Question

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).

Answer 1

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