(a) Explanation Show the T-s diagram as in Figure (1). Express Prandtl number at state 8. P r 8 = P 8 P 7 P r 7 (I) Here, pressure at state 8 is P 8 , pressure at state 7 is P 7 and Prandtl number at state 7 is P r 7 . Express Prandtl number at state 10. P r 10 = P 10 P 9 P r 9 (II) Here, pressure at state 10 is P 10 , pressure at state 9 is P 9 and Prandtl number at state 9 is P r 9 . Express enthalpy at state 1. h 1 = h f @10 kPa (III) Here, enthalpy of saturation liquid at pressure of 10 kPa is h f @10 kPa . Express specific volume at state 1. v 1 = v f @10 kPa (IV) Here, specific volume of saturation liquid at pressure of 10 kPa is v f @10 kPa . Express initial work input. w pI,in = v 1 ( P 2 − P 1 ) (V) Here, pressure at state 2 and 1 is P 2 and P 1 respectively. Express enthalpy at state 2. h 2 = h 1 + w pI,in (VI) Express quality at state 6. x 6 = s 6 − s f @10 kPa s f g @10 kPa (VII) Here, entropy at state 6 is s 6 , entropy at saturation liquid and evaporation at pressure of 10 kPa is s f @10 kPa and s f g @10 kPa respectively. Express enthalpy at state 6. h 6 = h f @10 kPa + x 6 h f g @10 kPa (VIII) Here, enthalpy at saturation liquid and evaporation at pressure of 10 kPa is h f @10 kPa and h f g @10 kPa respectively. Express the mass flow rate of air in the gas-turbine cycle from energy balance equation. m ˙ air = h 3 − h 2 h 10 − h 11 m ˙ s (IX) Here, enthalpy at state 10 is h 10 , enthalpy at state 11 is h 11 , enthalpy at state 3 and 2 is h 3 and h 2 respectively, and mass flow rate of steam is m ˙ s . Conclusion: Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 7 and Prandtl number at state 7 corresponding to temperature at state 7 of 310 K . h 7 = 310.24 kJ / kg P r 7 = 1.5546 Substitute 12 for P 8 P 7 and 1.5546 for P r 7 in Equation (I). P r 8 = 12 ( 1.5546 ) = 18.66 Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 8 corresponding to Prandtl number at state 8 of 18.66 using an interpolation method. Write the formula of interpolation method of two variables. y 2 = ( x 2 − x 1 ) ( y 3 − y 1 ) ( x 3 − x 1 ) + y 1 (X) Here, the variables denote by x and y is Prandtl number at state8 and enthalpy at state 8 respectively. Show the enthalpy at state 8 corresponding to Prandtl number as in Table (1). Prandtl number at state 8 P r 8 Enthalpy at state 8 h 8 ( kJ / kg ) 18.36 ( x 1 ) 628.07 ( y 1 ) 18.66 ( x 2 ) ( y 2 = ? ) 19.84 ( x 3 ) 638.63 ( y 3 ) Substitute 18 .36, 18 .66 and 19 .84 for x 1 , x 2 and x 3 respectively, 628 .07 kJ / kg for y 1 and 638 .63 kJ / kg for y 3 in Equation (X). y 2 = ( 18 .66 − 18 .36 ) ( 638 .63 kJ / kg − 628 .07 kJ / kg ) ( 19 .84 − 18 .36 ) + 628 .07 kJ / kg = 630 .18 kJ / kg = h 8 Thus, enthalpy at state corresponding to Prandtl number at state 8 of 18.66 is, h 8 = 630 .18 kJ / kg Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 9 and Prandtl number at state 9 corresponding to temperature at state 9 of 1400 K . h 9 = 1515.42 kJ / kg P r 9 = 450.5 Here, enthalpy at state 9 is h 9 and Prandtl number at state 9 is P r 9 . Substitute 1 12 for P 10 P 9 and 450.5 for P r 9 in Equation (II). P r 10 = 1 12 ( 450.5 ) = 37.54 Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 10 s corresponding to Prandtl number at state 10 of 37.54 using an interpolation method. Show the enthalpy at state 10 corresponding to Prandtl number as in Table (2). Prandtl number at state 10 P r 10 Enthalpy at state 10 h 10 ( kJ / kg ) 37.35 ( x 1 ) 767.29 ( y 1 ) 37.54 ( x 2 ) ( y 2 = ? ) 39.27 ( x 3 ) 778.18 ( y 3 ) Use excels and substitutes the values from Table (II) in Equation (X) to get, h 10 = 768.38 kJ / kg Here, enthalpy at state 10 is h 10 . Refer Table A-5, “saturated water-pressure table”, and write the properties at pressure of 10 kPa . h f @10 kPa = 191.81 kJ / kg v f @10 kPa = 0 (b) To determine The rate of total heat input. (c) To determine The thermal efficiency of the combined cycle.

