Explanation Show the T-s diagram as in Figure (1). Express enthalpy at state 4. h 4 = h g @ 1.4 MPa (I) Here, enthalpy of saturation vapor at pressure of 1 .4 MPa is h g @ 1.4 MPa . Express entropy at state 4. s 4 = s g @ 1.4 MPa (II) Here, entropy of saturation vapor at pressure of 1 .4 MPa is s g @ 1.4 MPa . Express quality at state 5 s . x 5 s = s 4 − s f @ 10 kPa s f g @ 10 kPa (III) Here, entropy of saturation liquid at pressure of 10 kPa is s f @ 10 kPa and entropy of evaporation at pressure of 10 kPa is s f g @ 10 kPa . Express enthalpy at state 5 s . h 5 s = h f @ 10 kPa + x 5 s h f g @ 10 kPa (IV) Here, enthalpy of saturation liquid at pressure of 10 kPa is h f @ 10 kPa and enthalpy of evaporation at pressure of 10 kPa is h f g @ 10 kPa . Express enthalpy at state 5 using isentropic efficiency of turbine. h 5 = h 4 − η T ( h 4 − h 5 s ) (V) Express work done on low pressure turbine. w turb,low = h 4 − h 5 (VI) Express mass flow rate of low pressure turbine. m ˙ lowturb = W ˙ turb,II w turb,low (VII) Here, rate of work produced by low pressure turbine is W ˙ turb,II . Express mass flow rate of steam at the inlet of the high-pressure turbine. m ˙ total = m ˙ lowturb + m ˙ steam,turb (VIII) Here, mass flow rate of steam at the turbine exit is m ˙ steam,turb . Express work done by high pressure turbine. w turb,high = W ˙ turb,I m ˙ total (IX) Here, rate of work done by high pressure turbine at exit is W ˙ turb,I . Express enthalpy at state 3. h 3 = w turb,high + h 4 (X) Express enthalpy at state 4 s . h 4 s = h 3 − ( h 3 − h 4 ) η T , high (XI) Here, thermal efficiency of high pressure turbine is η T , high . Express quality at state 4 s . x 4 s = h 4 s − h f @ 1.4 MPa h f g @ 1.4 MPa (XII) Here, enthalpy of saturation liquid at pressure of 1 .4 MPa is h f @ 1.4 MPa and enthalpy of evaporation at pressure of 1 .4 MPa is h f g @ 1.4 MPa . Express entropy at state 4 s . s 4 s = s f @ 1.4 MPa + x 4 s ( s f g @ 1.4 MPa ) (XIII) Here, entropy of saturation liquid at pressure of 1 .4 MPa is s f @ 1.4 MPa and entropy of evaporation at pressure of 1 .4 MPa is s f g @ 1.4 MPa . Due to throttling process, write the entropy at state 3 is equal to entropy at state 4 s . s 3 = s 4 s (XIV) Conclusion: Refer Table A-5, “saturated water-pressure table”, and write the properties at pressure of 1 .4 MPa and 10 kPa . h g @ 1.4 MPa = 2788.9 kJ / kg s g @ 1.4 MPa = 6.4675 kJ / kg ⋅ K s f @ 10 kPa = 0.6492 kJ / kg ⋅ K s f g @ 10 kPa = 7.4996 kJ / kg ⋅ K h f @ 10 kPa = 191.81 kJ / kg h f g @ 10 kPa = 2392.1 kJ / kg h f @ 1.4 MPa = 829.96 kJ / kg h f g @ 1.4 MPa = 1958.9 kJ / kg s f @ 1.4 MPa = 2.2835 kJ / kg ⋅ K s f g @ 1.4 MPa = 4.1840 kJ / kg ⋅ K Substitute 2788.9 kJ / kg for h g @ 1.4 MPa in Equation (I). h 4 = 2788.9 kJ / kg Substitute 6.4675 kJ / kg ⋅ K for s g @ 1.4 MPa in Equation (II). s 4 = 6.4675 kJ / kg ⋅ K Substitute 6.4675 kJ / kg ⋅ K for s 4 , 0.6492 kJ / kg ⋅ K for s f @ 10 kPa and 7.4996 kJ / kg ⋅ K for s f g @ 10 kPa in Equation (III). x 5 s = 6.4675 kJ / kg ⋅ K − 0.6492 kJ / kg ⋅ K 7.4996 kJ / kg ⋅ K = 0.7758 Substitute 191.81 kJ / kg for h f @ 10 kPa , 0.7758 for x 5 s and 2392.1 kJ / kg for h f g @ 10 kPa in Equation (IV)

CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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ISBN: 9781260048636

Author: CENGEL

Publisher: MCG

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Chapter 10.9, Problem 104RP

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The temperature, pressure, and the flow rate of steam at the inlet of the high-pressure turbine.

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Chapter 10.9, Problem 103RP

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CONNECT FOR THERMODYNAMICS: AN ENGINEERI

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