Homework 9 Solutions - ENGRD 2210 Thermodynamics PROBLEM...

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Unformatted text preview: ENGRD 2210 Thermodynamics PROBLEM SET 9 SOLUTIONS Problem 6.86: Problem Statement: The following figure provides the steady-state operating data for a well-insulated device having steam entering at one location and exiting at another. Neglecting kinetic and potential energy effects, determine: (a) the direction of flow (b) the power output or input, as appropriate, in kJ per kg of steam flowing Solution: (a) Normally directionality of flow would be determined using the 2 nd Law, as described in section 5.1 and 6.8. However, in this case, a direction of flow is assumed and the associated entropy production is evaluated to determine if the direction assumed is correct. Property Data: Sat. Vapor @ P = 100 kPa (Table A-3): h = h g (1 bar) = 2675.5 kJ/kg; s = s g (1 bar) = 7.3594 kJ/kg-K Superheated Vapor @ P = 1.0 MPa, T = 320 o C (Table A-4): h = 3093.9 kJ/kg; s = 7.1962 kJ/kg-K Entropy Balance to find entropy production, ?? : Assume inlet is the saturated vapor, State 1 in the diagram, at P = 100 kPa ?? ?? ? = ? ? ? + ? ? ? ? ? + ?? [Eqn 6.34] (1) Since the system is at steady-state ?? ?? ? = 0 . There is no heat into the system, so ? ? ? = 0 ? 1 2 Equation (1) reduces to ?? ? = ? ? ? ? (2) , since ? ? = ? = ? Plugging in known values gives ?? ? = (7.1962 7.3594) kJ/kg-K = -0.1632 kj/kg-K Since ?? ? must be greater than or equal to 0, the direction of flow is FROM P=1.0 MPa, T=320 o C TO Sat. Vapor at P=100kPa (or from 2 to 1 in the diagram) (b) Energy Rate Balance = ?? ?? + ? ( ? ? ? ? ) (3) Steady State ( = 0 ), no heat ( ?? = 0 ) gives ?? ? = ? ? ? ? (4) Where h in = 3093.9 kJ/kg = h @ 1.0 MPa, 320 o C and h out = 2675.5 kJ/kg = h @ 0.1 MPa ?? ? = (3093.9 2675.5) kJ/kg = 418.4 kJ/kg Problem 6.117: Problem Statement: The following figure shows a simple vapor power plant operating at steady state with water as the working fluid. Data at key locations are given on the figure. The mass flow rate of the water circulating through the components is 109 kg/s. Stray heat transfer and kinetic and potential energy effects can be ignored. Determine (a) the mass flow rate of the cooling water, in kg/s (b) the thermal efficiency (c) the rates of entropy production, each in kW/K, for the turbine, condenser, and pump (d) Using the results of part (c), place the components in rank order, beginning with the component contributing most to the inefficient operation of the overall system Solution: The following table shows the enthalpy and entropy values at every state State h (kJ/kg) s (kJ/kg-K) Source 1 3425.1 6.6622 Table A-4 2 2336.7 7.4651 Table A-3, x=0.9 3 173.9 0.5926 Table A-3 4 188.9 0.6061 Table A-5, interpolated @ 43 o C 5 83.96 0.2966 Table A-2 6 146.68 0.5053 Table A-2 (a) To find the mass flow rate of the cooling water, do an energy balance for the condenser: ? ??...
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This note was uploaded on 03/15/2010 for the course ENGRD 2210 at Cornell University (Engineering School).

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Homework 9 Solutions - ENGRD 2210 Thermodynamics PROBLEM...

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