9th Edition

ISBN: 9781266084584

Author: CENGEL

Publisher: MCG

See similar textbooks

Videos

Question

Chapter 10.9, Problem 108RP

(a)

To determine

The mass flow rate of air in the gas-turbine cycle.

(b)

To determine

The rate of total heat input.

(c)

To determine

The thermal efficiency of the combined cycle.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.9, Problem 107RP

Chapter 10.9, Problem 109RP

Students have asked these similar questions

Q Derive (continuity equation)? I want to derive clear mathematics.

motor supplies 200 kW at 6 Hz to flange A of the shaft shown in Figure. Gear B transfers 125 W of power to operating machinery in the factory, and the remaining power in the shaft is mansferred by gear D. Shafts (1) and (2) are solid aluminum (G = 28 GPa) shafts that have the same diameter and an allowable shear stress of t= 40 MPa. Shaft (3) is a solid steel (G = 80 GPa) shaft with an allowable shear stress of t = 55 MPa. Determine: a) the minimum permissible diameter for aluminum shafts (1) and (2) b) the minimum permissible diameter for steel shaft (3). c) the rotation angle of gear D with respect to flange A if the shafts have the minimum permissible diameters as determined in (a) and (b).

First monthly exam Gas dynamics Third stage Q1/Water at 15° C flow through a 300 mm diameter riveted steel pipe, E-3 mm with a head loss of 6 m in 300 m length. Determine the flow rate in pipe. Use moody chart. Q2/ Assume a car's exhaust system can be approximated as 14 ft long and 0.125 ft-diameter cast-iron pipe ( = 0.00085 ft) with the equivalent of (6) regular 90° flanged elbows (KL = 0.3) and a muffler. The muffler acts as a resistor with a loss coefficient of KL= 8.5. Determine the pressure at the beginning of the exhaust system (pl) if the flowrate is 0.10 cfs, and the exhaust has the same properties as air.(p = 1.74 × 10-3 slug/ft³, u= 4.7 x 10-7 lb.s/ft²) Use moody chart (1) MIDAS Kel=0.3 Q3/Liquid ammonia at -20°C is flowing through a 30 m long section of a 5 mm diameter copper tube(e = 1.5 × 10-6 m) at a rate of 0.15 kg/s. Determine the pressure drop and the head losses. .μ= 2.36 × 10-4 kg/m.s)p = 665.1 kg/m³

Chapter 10 Solutions

THERMODYNAMICS: ENG APPROACH LOOSELEAF

Ch. 10.9 - Why is the Carnot cycle not a realistic model for...Ch. 10.9 - Why is excessive moisture in steam undesirable in...Ch. 10.9 - A steady-flow Carnot cycle uses water as the...Ch. 10.9 - A steady-flow Carnot cycle uses water as the...Ch. 10.9 - Consider a steady-flow Carnot cycle with water as...Ch. 10.9 - Water enters the boiler of a steady-flow Carnot...Ch. 10.9 - What four processes make up the simple ideal...Ch. 10.9 - Consider a simple ideal Rankine cycle with fixed...Ch. 10.9 - Consider a simple ideal Rankine cycle with fixed...Ch. 10.9 - Consider a simple ideal Rankine cycle with fixed...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The mass flow rate of air in the gas-turbine cycle.

Solution Summary: The author explains the mass flow rate of air in the gas-turbine cycle.

Videos

The mass flow rate of air in the gas-turbine cycle.

The rate of total heat input.

The thermal efficiency of the combined cycle.

Want to see the full answer?

Chapter 10 Solutions