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EGR 334 Thermodynamics Chapter 4: Section 4-5 Lecture 13: Control Volumes and Energy Balance Quiz Today?

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Today’s main concepts: Be able to draw a graphic representation of a control volume with energy balance. Explain what flow work. Write the energy balance equation Identify some commons assumptions that simplify the energy balance equation. Use mass balance and energy balance to solve thermodynamic problems. Reading Assignment: Homework Assignment: Read Chap 4: Sections 6-9 (for next Wed.) From Chap 4: 25, 28, 31, 34

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3 Mass Rate Balance: Energy Rate Balance: For a Control Volume: The rate at which mass accumulates in the control volume is the difference between the rate of mass flow in and flow out. The rate at which energy accumulates in the control volume is the difference between the heat flow rate in and the power out of the system and the difference between the rate at which mass carries energy with it either into and out of the control volume.

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4 Sec 4.4: Conservation of Energy for a Control Volume E within the system net Q input net W output [ ] = + Energy Balance m within the system net m input net m output [ ] = + Mass Balance Net E from Mass transfer + [ ] Add a term to the energy balance. where then energy due to mass transfer is mass transfer energy expressed as a rate equation then the full 1 st law of Thermodynamics is given as

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5 Sec 4.4: Conservation of Energy for a Control Volume Two Types of Work Flow Work: work done BY the flow Control Volume Work: work done BY the control volume (W CV ) - work done by P V, turning a shaft, electricity (types of work we have dealt with thus far) [ ] Time rate of Energy transfer from the control volume at exit due to mass transfer (-) Work Done ON system as mass enters (+) Work Done On environment as mass exits Define Flow Work Work done by system due to rotating shafts, electrical effects, or boundary displacement (∫pdV)

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6 Sec 4.4: Conservation of Energy for a Control Volume Putting all these terms together including the separate work terms, W = W CV + W PAv Thus u “heat energy” h “heat energy + flow work” h i hehe Recall that enthalpy is defined h = u + p v then the form that will usually be your starting equation is

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7 Sec 4.5: Analyzing Control Volumes at Steady State For Steady State Conditions: Mass Balance at steady state: and Often W = 0 if No change in system volume (fixed container) No turbines/pumps or electrical devices Energy Balance at steady state: Often Q = 0 if Small surface area System is insulated T between system & environment is small Common Simplifying Assumptions:

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8 Example 1: (Problem 4.27) Air at 600 kPa, 330 K enters a well-insulated, horizontal pipe having a diameter of 1.2 cm and exits at 120 kPa, 300 K. Applying the ideal gas model for air, determine at steady state (a) the inlet and exit velocities, each in m/s (b) the mass flow rate, in kg/s. 600 kPa 330 K 120 kPa 300 K d =1.2 cm Sec 4.5: Analyzing Control Volumes at Steady State stateInletExit P (kPa)600120 T (K)330300 h (kJ/kg)330.34300.19 Identify State Properties stateInletExit P (kPa)600120 T (K)330300 h (kJ/kg) Look up values for h on Table A-22 stateInletExit p [kPa] T [K] h [kJ/kg]

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9 Assumptions: System at steady state No significant heat loss No work done by system No large change in elevation Sec 4.5: Analyzing Control Volumes at Steady State Energy balance: stateInletExit P (kPa)600120 T (K)330300 h (kJ/kg)330.34300.19 Table A-22 is reduced with assumptions to: Steady state also implies that mass flow rate is equals mass flow rate out.

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10 Next find mass flow rate Sec 4.5: Analyzing Control Volumes at Steady State Apply continuity equation and mass balance: stateInletExit P (kPa)600120 T (K)330300 h (kJ/kg)330.34300.19 600 kPa 330 K 120 kPa 300 K d =1.2 cm therefore: applying Ideal Gas Equation:

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11 Finally combine what we know Sec 4.5: Analyzing Control Volumes at Steady State from Energy Balance: from Mass balance: stateInletExit P (kPa)600120 T (K)330300 h (kJ/kg)330.34300.19 600 kPa 330 K 120 kPa 300 K d =1.2 cm therefore: solving for velocities:

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12 Sec 4.5: Analyzing Control Volumes at Steady State to find the mass flow rate: from table 3.1: R = 0.2870 kJ/kg-K stateInletExit P (kPa)600120 T (K)330300 h (kJ/kg)330.34300.19 600 kPa 330 K 120 kPa 300 K d =1.2 cm

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13 Example 2: (Problem 4.29) Refrigerant 134a flows at a steady state through horizontal pipe with an inside diameter of 4cm. It enters as -8 deg C saturated vapor with a mass flow rate of 17 kg/min. It exits at a pressure of 2 bar. If the heat transfer rate to the refrigerant is 3.4 kW, determine a) the exit temperature b) the inlet and outlet velocities Sec 4.5: Analyzing Control Volumes at Steady State statecInlet (sat. vapor) Exit p [bar]2.17042 T [ o C]-8 v [m 3 /kg]0.0919 h [kJ/kg]242.54 Identify State Properties Look up values for h on Table A-10 -8 o C 17 kg/min 2 bar d= 4 cm. Q = 3.4 kW statecInlet (sat. vapor) Exit p [bar] T [ o C] v [m 3 /kg] h [kJ/kg]

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14 Sec 4.5: Analyzing Control Volumes at Steady State stateInlet (sat. vapor) Exit p [bar] 2.17042 T [ o C]-8 v [m 3 /kg]0.0919 h [kJ/kg]242.54 Using continuity equation at the inlet Find the volumetric flow rate -8 o C 17 kg/min 2 bar d= 4 cm. Q = 3.4 kW Find the inlet velocity

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15 Sec 4.5: Analyzing Control Volumes at Steady State stateInlet (sat. vapor) Exit p [bar] 2.17042 T [ o C]-8 v [m 3 /kg]0.0919 h [kJ/kg]242.54 for steady state: mass balance is given as For steady state the energy balance is given as: -8 o C 17 kg/min 2 bar d= 4 cm. Q = 3.4 kW for steady state, no work done, and a horizontal pipe:

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16 Sec 4.5: Analyzing Control Volumes at Steady State stateInlet (sat. vapor) Exit p [bar] 2.17042 T [ o C]-8 v [m 3 /kg]0.0919 h [kJ/kg]242.54 combining the mass balance and energy balance equations: or -8 o C 17 kg/min 2 bar d= 4 cm. Q = 3.4 kW gives

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17 Sec 4.5: Analyzing Control Volumes at Steady State stateInlet (sat. vapor) Exit p [bar]2.17042 T [ o C]-8 v [m 3 /kg]0.0919 h [kJ/kg]242.54 This equation contains two unknowns, h e and v e. Note that both are properties of the exit state which can be both considered functions of an unknown temperature, T B and a known pressure, p B = 2 bar. One way to solve this would be to guess a temperature of the outlet….then look up the corresponding values of h and v….put them back in the equation and see if they give the correct value. This iterative approach is exactly the type of solution that IT is very good at….so anther way to solve this is to use IT to define the h and v in terms of an unknown T B and let it iteratively find the temperature for us. -8 o C 17 kg/min 2 bar d= 4 cm. Q = 3.4 kW

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18 Outlet temperature T e = -10.09 o C Inlet velocity V i = 20.7 m/s Outlet velocity Results: c

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19 end of Lecture 13 Slides

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