2 Today’s main concepts: Be able to set up mass and energy balance models for Turbines Pumps Compressors Boilers Heat Exchangers Nozzles Diffusers ThrottleReading Assignment:Read Chapter 4, Sections 11-12Homework Assignment:Problems from Chap 4: 75, 83, 87, 90
3 Modeling applications with Control Volumes: Many important applications involve one inlet, one exit control volumes at steady state. Other systems include feature additional inlet and outlet streams. One very common device, the heat exchanger, requires multiple streams of mass flow to be modeled.Mass Rate Balance: multi-path, steady stateEnergy Rate Balance: 1 path, steady state
4 Control Volume Applications: NozzlesDiffuserTurbineBoilerCompressorPumpThrottling ValveHeat Exchanger
5 Heat ExchangersDirect contact: A mixing chamber in which hot and cold streams are mixed directly.Tube-within-a-tube counterflow: A gas or liquid stream is separated from another gas or liquid by a wall through which energy is conducted. Heat transfer occurs from the hot stream to the cold stream as the streams flow in opposite directions.
6 Direct Contact Heat Exchangers: Sec 4.9: Heat ExchangersDirect Contact Heat Exchangers:In direct contact heaters, mass streams combineand exchange heat by mass transport as well as by convection and conduction.Examples:Swamp CoolersHot-Cold waterfaucetCooling Towers
7 Separate Flow (Tube) Heat Exchangers: Sec 4.9: Heat ExchangersSeparate Flow (Tube) Heat Exchangers:Heat transfer occurs by conduction through a material that separates individual flow streamsParallel Flow: Streams enter and leave along same directionsCross Flow: Stream pass along direction perpendicular to each otherCounter Flow: Streams enter and leave opposite each other.
8 Tube Heat Exchanger are all around you. StillRadiatorHome air conditioningIndustrial Heating BoilerGeothermalHeatingRefrigeratorcondenserIndustrialSystemsPower Plant
9 Designed to maximize “contact” between the two fluids Sec 4.9: Heat ExchangersA heat exchanger attempts to maximize the heat transfer between separate fluid streams.Designed to maximize “contact” between the two fluidsMaximize time of contact and Maximize area of contact
10 For Separate Stream Heat Exchangers: Sec 4.9: Heat ExchangersFor Separate Stream Heat Exchangers:Mass Balance Model:andEnergy Balance Model:Steady StateV0Horizontal Section(or very short vertical)Insulatedfrom surroundingsNo workFirst FluidSecond Fluid
11 Assumptions Steady State KE= PE = 0 QCV= 0 Sec 4.9: Heat Exchangers Example: (4.81) A feedwater heater operates at steady state with liquid water entering at inlet #1 at 7 bar, 42 °C and a mass flow rate of 70 kg/s. A separate stream of water enters at inlet 2 as a two phase liquid-vapor mixture at 7 bar with a quality of 98%. Saturated liquid at 7 bar exits the feedwater heater at #3. Ignoring heat transfer with surroundings and neglecting KE and PE effects, determine the mass flow rate in kg/s at inlet #2.state123m (kg/s)70x0.98p (bar)7T (°C)42165h (kJ/kg)state123m (kg/s)xp (bar)T (°C)h (kJ/kg)state123m (kg/s)70x0.98p (bar)7T (°C)42165h (kJ/kg)175.9697.22#27 bar98%#17 bar42 °C70 kg/s#37 barSat. Liq.Look up h values onTable A-2 and A-3AssumptionsSteady StateKE= PE = 0QCV= 0
12 Example: (4.81) Determine the mass flow rate in kg/s at inlet #2. Sec 4.9: Heat ExchangersExample: (4.81) Determine the mass flow rate in kg/s at inlet #2.From Table A-3state123m (kg/s)70x0.98p (bar)7T (°C)42165h (kJ/kg)175.9697.22Mass Balance:Energy Balance:OutInCombining
13 Using energy balance: for the heat exchanger Sec 4.9: Heat ExchangersExample: (4.85) A parallel flow heat exchanger has separate streams of air and water. Streams undergo no significant change of pressure. Ignore heat loss to the surroundings and changes in KE and PE. Apply ideal gas to the air stream. If both streams leave at the same temperature, determine the exit temperature.state1234H2OAirm (kg/s)105p (bar)T99.63 oC1200 Kh (kJ/kg)2675.5state1234H2OAirm (kg/s)105p (bar)T1200 Kh (kJ/kg)from Table A-3 at psat = 1 bar, T1= Tsat = C and h1=hg= kJ/kgUsing energy balance: for the heat exchangerReducing this with assumptions
14 Sec 4.9: Heat ExchangersExample: (4.85) A parallel flow heat exchanger has separate streams of air and water. Streams have no significant change of pressure. Ignore heat loss to the surroundings and changes in KE and PE. Apply ideal gas to the air stream. If both stream leave at the same temperature, determine the exit temperature.state1234H2OAirm (kg/s)105p (bar)T99.63oC1200 Kh (kJ/kg)2675.5Using ideal gas modelNote: since h2 depends upon T2=T4 and p2 = 1 bar, it could be found from Table A-4 if the exit temperature were known. This problem can be solved by picking a value of T2 and then looking up h2 and checking to see if it satisfies the equation. It can also be easily solved using IT’s iterative solver.
15 Then the heat transfer between the two fluids is Sec 4.9: Heat ExchangersSolution using ITExit TemperatureThen the heat transfer between the two fluids isstate1234H2OAirm (kg/s)105p (bar)T99.63oC2881200 Kh (kJ/kg)2675.53049
16 Throttling DevicesThrottling Device: a device that achieves a significant reduction in pressure by introducing a restriction into a line through which a gas or liquid flows. Means to introduce the restriction include a partially opened valve or a porous plug.
17 Throttling Device: Reduces Pressure Sec 4.10: Throttling DevicesThrottling Device: Reduces PressureMass Balance:Typical Energy Balance simplifications,Steady Statev0Horizontal Section(or very short vertical)No heat transferNo work
18 Enthalpy of saturated liquid at 80°F Sec 4.10: Throttling DevicesExample: (4.92) Refrigerant 134a enters the expansion valve of an air conditioning unit at 140 psi, 80 °F and exits at 50 psi. If the refrigerant undergoes a throttling process, what are the temperature, in °F, and the quality at the exit valve?stateinletexitx?P (psi)14050T (°F)80h (BTU/lb)37.27AssumptionsSteady StateKE= PE = 0QCV= WCV = 0Enthalpy of saturated liquid at 80°F140 psi80 °F50 psi
19 Example: (4.92) Refrigerant 134a Sec 4.10: Throttling DevicesExample: (4.92) Refrigerant 134astateinletexitx?P (psi)14050T (°F)8040.27h (BTU/lb)37.27At 140 psi, the saturation T is °F, so this is a super-cooled liquid. There are no tables for super-cooled liquid, so neglect pressure effects.From Table A-10E