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Maslowich
2 years ago
14

A mixture of air and methane is formed in the inlet manifold of a natural gas-fueled internal combustion engine. The mole fracti

on of the methane is 15 percent. This engine is operated at 3000 rpm and has a 5-L displacement. Determine the mass flow rate of this mixture in the manifold where the pressure and temperature are 80 kPa and 208C.

Engineering
1 answer:
german2 years ago
4 0

Answer:

The mass flow rate of the mixture in the manifold is 6.654 kg/min

Explanation;

In this question, we are asked to calculate mass flow rate of the mixture in the manifold

Please check attachment for complete solution and step by step explanation.

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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");

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3 0
2 years ago
Sharon is designing a house in an area that receives a lot of rainfall all year. Which material should she use to stick the wood
kakasveta [241]

Explanation:

She is passionate about architecture, typography, and black & white film ... Since moving to Texas, I've heard a lot of people say, "If you don't like ... Oc, 3.74, 56, 80 ... Not only does the weather have to be clear to pour the concrete, but it ... system that goes within the slab) is complete, any additional rain will

4 0
2 years ago
2An oil pump is drawing 44 kW of electric power while pumping oil withrho=860kg/m3at a rate of 0.1m3/s.The inlet and outlet diam
Natasha2012 [34]

Answer:

\eta = 91.7%

Explanation:

Determine the initial velocity

v_1 = \frac{\dot v}{A_1}

    = \frac{0.1}{\pi}{4} 0.08^2

     = 19.89 m/s

final velocity

v_2 =\frac{\dot v}{A_2}

      = \frac{0.1}{\frac{\pi}{4} 0.12^2}

      =8.84 m/s

total mechanical energy is given as

E_{mech} = \dot m (P_2v_2 -P_1v_1) + \dot m \frac{v_2^2 - v_1^2}{2}

\dot v = \dot m v                       ( v =v_1 =v_2)

E_{mech} = \dot mv (P_2 -P_1) + \dot m \frac{v_2^2 - v_1^2}{2}

                = mv\Delta P + \dot m  \frac{v_2^2 -v_1^2}{2}

                 = \dot v \Delta P  + \dot v \rho \frac{v_2^2 -v_1^2}{2}

              = 0.1\times 500 + 0.1\times 860\frac{8.84^2 -19.89^2}{2}\times \frac{1}{1000}

E_{mech} = 36.34 W

Shaft power

W = \eta_[motar} W_{elec}

    =0.9\times 44 =39.6

mechanical efficiency

\eta{pump} =\frac{ E_{mech}}{W}

=\frac{36.34}{39.6} = 0.917  = 91.7%

8 0
2 years ago
Air enters a nozzle steadily at 280 kPa and 77°C with a velocity of 50 m/s and exits at 85 kPa and 320 m/s. The heat losses from
Arlecino [84]

The input values are the following

\left.T_{1}=350 K\right\ then

h_{1}=350.49 \frac{k j}{k g}, s_{1}=1.85708 \frac{K j}{K g \cdot K}

By using the energy equilibrium

\dot{E}_{i n}-\dot{E}_{o u t}=\Delta \dot{E}_{s y s t e m}=0 , \dot{E}_{i n}=\dot{E}_{o u t}

we haveT_{2}=297.2 K

eq (1)  \dot{m}\left(h_{1}+\frac{V_{1}^{2}}{2}\right)=\dot{m}\left(h_{2}+\frac{V_{2}^{2}}{2}\right)+\dot{Q}_{o u t} ∴

0=q+h_{2}-h_{1}+\frac{V_{2}^{2}-V_{1}^{2}}{2}

Now, for specific energy h2:

h_{2}=h_{1}-q_{o u t}-\frac{V_{2}^{2}-V_{1}^{2}}{2}

By replacing the eq (1) we have

h_{2}=350.49 \frac{k j}{k g}-3.2-\frac{\left(320 \frac{m}{s}\right)^{2}-\left(50 \frac{m}{m}\right)^{2}}{2}\left(\frac{1 \frac{k j}{k g}}{1000 \frac{m^{2}}{s^{2}}}\right)

h_{2}=297.34 \frac{k j}{k g}

By using a standard ideal-gas properties of air we have

T_{2}=297.2 K

6 0
2 years ago
A water pump increases the water pressure from 15 psia to 70 psia. Determine the power input required, in hp, to pump 1.5 ft3/s
svetoff [14.1K]

Answer:

11.52 hp

Explanation:

<u><em>Givens: </em></u>

p_1  = 15 pisa

p_2  = 70 pisa

V_ol=1.5 ft^3/s

<u><em>Solution:  </em></u>

Note: m = p x V_ol (assuming in compressible flow —> p =const)  

The total change in the system mechanical energy can be calculated as follows,  

Δ e= (p_2 - p_1 ) /p

The power needed can be calculated as follows

P = W =mΔ e  = p x  V_ol x(p_2 - p_1 ) /p

  = V_ol x (p_2 - p_1 )

  = 44 pisa. ft^3/s

  = 44 x (1 btu/5.404pisa. ft^3) x (1 hp/0.7068btu/s)

  = 11.52 hp

3 0
2 years ago
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