Ask question

Ask question

All categories

2 years ago

15

Technician A says that if fuel pump pressure is correct, fuel pump volume will be correct as well. Technician B says that a fuel

pump may produce specified pressure but below specified volume. Which technician is correct

1 answer:

guajiro [1.7K]2 years ago

5 0

Answer:

Technician B only

Explanation:

hope this helps :)

You might be interested in

Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabati

Artyom0805 [142]

Answer:

A) W' = 178.568 KW

B) ΔS = 2.6367 KW/k

C) η = 0.3

Explanation:

We are given;

Temperature at state 1;T1 = 360 °C

Temperature at state 2;T2 = 160 °C

Pressure at state 1;P1 = 10 bar

Pressure at State 2;P2 = 1 bar

Volumetric flow rate;V' = 0.8 m³/s

A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;

Specific volume;v1 = 0.287322 m³/kg

Mass flow rate of water vapour at turbine is defined by the formula;

m' = V'/v1

So; m' = 0.8/0.287322

m' = 2.784 kg/s

Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific enthalpy;h1 = 3179.46 KJ/kg

Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific enthalpy;h2 = 3115.32 KJ/kg

Now, since stray heat transfer is neglected at turbine, we have;

-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2 - h1)

Plugging in relevant values, the work of the turbine is;

W' = -2.784(3115.32 - 3179.46)

W' = 178.568 KW

B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific entropy: s1 = 7.3357 KJ/Kg.k

Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific entropy; s2 = 8.2828 KJ/kg.k

The amount of entropy produced is defined by;

ΔS = m'(s2 - s1)

ΔS = 2.784(8.2828 - 7.3357)

ΔS = 2.6367 KW/k

C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;

h2s = 2966.14 KJ/Kg

Energy equation for turbine at ideal process is defined as;

Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Again, Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2s - h1)

W' = -2.784(2966.14 - 3179.46)

W' = 593.88 KW

the isentropic turbine efficiency is defined as;

η = W_actual/W_ideal

η = 178.568/593.88 = 0.3

8 0

3 years ago

Engineering controls are the physical changes that employers make to the work environment or to equipment that make it safer to

____ [38]

Answer:

Engineering Controls. The best engineering controls to prevent heat-related illness is to make the work environment cooler and to reduce manual workload with mechanization. A variety of engineering controls can reduce workers' exposure to heat: Air conditioning, Increased general ventilation , Cooling fans , Local exhaust ventilation at points of high heat production or moisture, Reflective shields to redirect radiant heat , Insulation of hot surfaces Elimination of steam leaks , Cooled seats or benches for rest breaks , Use of mechanical equipment to reduce manual work, Misting fans that produce a spray of fine water droplets.

Hope this helped you!

Explanation:

5 0

3 years ago

Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,

Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of 195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2) There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

= 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0

3 years ago

Same rule: If both players spend the same number of coins, player 2 gains 1 coin. Off-by-one rule: If the players do not spend t

Galina-37 [17]

Answer:

Check the explanation

Explanation:

1 -

public int getPlayer2Move(int round)

{

int result = 0;

//If round is divided by 3

if(round%3 == 0) {

result= 3;

}

//if round is not divided by 3 and is divided by 2

else if(round%3 != 0 && round%2 == 0) {

result = 2;

}

//if round is not divided by 3 or 2

else {

result = 1;

}

return result;

}

2-

public void playGame()

{

//Initializing player 1 coins

int player1Coins = startingCoins;

//Initializing player 2 coins

int player2Coins = startingCoins;

for ( int round = 1 ; round <= maxRounds ; round++) {

//if the player 1 or player 2 coins are less than 3

if(player1Coins < 3 || player2Coins < 3) {

break;

}

//The number of coins player 1 spends

int player1Spends = getPlayer1Move();

//The number of coins player 2 spends

int player2Spends = getPlayer2Move(round);

//Remaining coins of player 1

player1Coins -= player1Spends;

//Remaining coins of player 2

player2Coins -= player2Spends;

//If player 2 spends the same number of coins as player 2 spends

if ( player1Spends == player2Spends) {

player2Coins += 1;

continue;

}

//positive difference between the number of coins spent by the two players

int difference = Math.abs(player1Spends - player2Spends) ;

//if difference is 1

if( difference == 1) {

player2Coins += 1;

continue;

}

//If difference is 2

if(difference == 2) {

player1Coins += 2;

continue;

}

}

// At the end of the game

//If player 1 coins is equal to player two coins

if(player1Coins == player2Coins) {

System.out.println("tie game");

}

//If player 1 coins are greater than player 2 coins

else if(player1Coins > player2Coins) {

System.out.println("player 1 wins");

}

//If player 2 coins is grater than player 2 coins

else if(player1Coins < player2Coins) {

System.out.println("player 2 wins");

}

}

3 0

3 years ago

(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in leng

igor_vitrenko [27]

Answer:

a) $\sigma = 12.2$ (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

$\sigma = \frac{I l}{V \pi r^2}$

$\sigma = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}$

$\sigma = 12.2$ (Ω-m)^{-1}[/tex]

b)

$Resistance = \frac{l}{\sigma A}$

$= \frac{l}{ \sigma \pi r^2}$

$= \frac{57 \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}$

Resistance = 121.4 Ω

8 0

3 years ago