Explanation The given cycle is the simple Rankine cycle with water as the working fluid and operated at the boiler and condenser pressures of 5 MPa and 10 kPa respectively. The simple Rankine cycle involves following processes and following conditions: Processes: 1. Process 1 to 2 – Isentropic compression in a Pump 2. Process 2 to 3 – Heat addition at constant pressure in a Boiler 3. Process 3 to 4 – Isentropic expansion in a Turbine 4. Process 4 to 1 – Heat rejection at constant pressure in a Condenser Conditions: The pressure is constant for the process 2 to 3 and the process 4 to 1. P 2 = P 3 and P 4 = P 1 The entropy is constant for the process 1 to 2 and the process 3 to 4. s 1 = s 2 and s 3 = s 4 Here, water is the working fluid and neglect the pump work. Hence, the net power output of the plant is equal to the turbine output. Write the formula for net power output of the plant. W ˙ net,out = W ˙ T, out = m ˙ ( h 3 − h 4 ) (I) Here, the mass flow rate is m ˙ , the enthalpy is h and the subscripts 3 and 4 indicates the process states and condition of steam. Write the formula for mass flow rate for the given plant. m ˙ = Q ˙ in q in (II) Write the formula for heat input of the plant ( q in ) . q in = h 3 − h 2 (III) Write the formula for quality of water at the exit of the turbine (state 4) is expressed as follows. x 4 = s 4 − s f s f g (IV) The enthalpy and entropy at state 4 is expressed as follows. h 4 = h f , 4 + x 4 h f g , 4 (V) Here, the enthalpy is h , the entropy is s , the quality of the water is x , the suffix f indicates the fluid condition, the suffix f g indicates the change of vaporization phase; the suffix 4 indicates the state 4. At state 1: The water exits the condenser as a saturated liquid at the pressure of 10 kPa . Hence, the enthalpy at state 1 is as follows. h 1 = h f @ 10 kPa Refer Table A-5, “Saturated water-Pressure table”. The enthalpy ( h 1 ) at state 1 corresponding to the pressure of 10 kPa is 191.81 kJ/kg . At state 3: (Turbine inlet) Refer Table A-6, “Superheated water”. The enthalpy ( h 3 ) and entropy ( s 3 ) at state 3 corresponding to the pressure of 5 MPa and the temperature of 600 ° C is 3666.9 kJ/kg and 7.2605 kJ/kg ⋅ K respectively. At state 4: (Condenser inlet) Refer Table A-5, “Saturated water-Pressure table”. Obtain the following properties corresponding to the pressure of 10 kPa . h f , 4 = 191.81 kJ/kg h f g , 4 = 2392.1 kJ/kg s f , 4 = 0.6492 kJ/kg ⋅ K s f g , 4 = 7.4996 kJ/kg ⋅ K Conclusion: When the pump work is neglected, the enthalpy at state 1 and state 2 is equal

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Chapter 10.9, Problem 123FEP

To determine

The power output of this plant from the following:

(a) 93 kW

(b) 118 kW

(c) 190 kW

(d) 216 kW

(e) 300 kW

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Chapter 10.9, Problem 122FEP

Chapter 10.9, Problem 124FEP

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A spring package with two springs and an external force, 200N. The short spring has a loin of 35 mm. Constantly looking for spring for short spring so that total compression is 35 mm (d). Known values: Long spring: Short spring:C=3.98 N/mm Lo=65mmLo=87.4mmF=c·fTotal compression is same for both spring. 200 = (3.98(c1) × 35) + (c₂ × 35) 200 = 139.3 + 35c₂ 200 - 139.3 = 35c₂ 60.7 = 35c₂ c₂ = 60.7/35 Short spring (c₂) = 1.73 N/mm According to my study book, the correct answer is 4.82N/mm What is wrong with the calculating?

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Chapter 10 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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The power output of this plant from the following: (a) 93 kW (b) 118 kW (c) 190 kW (d) 216 kW (e) 300 kW

The power output of this plant from the following: (a) 93 kW (b) 118 kW (c) 190 kW (d) 216 kW (e) 300 kW

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Chapter 10 Solutions

The power output of this plant from the following:

(a) 93 kW

(b) 118 kW

(c) 190 kW

(d) 216 kW

(e) 300 kW