3. Steam enters a turbine operating at steady state at 500...

3. Steam enters a turbine operating at steady state at 500 lbf/in.², 600°F, and exits as saturated vapor at 2 lbf/in.². The power developed by the turbine is 2.5×10? Btu/h and the steam flow rate is 15,000 lb/h. Heat transfer occurs at a location on the surface of the turbine where the average temperature is 300°F. Neglecting kinetic and potential energy effects. a. Determine heat transfer rate, in Btu/h. b. Determine the rate of entropy production, in Btu/h°R. c. Draw a T-s diagram. 500 lbf/in.² 600 °F 15,000 lb/h (1) Wcv = 2.5 x 10? Btu/h 2 lbf/in.² Sat. vapor (2) Tb = 300 °F = 760 °R Qcv

Transcript

00:01 Hello students in this question steam enters a turbine operating in a steady state so p1 is equal to 500 pounds per inch square and t1 is equal to 600 degrees fahrenheit and the final state p2 is equal to 2 pounds per inch square and t2 is equal to 300 degrees fahrenheit the scene enters and leaves the system in such a way the power developed by the turbine is given it by to be equal to 2 .5 times 5 1 times 10 to the power 6 btu per hour now the steam flow rate m dot is equal to 15 ,000 pounds per hour to determine the heat transfer rate, right? so we can use the equation through the system the heat transfer rate can be calculated as q is equal to m dot times h2 minus h1 plus w that is the the equation for the heat transfer rate if you look it up on steam table you get h1 is equal to thousand b2 and h2 will be equal to 1400 btu per pound btu per pound so let's substitute the values and find the q the heat transfer rate so m dot is equal to 15 ,000 right times 400 that is delta h plus w it is a work done at work done will be equal to let's assume that it will be for one hour then it will be equal to the power so that is two point five one times ten to the power six so q will be equal to that is equal to eighty five one zero zero zero zero -zero btu that is the unit now.

02:18 We need to determine the specific rate of entropy production.

02:22 So s that can be second part so we can write down the equation specific the rate of entropy production that is s dot is equal to m dot times s 2 minus s 1 plus q by t 0 so here we need to look it up the s 2 the specific uh entropy is s 1 and s 2 the moisture moeller models chart we have to look it up from steam table.

02:52 Also, we can look it up and once we look at that up we get so we can take the s 1 and s 2 values on this team table so s dot will be equal to m dot that m dot which is given to be equal to thousand five hundred so the fifteen thousand times s 2 which is equal to 1 .5 minus 1 .8 plus q which is equal to eighty five one divided by t 0 t 0 is equal to 760.

03:28 So s dot will be equal to six six nine seven point three six entropy per unit time that is the stored value so we have found out everything and now to plot the diagram we have the data which is s 1 is equal to 1 .8 and s 2 is equal to 1 .5 in t 1 is given as 1060 and t 2 is converted as 760 so we can plot the point right.

04:04 This is the diagram so plot the point t 1 s 1 at t 1 s 1 so t and s is there so t 1 is equal to 1060.

04:17 So it starts here and t 2 is equal to 760 1060 so s 1 is equal to this is t 1 s 1 is equal to 81 point 8 so s 1 is equal to 1 .8 and s 2 is equal to 1 .5...

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