Chapter7_A - Chapter 7 - Section A - Mathcad Solutions 7.1...

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Interpolation in Table F.2 at P = 525 kPa and S = 7.1595 kJ/(kg*K) yields: H 2 2855.2 kJ kg := V 2 531.21 cm 3 gm := mdot 0.75 kg sec := With the heat, work, and potential-energy terms set equal to zero and with the initial velocity equal to zero, Eq. (2.32a) reduces to: H u 2 2 2 + 0 = Whence u 2 2 H 2 H 1 () := u 2 565.2 m sec = Ans. By Eq. (2.27), A 2 mdot V 2 u 2 := A 2 7.05cm 2 = Ans. 7.5 The calculations of the preceding problem may be carried out for a series of exit pressures until a minimum cross-sectional area is found. The corresponding pressure is the minimum obtainable in the converging nozzle. Initial property values are as in the preceding problem. Chapter 7 - Section A - Mathcad Solutions 7.1 u 2 325 m sec := R 8.314 J mol K := molwt 28.9 gm mol := C P 7 2 R molwt := With the heat, work, and potential-energy terms set equal to zero and with the initial velocity equal to zero, Eq. (2.32a) reduces to H u 2 2 2 + 0 = But HC P T = Whence T u 2 2 2C P := T 52.45 K = Ans. 7.4 From Table F.2 at 800 kPa and 280 degC: H 1 3014.9 kJ kg := S 1 7.1595 kJ kg K := 220
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Ans. Ap min () 7.021cm 2 = Ans. p min 431.78kPa = p min Find p min := p min min d d 0 cm 2 kPa = Given (guess) p min 400 kPa := AP ( ) interp s p , a 2 , P , := s cspline P A 2 , := a 2 i A 2 i := p i P i := i1 5 .. := Fit the P vs. A2 data with cubic spline and find the minimum P at the point where the first derivative of the spline is zero. A 2 7.05 7.022 7.028 7.059 7.127 cm 2 = u 2 565.2 541.7 518.1 494.8 471.2 m sec = A 2 mdot V 2 u 2 ⎯⎯⎯ := u 2 2 H 2 H 1 ⎯⎯⎯⎯⎯⎯ := mdot 0.75 kg sec := V 2 531.21 507.12 485.45 465.69 447.72 cm 3 gm := H 2 2855.2 2868.2 2880.7 2892.5 2903.9 kJ kg := P 400 425 450 475 500 kPa := Interpolations in Table F.2 at several pressures and at the given entropy yield the following values: S 2 S 1 = S 1 7.1595 kJ kg K := H 1 3014.9 kJ kg := 221
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Show spline fit graphically: p 400 kPa 401 kPa , 500 kPa .. := 400 420 440 460 480 500 7.01 7.03 7.05 7.07 7.09 7.11 7.13 A 2 i cm 2 Ap () cm 2 P i kPa p kPa , 7.9 From Table F.2 at 1400 kPa and 325 degC: H 1 3096.5 kJ kg := S 1 7.0499 kJ kg K := S 2 S 1 := Interpolate in Table F.2 at a series of downstream pressures and at S = 7.0499 kJ/(kg*K) to find the minimum cross-sectional area. P 800 775 750 725 700 kPa := H 2 2956.0 2948.5 2940.8 2932.8 2924.9 kJ kg := V 2 294.81 302.12 309.82 317.97 326.69 cm 3 gm := u 2 2 H 2 H 1 ⎯⎯⎯⎯⎯⎯ := A 2 V 2 u 2 mdot = 222
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S vap 1.6872 Btu lb m rankine := S liq 0.3809 Btu lb m rankine := H vap 1167.1 Btu lb m := H liq 228.03 Btu lb m := From Table F.4 at 35(psi), we see that the final state is wet steam: H 2 1154.8 Btu lb m = H 2 H 1 H + := H 78.8 Btu lb m = H u 1 2 u 2 2 2 := By Eq. (2.32a), S 1 1.6310 Btu lb m rankine := H 1 1233.6 Btu lb m := From Table F.4 at 130(psi) and 420 degF: u 2 2000 ft sec := u 1 230 ft sec := 7.10 x 0.966 = x S 1 S liq S vap S liq := S vap 7.2479 kJ kg K := S liq 1.4098 kJ kg K := At the nozzle exit, P = 140 kPa and S = S1, the initial value.
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This note was uploaded on 04/18/2011 for the course CEM 497 taught by Professor Ku during the Spring '11 term at Punjab Engineering College.

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Chapter7_A - Chapter 7 - Section A - Mathcad Solutions 7.1...

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