}{R}}

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

V=\sqrt{\dfrac{GM}{R}}

V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}

V=7059.44 m/s

iii)

We know that centripetal fore given as

F=\dfrac{mV^2}{R}

Here given that m= 200 kg

R= 8000 km

so now by putting the values

F=\dfrac{mV^2}{R}

F=\dfrac{200\times 7059.44^2}{8000\times 10^3}

F=1245.8 N

3 0
3 years ago
"Carbon 14 (C-14), a radioactive isotope of carbon, has a half-life of 5730 ± 40 years. Measuring the amount of this isotope lef
igor_vitrenko [27]

Answer:

The age of the bones is approximately 14172 years.

Explanation:

The age of the bones can be determinated using the following decay equation:

N_{(t)} = N_{0}e^{-\lambda t}   (1)

<u>Where:</u>

N(t): is the quantity of C-14 at time t

No: is the initial quantity of C-14  

λ: is the decay rate      

t: is the time

First, we need to find λ:

\lambda = \frac{ln(2)}{t_{1/2}}

<u>Where:</u>

t(1/2): is the half-life of C-14 = 5730 y

\lambda = \frac{ln(2)}{5730 y} = 1.21 \cdot 10^{-04} y^{-1}

Now, we can calculate the age of the bones by solving equation (1) for t:

t = \frac{-ln(\frac{N_{(t)}}{N_{0}})}{\lambda}

We know that the bones have lost 82% of the C-14 they originally contained, so:

N_{t} = (1 - 0.82)N_{0} = 0.18N_{0}

t = \frac{-ln(0.18)}{1.21 \cdot 10^{-04} y^{-1}}

t = 14172 y

Therefore, the age of the bones is approximately 14172 years.

I hope it helps you!

3 0
3 years ago
When considering free energy change, biochemists usually define a standard state, the biochemical standard state, which is modif
Softa [21]

Answer:

Maximum work under this condition (∆G) = Maximum work under Standard Condition (∆G°) + Activities defining this condition

Explanation:

In this equation, the term DGo provides us with a value for the maximal work we could obtain from the reaction starting with all reactants and products in their standard states, and going to equilibrium. The term DG' provides us with a value for the maximal work we could obtain under the conditions defined by the activities in the logarithmic term. The logarithmic term can be seen as modifying the value under standard conditions to account for the actual conditions. In describing the work available in metabolic processes, we are concerned with the actual conditions in the reaction medium (whether that is a test-tube, or the cell cytoplasm); the important term is therefore DG'. If we measure the actual activities (in practice, we make do with concentrations), and look up a value for DGo in a reference book, we can calculate DG' from the above equation.

Values for DGo provide a useful indication through which we can compare the relative work potential from different processes, because they refer to a standard set of conditions.

Therefore both phrases describe the Biochemical and Chemical Standard State

3 0
3 years ago
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Maksim231197 [3]

Its FALSE  123456789012345678987653213e4r5678

6 0
4 years ago
Read 2 more answers
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
sergiy2304 [10]

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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