# EGR 334 Thermodynamics Chapter 4: Section 4-5

## Presentation on theme: "EGR 334 Thermodynamics Chapter 4: Section 4-5"— Presentation transcript:

EGR 334 Thermodynamics Chapter 4: Section 4-5
Test #1 Next Monday Lecture 13: Control Volumes and Energy Balance Quiz Today?

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. Test #1 Next Monday Reading Assignment: Read Chap 4: Sections 6-9 (for next Wed.) Homework Assignment: From Chap 4: 25, 28, 31, 34

For a Control Volume: Mass Rate Balance: The rate at which mass accumulates in the control volume is the difference between the rate of mass flow in and flow out. Energy Rate Balance: 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.

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

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

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

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

600 kPa 120 kPa 330 K 300 K d =1.2 cm Example 1: (Problem 4.27)
Sec 4.5: Analyzing Control Volumes at Steady State 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 Identify State Properties state Inlet Exit P (kPa) 600 120 T (K) 330 300 h (kJ/kg) 330.34 300.19 state Inlet Exit P (kPa) 600 120 T (K) 330 300 h (kJ/kg) state Inlet Exit p [kPa] T [K] h [kJ/kg] Look up values for h on Table A-22

No significant heat loss No work done by system
Sec 4.5: Analyzing Control Volumes at Steady State Assumptions: System at steady state No significant heat loss No work done by system No large change in elevation state Inlet Exit P (kPa) 600 120 T (K) 330 300 h (kJ/kg) 330.34 300.19 Table A-22 Energy balance: is reduced with assumptions to: Steady state also implies that mass flow rate is equals mass flow rate out.

Next find mass flow rate
Sec 4.5: Analyzing Control Volumes at Steady State Next find mass flow rate state Inlet Exit P (kPa) 600 120 T (K) 330 300 h (kJ/kg) 330.34 300.19 600 kPa 330 K 120 kPa 300 K d =1.2 cm Apply continuity equation and mass balance: therefore: applying Ideal Gas Equation:

Finally combine what we know
Sec 4.5: Analyzing Control Volumes at Steady State Finally combine what we know state Inlet Exit P (kPa) 600 120 T (K) 330 300 h (kJ/kg) 330.34 300.19 600 kPa 330 K 120 kPa 300 K d =1.2 cm from Energy Balance: from Mass balance: therefore: solving for velocities:

Sec 4.5: Analyzing Control Volumes at Steady State
Inlet Exit P (kPa) 600 120 T (K) 330 300 h (kJ/kg) 330.34 300.19 600 kPa 330 K 120 kPa 300 K d =1.2 cm to find the mass flow rate: from table 3.1: R = kJ/kg-K

2 bar d= 4 cm Example 2: (Problem 4.29)
Sec 4.5: Analyzing Control Volumes at Steady State 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 -8 oC 17 kg/min 2 bar d= 4 cm . Q = 3.4 kW Identify State Properties statec Inlet (sat. vapor) Exit p [bar] 2.1704 2 T [oC] -8 v [m3/kg] 0.0919 h [kJ/kg] 242.54 statec Inlet (sat. vapor) Exit p [bar] T [oC] v [m3/kg] h [kJ/kg] Look up values for h on Table A-10

Sec 4.5: Analyzing Control Volumes at Steady State
-8 oC 17 kg/min 2 bar d= 4 cm . Q = 3.4 kW state Inlet (sat. vapor) Exit p [bar] 2.1704 2 T [oC] -8 v [m3/kg] 0.0919 h [kJ/kg] 242.54 Using continuity equation at the inlet Find the volumetric flow rate Find the inlet velocity

Sec 4.5: Analyzing Control Volumes at Steady State
-8 oC 17 kg/min 2 bar d= 4 cm . Q = 3.4 kW state Inlet (sat. vapor) Exit p [bar] 2.1704 2 T [oC] -8 v [m3/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: for steady state, no work done, and a horizontal pipe:

Sec 4.5: Analyzing Control Volumes at Steady State
-8 oC 17 kg/min 2 bar d= 4 cm . Q = 3.4 kW state Inlet (sat. vapor) Exit p [bar] 2.1704 2 T [oC] -8 v [m3/kg] 0.0919 h [kJ/kg] 242.54 combining the mass balance and energy balance equations: gives or

Sec 4.5: Analyzing Control Volumes at Steady State
-8 oC 17 kg/min 2 bar d= 4 cm . Q = 3.4 kW state Inlet (sat. vapor) Exit p [bar] 2.1704 2 T [oC] -8 v [m3/kg] 0.0919 h [kJ/kg] 242.54 This equation contains two unknowns, he and ve. Note that both are properties of the exit state which can be both considered functions of an unknown temperature, TB and a known pressure, pB = 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 TB and let it iteratively find the temperature for us.

c Results: Outlet temperature Te = -10.09 oC Inlet velocity
Vi = 20.7 m/s Outlet velocity

end of Lecture 13 Slides Test #1 Next Monday