Physics, 21.12.2019 02:31 marley462847

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).

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Answer from: Jana1517

(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

$\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*<img src=$ " alt="\frac{1}{2*32.2*778}" />" /> = -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

Answer from: Quest

The horse latitudes-high-pressure zones about 30 degrees north and 30 degrees south of the equator.-warm air traveling away from the equator cools and sinks-weather tends to be clear and drythe doldrums-low-pressure near equator-air rises and spreads out toward the poles-rising moist air produces clouds and heavy rainthe westerlies-blow from the west-move from the horse latitudes toward the poles.-bring storms across much of the united statesthe tradewinds-blow from the east-move from the horse latitudes toward the equator -strong, steady winds

Answer from: Quest

the grams of oxygen gas that can be produced is 20.6 grams of o₂

calculation

2liclo₃ → 2 licl + 3 o₂

step 1: find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5 +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281 moles

step 2: use the mole ratio to calculate the moles of o2

from equation above, liclo₃: o₂ is 2: 3

therefore the moles of o₂=0.4281 moles x3/2 =0.6422 moles

step 3: find mass of o₂

mass= moles x molar mass

from periodic table the molar mass of o₂ = 16 x2= 32 g/mol

mass=0.6422 moles x 32 g/mol = 20.6 grams of o₂

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