This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ENGRD 2210 Thermodynamics PROBLEM SET 9 SOLUTIONS Problem 6.86: Problem Statement: The following figure provides the steadystate operating data for a wellinsulated 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 A3): h = h g (1 bar) = 2675.5 kJ/kg; s = s g (1 bar) = 7.3594 kJ/kgK Superheated Vapor @ P = 1.0 MPa, T = 320 o C (Table A4): h = 3093.9 kJ/kg; s = 7.1962 kJ/kgK 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 steadystate ?? ?? ? = 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/kgK = 0.1632 kj/kgK 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/kgK) Source 1 3425.1 6.6622 Table A4 2 2336.7 7.4651 Table A3, x=0.9 3 173.9 0.5926 Table A3 4 188.9 0.6061 Table A5, interpolated @ 43 o C 5 83.96 0.2966 Table A2 6 146.68 0.5053 Table A2 (a) To find the mass flow rate of the cooling water, do an energy balance for the condenser: ? ??...
View
Full
Document
This note was uploaded on 03/15/2010 for the course ENGRD 2210 at Cornell University (Engineering School).
 '08
 TORRANCE

Click to edit the document details