Question: Taking a control volume enclosing just the turbine, evaluate the turbine inlet temperature, in F. (please show work) Schematic and Given Data: Engineering Model: 1.

Taking a control volume enclosing just the turbine, evaluate the turbine inlet temperature, in F. (please show work)  Taking a control volume enclosing just the turbine, evaluate the turbine
inlet temperature, in F. (please show work) Schematic and Given Data: Engineering

Schematic and Given Data: Engineering Model: 1. The control volume shown on the accompanying figure is at steady state. 2. Heat transfer is negligible, and changes in kinetie and potential encrgy can be ignored. 3. There is no pressure drop lor water flowing through the bteam generator. 4. The combustion products are modeled as air as an ideal gas. Flg=EA,10 Analysis: (a) The power developed by the turbine is determined from a control volume enclosing both the steam generator and the turbinc. Since the gas and water streams do not mix, mass rate balances for each of the streams reduce, respectively, to give m1=ms,m1=ms For this control solume, the appropriate form of the steady-state energy rate balance is Eq. 4.18, which reads 0=QbWi2+m1(h1+22+gz1V12)+m3(h3+2V22+gt1V32)m2(h2)m3(h5+2V52+gz5) The underlined terms drop out by assumption 2. With these simplifications together with the above mass flow rate relations, the energy rate balance becomes Ws=m1(h1h2)+m3(h3h3) The mass flow nite m, can be evaluated with given data at inlet I and the ideal gas equation of state m1=v1(AV)1=(R/M)T1(AV)iPi=(28.97lbR1545ftlbf)(860R)(2103ft3/min)(14.7lbf/in2)1ft2144in.2=9230,6Jb/min The specific enthalpies h1 and h2 cin be found from Table A.22E: At 860R,h1=206.46Btu/b, and at 720Rh2= 172.39 Btuib. At state 3, water is a liquid. Using Eq. 3.14 and saturated liquid data from Table A-2F, h3= h1(T3)=70 Btwilh State S. is a two-phase liquid-vapor mixture. With data from Table A-3E and the given quality? h3=h55+x9(hch5b)=69.74+0.93(1036.0)=1033.2Btwlb Substituting values into the expression for Wov Wc==(9230.6minlb)(206.46172.39)lbBtu+(275minIb)(701033.2)tbBtu49610minBtu (1) Alternatively, to determine h4 a control volume enclosing just the turbine can be considered. (2) The decision about implementing this solution to the problem of utilizing the hot combustion products discharged from an industrial process would necessarily rest on the outcome of a detailed economic evaluation, including the cost of purchasing and operating the steam generator, turbine, and auxiliary equipment. Taking a control volume enclosing just the turbine, evaluate the turbine inlet temperature, in F. Ans. 354F. Schematic and Given Data: Engineering Model: 1. The control volume shown on the accompanying figure is at steady state. 2. Heat transfer is negligible, and changes in kinetie and potential encrgy can be ignored. 3. There is no pressure drop lor water flowing through the bteam generator. 4. The combustion products are modeled as air as an ideal gas. Flg=EA,10 Analysis: (a) The power developed by the turbine is determined from a control volume enclosing both the steam generator and the turbinc. Since the gas and water streams do not mix, mass rate balances for each of the streams reduce, respectively, to give m1=ms,m1=ms For this control solume, the appropriate form of the steady-state energy rate balance is Eq. 4.18, which reads 0=QbWi2+m1(h1+22+gz1V12)+m3(h3+2V22+gt1V32)m2(h2)m3(h5+2V52+gz5) The underlined terms drop out by assumption 2. With these simplifications together with the above mass flow rate relations, the energy rate balance becomes Ws=m1(h1h2)+m3(h3h3) The mass flow nite m, can be evaluated with given data at inlet I and the ideal gas equation of state m1=v1(AV)1=(R/M)T1(AV)iPi=(28.97lbR1545ftlbf)(860R)(2103ft3/min)(14.7lbf/in2)1ft2144in.2=9230,6Jb/min The specific enthalpies h1 and h2 cin be found from Table A.22E: At 860R,h1=206.46Btu/b, and at 720Rh2= 172.39 Btuib. At state 3, water is a liquid. Using Eq. 3.14 and saturated liquid data from Table A-2F, h3= h1(T3)=70 Btwilh State S. is a two-phase liquid-vapor mixture. With data from Table A-3E and the given quality? h3=h55+x9(hch5b)=69.74+0.93(1036.0)=1033.2Btwlb Substituting values into the expression for Wov Wc==(9230.6minlb)(206.46172.39)lbBtu+(275minIb)(701033.2)tbBtu49610minBtu (1) Alternatively, to determine h4 a control volume enclosing just the turbine can be considered. (2) The decision about implementing this solution to the problem of utilizing the hot combustion products discharged from an industrial process would necessarily rest on the outcome of a detailed economic evaluation, including the cost of purchasing and operating the steam generator, turbine, and auxiliary equipment. Taking a control volume enclosing just the turbine, evaluate the turbine inlet temperature, in F. Ans. 354F

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Chemical Engineering Questions!