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irinina [24]
3 years ago
12

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

. At the turbine exit the conditions are 15Ibf/in.^2, 600ºR, and 100ft/s. Heat transfer from the turbine to its surroundings takes place at an average surface temperature of 620ºR. The rate of heat transfer is 2btu/Ib of air passing through the turbine. For the turbine, determine: a. The work developed (Btu/Ib); and b. The exergy destruction (Btu/lb).
Physics
1 answer:
Arturiano [62]3 years ago
5 0

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

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2 years ago
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Answer:

191.36 N/m

Explanation:

From the question,

The Potential Energy of the safe = Energy of the spring when it was compressed.

mgh = 1/2ke²............... Equation 1

Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression

Making k the subject of the equation,

k =2mgh/e²................ Equation 2

Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m

Constant: g = 9.8 m/s²

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k = 2(1100)(9.8)(0.0024)/0.52²

k = 51.744/0.2704

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3 0
3 years ago
A ball is thrown straight down with a speed of Va2.00 from a building 40.00 meters high. How much S time passes before the ball
-Dominant- [34]

Answer:

Option C is the correct answer.

Explanation:

Considering vertical motion of ball:-

Initial velocity, u =  2 m/s

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We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    40 = 2 x t + 0.5 x 9.81 x t²

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Displacement is the shortest distance covered by an object. It is basically the change in position of object. It is a vector quantity i.e direction as well as magnitude.

When an object is travelling in a straight line and stops at the end point, then both distance and displacement are same.

When an object is travelling in a straight line and then changes its direction or we can say come backwards then the magnitude of distance and displacement are different.

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Answer:

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(b) - 6.8 x 10^-4 m/s^2

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initial velocity of star, u = 20.7 km/s

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