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Engineering and Tech / Fundamentals of Thermodynamcs 8 / Chapter 4 / Problem 198EUP

Fundamentals of Thermodynamcs | 8th Edition | ISBN: 9781118131992 | Authors: Claus Borgnakke, Richard E. Sonntag

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Solved: The following data are for a simple steam power

Chapter 4, Problem 198EUP

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QUESTION:

Problem 198EUP

The following data are for a simple steam power plant as shown in Fig. P4.118:

State	1	2	3	4	5	6	7
P lbf/in.2	900	890	860	830	800	1.5	1.4
T F		115	350	920	900		110
h, Btu/lbm		85.3	323	1468	1456	1029	78

State 6 has x6 = 0.92 and a velocity of 600 ft/s. The rate of steam flow is 200 000 lbm/h, with 400-hp input to the pump. Piping diameters are 8 in. from the steam generator to the turbine and 3 in. from the condenser to the steam generator. Determine the power output of the turbine and the heat transfer rate in the condenser.

FIGURE P4.118

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QUESTION:

Problem 198EUP

The following data are for a simple steam power plant as shown in Fig. P4.118:

State	1	2	3	4	5	6	7
P lbf/in.2	900	890	860	830	800	1.5	1.4
T F		115	350	920	900		110
h, Btu/lbm		85.3	323	1468	1456	1029	78

State 6 has x6 = 0.92 and a velocity of 600 ft/s. The rate of steam flow is 200 000 lbm/h, with 400-hp input to the pump. Piping diameters are 8 in. from the steam generator to the turbine and 3 in. from the condenser to the steam generator. Determine the power output of the turbine and the heat transfer rate in the condenser.

FIGURE P4.118

ANSWER:

Solution 198 EUP:

Step 1 of 2:-

Here we have to determine the power output of the turbine and the heat transfer from the conser.

For the turbine,

${A}_{5}=\pi{}{{D}_{5}}^{2}\ /4=0.349\ f{t}^{2}$ .

${v}_{5}=0.964\ f{t}^{3}/lbm$ .

So, ${V}_{5}=\dot{m}{v}_{5}/\ {A}_{5}$

$=\frac{200000\times{}0.964}{3600\times{}0.349}=153ft/s$ .

And $W=\left({{h}_{5}+0.5{{V}_{5}}^{2}}\right)-\left({{h}_{6}+0.5{{V}_{6}}^{2}}\right)$

$=1456-1029-\frac{60{0}^{2}-15{3}^{2}}{2\times{}25037}$

$=420.2\ Btu/lbm$

We know that,

$1\ Btu/lbm=25037\ f{t}^{2}/{s}^{2}$

And $1\ hp=2544\ Btu/hr$ .

So,

The power output will be,

$\dot{W}=\frac{420.2\times{}200000}{2544}$

$=33000\ hp$ .

So the power output will be $33000\ hp$ .

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