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julia-pushkina [17]
3 years ago
13

Suppose that the weight (in pounds) of an airplane is a linear function of the amount of fuel (in gallons) in its tank. When car

rying 18 gallons of fuel, the airplane weighs 1999 pounds. When carrying 46 gallons of fuel, it weighs 2153 pounds. How much does the airplane weigh if it is carrying 58 gallons of fuel?
Physics
1 answer:
Inessa [10]3 years ago
6 0

Answer:

W=2219pounds

Explanation:

If the weight  is a linear function of the amount of fuel, the following correlation is fulfilled :

\frac{2153pounds-1999pounds}{46gallons-18gallons} = \frac{W-1999pounds}{58gallons-18galons}

we solve the equation:

W=2219pounds

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How far would a jet going 155 m/s travel in 9 s?
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Air enters a turbine operating at steady state with a pressure of 75 Ibf/in.^2, a temperature of 800º R and velocity of 400 ft/s
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Answer:

(a) W/m = 49.334 Btu/lb

(b) \frac{E_{d} }{m} = 22.12 Btu/lb

Explanation:

For the given problem, it can be assumed that the system is operating at steady state and the effects of potential energy can be neglected.

(a) Using the thermodynamic table for air.

At the temperature (T_{1})of 800 ºR and pressure (P_{1}) of 75 Ibf/in.^2, we can deduce that:

Specific enthalpy (h_{1}) = 191.81 BTu/lb

Specific entropy (s_{1}) = 0.6956 Btu/(lb.ºR)

At the temperature (T_{2})of 600 ºR and pressure (P_{2}) of 15 Ibf/in.^2, we can deduce that:

Specific enthalpy (h_{2}) = 143.47 BTu/lb

Specific entropy (s_{2}) = 0.6261 Btu/(lb.ºR)

The work done can be calculated using energy rate equation:

\frac{W}{m} = \frac{Q}{m} + (h_{1} - h_{2}) + \frac{V_{1}^{2} - V_{2}^{2}}{2}

Q/m = heat transfer = -2 Btu/lb

V_{1} = 400 ft/s

V_{2} = 100 ft/s

\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*[tex]\frac{1}{2*32.2*778}[/tex] = -2 + 48.34 + 29.938 = 49.334 Btu/lb

(b) To calculate the exergy destruction, we will use the equation for exergy rate:

\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}]

The equation above is further simplified to:

\frac{Ed}{m} = T_{o}[(s_{2} -s_{1}) - Rln\frac{P_{2} }{P_{1} } - \frac{Q/m}{T_{b} }]

Using a reference temperature (To) = 500 °R

Average surface temperature (Tb = 620°R

\frac{Ed}{m} = 500*[(0.6261 -0.6956) - (1.986/28.97)ln\frac{15 }{75 } - \frac{-2}{620}}]

\frac{E_{d} }{m} = 500*[-0.0695 +0.068688*1.609 +0.003225] = 22.12 Btu/lb

5 0
4 years ago
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Answer:

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Explanation:

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       Q +(-Q) =0.

  • If we now remove sphere B, and place it far away, there will be a charge redistribution within sphere A, making to disappear the separation between Q and -Q.
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