# SM_7A - Chapter 7 - Section A - Mathcad Solutions 7.1 u2 :=...

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S 1 7.1595 kJ kg K = 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 - ( 29 = 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 H C P T = Whence T u 2 2 - 2 C P = T 52.45 - K = Ans. 7.4 From Table F.2 at 800 kPa and 280 degC: H 1 3014.9 kJ kg = 186

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Ans. A p min ( 29 7.021cm 2 = Ans. p min 431.7kPa = p min Find p min ( 29 = p min A p min ( 29 d d 0 cm 2 kPa = Given (guess) p min 400 kPa = A P ( ) interp s p , a 2 , P , ( 29 = s cspline P A 2 , ( 29 = a 2 i A 2 i = p i P i = i 1 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 - ( 29  = 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 = 187
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 A p ( ) 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

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## This note was uploaded on 01/29/2011 for the course CHEM 101 taught by Professor Brown during the Spring '10 term at The University of Akron.

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SM_7A - Chapter 7 - Section A - Mathcad Solutions 7.1 u2 :=...

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