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**EGR 334 Thermodynamics Chapter 4: Section 9-10**

Lecture 17: Control Volume Applications: Day 2 Quiz Today?

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**Today’s main concepts:**

Be able to set up mass and energy balance models for Turbines Pumps Compressors Boilers Heat Exchangers Nozzles Diffusers Throttle Reading Assignment: Read Chapter 4, Sections 11-12 Homework Assignment: Problems from Chap 4: 75, 83, 87, 90

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**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 state Energy Rate Balance: 1 path, steady state

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**Control Volume Applications:**

Nozzles Diffuser Turbine Boiler Compressor Pump Throttling Valve Heat Exchanger

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Heat Exchangers Direct 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.

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**Direct Contact Heat Exchangers:**

Sec 4.9: Heat Exchangers Direct Contact Heat Exchangers: In direct contact heaters, mass streams combine and exchange heat by mass transport as well as by convection and conduction. Examples: Swamp Coolers Hot-Cold water faucet Cooling Towers

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**Separate Flow (Tube) Heat Exchangers:**

Sec 4.9: Heat Exchangers Separate Flow (Tube) Heat Exchangers: Heat transfer occurs by conduction through a material that separates individual flow streams Parallel Flow: Streams enter and leave along same directions Cross Flow: Stream pass along direction perpendicular to each other Counter Flow: Streams enter and leave opposite each other.

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**Tube Heat Exchanger are all around you.**

Still Radiator Home air conditioning Industrial Heating Boiler Geothermal Heating Refrigerator condenser Industrial Systems Power Plant

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**Designed to maximize “contact” between the two fluids **

Sec 4.9: Heat Exchangers A heat exchanger attempts to maximize the heat transfer between separate fluid streams. Designed to maximize “contact” between the two fluids Maximize time of contact and Maximize area of contact

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**For Separate Stream Heat Exchangers:**

Sec 4.9: Heat Exchangers For Separate Stream Heat Exchangers: Mass Balance Model: and Energy Balance Model: Steady State V0 Horizontal Section (or very short vertical) Insulated from surroundings No work First Fluid Second Fluid

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**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. state 1 2 3 m (kg/s) 70 x 0.98 p (bar) 7 T (°C) 42 165 h (kJ/kg) state 1 2 3 m (kg/s) x p (bar) T (°C) h (kJ/kg) state 1 2 3 m (kg/s) 70 x 0.98 p (bar) 7 T (°C) 42 165 h (kJ/kg) 175.9 697.22 #2 7 bar 98% #1 7 bar 42 °C 70 kg/s #3 7 bar Sat. Liq. Look up h values on Table A-2 and A-3 Assumptions Steady State KE= PE = 0 QCV= 0

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**Example: (4.81) Determine the mass flow rate in kg/s at inlet #2.**

Sec 4.9: Heat Exchangers Example: (4.81) Determine the mass flow rate in kg/s at inlet #2. From Table A-3 state 1 2 3 m (kg/s) 70 x 0.98 p (bar) 7 T (°C) 42 165 h (kJ/kg) 175.9 697.22 Mass Balance: Energy Balance: Out In Combining

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**Using energy balance: for the heat exchanger**

Sec 4.9: Heat Exchangers Example: (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. state 1 2 3 4 H2O Air m (kg/s) 10 5 p (bar) T 99.63 oC 1200 K h (kJ/kg) 2675.5 state 1 2 3 4 H2O Air m (kg/s) 10 5 p (bar) T 1200 K h (kJ/kg) from Table A-3 at psat = 1 bar, T1= Tsat = C and h1=hg= kJ/kg Using energy balance: for the heat exchanger Reducing this with assumptions

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Sec 4.9: Heat Exchangers Example: (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. state 1 2 3 4 H2O Air m (kg/s) 10 5 p (bar) T 99.63oC 1200 K h (kJ/kg) 2675.5 Using ideal gas model Note: 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.

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**Then the heat transfer between the two fluids is**

Sec 4.9: Heat Exchangers Solution using IT Exit Temperature Then the heat transfer between the two fluids is state 1 2 3 4 H2O Air m (kg/s) 10 5 p (bar) T 99.63oC 288 1200 K h (kJ/kg) 2675.5 3049

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Throttling Devices Throttling 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.

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**Throttling Device: Reduces Pressure**

Sec 4.10: Throttling Devices Throttling Device: Reduces Pressure Mass Balance: Typical Energy Balance simplifications, Steady State v0 Horizontal Section (or very short vertical) No heat transfer No work

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**Enthalpy of saturated liquid at 80°F**

Sec 4.10: Throttling Devices Example: (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? state inlet exit x ? P (psi) 140 50 T (°F) 80 h (BTU/lb) 37.27 Assumptions Steady State KE= PE = 0 QCV= WCV = 0 Enthalpy of saturated liquid at 80°F 140 psi 80 °F 50 psi

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**Example: (4.92) Refrigerant 134a**

Sec 4.10: Throttling Devices Example: (4.92) Refrigerant 134a state inlet exit x ? P (psi) 140 50 T (°F) 80 40.27 h (BTU/lb) 37.27 At 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

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end of Lecture 17 Slides

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