What types of issues MAY occur to slow or prevent the best outcome?

Water enters the pump of a steam power plant as saturated liquid at 20 kPa at a rate of 45 kg/s and exits at 6 MPa. Neglecting t

Natasha2012 [34]

Answer:

$\dot W_{in} = 273.69\,kW$

Explanation:

The pump is modelled after the First Law of Thermodynamics. A reversible process means that fluid does not report any positive change in entropy:

$\dot W_{in} + \dot m \cdot (h_{in}-h_{out}) = 0$

The properties of the fluid at entrance and exit are, respectively:

Inlet (Saturated Liquid)

$P = 20\,kPa$

$T = 60.06\,^{\textdegree}C$

$h = 251.42\,\frac{kJ}{kg}$

$s = 0.8320\,\frac{kJ}{kg\cdot K}$

Outlet (Subcooled Liquid)

$P = 6000\,kPa$

$T = 60.06\,^{\textdegree}C$

$h = 257.502\,\frac{kJ}{kg}$

$s = 0.8320\,\frac{kJ}{kg\cdot K}$

The power input to the pump is computed hereafter:

$\dot W_{in} = \left(45\,\frac{kg}{s} \right)\cdot \left(257.502\,\frac{kJ}{kg} -251.42\,\frac{kJ}{kg} \right)$

$\dot W_{in} = 273.69\,kW$

8 0

4 years ago

The supplement file* that enclosed to this homework consists Time Versus Force data. The first column in the file stands for tim

AysviL [449]

Answer:

A.) 1mv = 2000N

B.) Impulse = 60Ns

C.) Acceleration = 66.67 m/s^2

Velocity = 4 m/s

Displacement = 0.075 metre

Absorbed energy = 60 J

Explanation:

A.) Using a mathematical linear equation,

Y = MX + C

Where M = (2000 - 0)/( 898 - 0 )

M = 2000/898

M = 2.23

Let Y = 2000 and X = 898

2000 = 2.23(898) + C

2000 = 2000 + C

C = 0

We can therefore conclude that

1 mV = 2000N

B.) Impulse is the product of force and time.

Also, impulse = momentum

Given that

Mass M = 30kg

Velocity V = 2 m/s

Impulse = M × V = momentum

Impulse = 30 × 2 = 60 Ns

C.) Force = mass × acceleration

F = ma

Substitute force and mass into the formula

2000 = 30a

Make a the subject of formula

a = 2000/30

acceleration a = 66.67 m/s^2

Since impulse = 60 Ns

From Newton 2nd law,

Force = rate of change in momentum

Where

change in momentum = -MV - (- MU)

Impulse = -MV + MU

Where U = initial velocity

60 = -60 + MU

30U = 120

U = 120/30

U = 4 m/s

Force = 2000N

Impulse = Ft

Substitute force and impulse to get time

60 = 2000t

t = 60/2000

t = 0.03 second

Using third equation of motion

V^2 = U^2 + 2as

Where S = displacement

4^2 = 2^2 + 2 × 66.67S

16 = 4 + 133.4S

133.4S = 10

S = 10/133.4

S = 0.075 metre

D.) Energy = 1/2 mV^2

Energy = 0.5 × 30 × 2^2

Energy = 15 × 4 = 60J

5 0

3 years ago

If the reading of mercury manometer was 728 mmHg, what is the reading for another liquid such as water in mH20 units?

vodka [1.7K]

Answer:

mH275 units

Explanation:

that was true

4 0

3 years ago

An experiment compares the initial speed of bullets fired from two handguns: a 9 mm and a 0.44 caliber. The guns are fired into

olasank [31]

Answer:

1.176

Explanation:

When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.

v2 * (M + mb) = v1 * mb

Where

v1: muzzle velocity of the bullet

M: mass of the bob

mb: mass of the bullet

v2: mass of the bob with the bullet after being hit

v2 = v1 * mb / (M + mb)

Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.

Ek = 1/2 (M + mb) * v2^2

Ep = (M + mb) * g * h

Ek = Ep

1/2 (M + mb) * v2^2 = (M + mb) * g * h

1/2 * (v1 * mb / (M + mb))^2 = g * h

1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h

v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)

$v2 = \sqrt{\frac{g *h * (M+ mb)^2}{\frac{1}{2} * mb^2}}$

The height h that it reaches is related to the length L of the pendulum arm and the angle it forms with the vertical.

h = L * (1 - cos(a))

$v2 = \sqrt{\frac{g * L * (1 - cos(a)) * (M+ mb)^2}{\frac{1}{2} * mb^2}}$

For the 9 mm:

$v2 = \sqrt{\frac{9.81 * L * (1 - cos(4.3)) * (10+ 0.006)^2}{\frac{1}{2} * 0.006^2}} = \sqrt{L} * 391$

For the 0.44 caliber:

$v2 = \sqrt{\frac{9.81 * L * (1 - cos(10.1)) * (10+ 0.012)^2}{\frac{1}{2} * 0.012^2}} = \sqrt{L} * 460$

The ratio is 460 / 391 = 1.176

6 0

3 years ago

Create a C language program that can be used to construct any arbitrary Deterministic Finite Automaton corresponding to the FDA

otez555 [7]

Answer:

see the explanation

Explanation:

/* C Program to construct Deterministic Finite Automaton */

#include <stdio.h>

#include <DFA.h>

#include <stdlib.h>

#include <math.h>

#include <string.h>

#include <stdbool.h>

struct node{

struct node *initialStateID0;

struct node *presentStateID1;

};

printf("Please enter the total number of states:");

scanf("%d",&count);

//To create the Deterministic Finite Automata

DFA* create_dfa DFA(){

q=(struct node *)malloc(sizeof(struct node)*count);

dfa->initialStateID = -1;

dfa->presentStateID = -1;

dfa->totalNumOfStates = 0;

return dfa;

}

//To make the next transition

void NextTransition(DFA* dfa, char c)

{

int tID;

for (tID = 0; tID < pPresentState->numOfTransitions; tID++){

if (pPresentState->transitions[tID].condition(c))

{

dfa->presentStateID = pPresentState->transitions[tID].toStateID;

return;

}

dfa->presentStateID = pPresentState->defaultToStateID;

}

//To Add the state to DFA by using number of states

void State_add (DFA* pDFA, DFAState* newState)

{

newState->ID = pDFA->numOfStates;

pDFA->states[pDFA->numOfStates] = newState;

pDFA->numOfStates++;

}

void transition_Add (DFA* dfa, int fromStateID, int(*condition)(char), int toStateID)

{

DFAState* state = dfa->states[fromStateID];

state->transitions[state->numOfTransitions].toStateID = toStateID;

state->numOfTransitions++;

}

void reset(DFA* dfa)

{

dfa->presentStateID = dfa->initialStateID;

}

5 0

4 years ago