The First Law for Open Systems - Thermodynamics | ENSC 2213, Study notes of Thermodynamics

Download The First Law for Open Systems - Thermodynamics | ENSC 2213 and more Study notes Thermodynamics in PDF only on Docsity! ENSC 2213 THERMODYNAMICS 1 ENSC 2213 THERMODYNAMICS Chapter 4 THE FIRST LAW FOR OPEN SYSTEMS AJ Johannes, Ph.D., P.E. Professor of Chemical Engineering School of Chemical Engineering Oklahoma State University Copies of Class Lecture Notes Portions of these notes were originally compiled by Professor Robert L. Robinson, Jr. ENSC 2213 THERMODYNAMICS 2 THE FIRST LAW FOR OPEN SYSTEMS A. Topics for Discussion 1. Develop the Open System Mass Balance 2. Develop the First Law for Open Systems B. The Open System Mass Balance Consider mass flowing into and out of a system at various points on the system (control volume, cv) boundary (control surface, cs): ENSC 2213 THERMODYNAMICS 5 Let’s focus only on the mass flow across the boundary (disregarding other heat and work effects). Control Volume ---------- Closed System ………. (E1)closed = (E1)cv + (E1)δm (E2)closed = (E2)cv so (E2 – E1)cv = (E2 – E1)closed + (E1)δm ENSC 2213 THERMODYNAMICS 6 2 1 closed 1 2 2 1 closed i i i i i 2 1 closed i i i 2 1 cv 1 δm i i i 1 δm i For theclosed system, we know how to write the first law for the processshown: (E E ) W the boundary moves so (E E ) [PdV ] but dV v δm or (E E ) P v δm so (E E ) (E ) P v δm but (E ) (u g − = − − =− =− − = − = + = + 21i i i2 21 2 1 cv i i i i i i2 21 i i i i2 z V )δm so (E E ) (u P v gz V )δm (h gz V )δm + − = + + + = + + ENSC 2213 THERMODYNAMICS 7 Thus, the mass coming into the control volume increases the energy in the CV by the amount of its k.e., its p.e., and it's hi (ENTHALPY, pronounced “N-THALPY”). The fact that it brings its specific energy plus a “pivi” term arises from the fact that it compresses the fluid already in the system in order to get in. ENSC 2213 THERMODYNAMICS 10 i em , m D. Steady-State, Steady-Flow Systems (SSSF) In many practical situations, systems operate for long periods where no changes in conditions occur within the process (e.g., process temperature, pressure, and flow rates are constant). are constant and E2 = E1 Then So the “SSSF” 1st Law in rate form is: 2 21 1 1 2 i i i i 1 2 e e e e2 2Q m (h gz V ) W m (h gz V )+ + + = + + + 2 21 1 i i i i e e e e2 2Q m (h gz V ) W m (h gz V )+ + + = + + + ENSC 2213 THERMODYNAMICS 11 E. Some Useful Flow Relations • Volumetric Flow ( )volumeQ VA HereQ heat time VA Q massm VA v v time ⎛ ⎞= ≠⎜ ⎟ ⎝ ⎠ ⎛ ⎞=ρ = = ⎜ ⎟ ⎝ ⎠ ENSC 2213 THERMODYNAMICS 12 F. Summary of First Law Relations CV ⇒ Control volume (system) CS ⇒ Control surface (system boundary) For any process: dECV = (δEin)CS - (δEout)CS Accumulation Flow of Flow of of energy energy in energy out inside CV across CS across CS 2CV CV CV CV c c 2 in CS i i i e c c CS 2 out CS e e e e c c CS mg m(dE) d U z V g 2g g 1(δE ) δm h z V δQ g 2g g 1(δE ) δm h z V δW g 2g ⎛ ⎞ = + +⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = + + +⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = + + +⎜ ⎟ ⎝ ⎠ ENSC 2213 THERMODYNAMICS 15 APPLICATIONS OF FIRST LAW TO OPEN SYSTEMS A. Topics for Discussion 1. Develop applicable forms of 1st law for specific processes 2. Solve example problems B. General Forms of 1st Law 1. Closed Systems Mass Balance: mi = me = 0 m2 = m1 1st Law: 2 22 2 1 1 2 2 2 1 1 1 c c c cCV CV 1 2 1 2 m g m m g mU z V U z V g 2g g 2g Q W ⎛ ⎞ ⎛ ⎞ + + − + +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = − ENSC 2213 THERMODYNAMICS 16 2. Steady-State Open Systems (SSSF) MB: EB: CV i CS e CS CV d (m) 0 (m ) (m ) dt d (E) 0 dt = ⇒ ∑ =∑ = ⇒ 2 i i i i c c CS 2 e e e e c c CS g 1Q m h z V g 2g g 1W m h z V g 2g ⎛ ⎞ +∑ + + =⎜ ⎟ ⎝ ⎠ ⎛ ⎞ +∑ + +⎜ ⎟ ⎝ ⎠ ENSC 2213 THERMODYNAMICS 17 C. What Can We Calculate Using 1st Law? Closed Systems • 1Q2: only from 1st law (until we study heat transfer) • 1W2: from 1st law or definition of work • Initial state of system / Final state of system Open (SSSF) Systems • Q : only from 1st law • : from 1st law or definition of work • Inlet conditions / Outlet conditions Calculation Tools • 1st Law • P, v, T, u, h relations • Description of process • Mass balance • W m VA=ρ ENSC 2213 THERMODYNAMICS 20 4. Throttling Process This applies to a throttling valve or diffuser Valve Diffuser Then So hi = he Note: for a Diffuser, the k.e. change may not be negligible! W 0 Q 0(often the case : insulated) k.e.& p.e. effects are often negligble • = • = • i i e e i e m h m h but m m (mass balance) = = ENSC 2213 THERMODYNAMICS 21 E. Comparison of Thermal, Kinetic Energy and Potential Energy Effects Example: Heating one lbm of liquid water 10ºF increases its internal energy (or enthalpy) about 10 BTU. Comparable P.E. effect is Comparable K.E. effect is This illustrates why K.E. & P.E. effects are often neglected. c mg10 BTU z g z 7,780 ft = Δ ⇒ Δ = 2 2 2 1 c 2 m10 BTU (V V ) 2g V 708 ft / sec = − ⇒ = ENSC 2213 THERMODYNAMICS 22 F. Behavior of Properties for Compressed Liquids Consider the following data for water: @ T = 50ºF p, psia v, ft3/lbm u, BTU/lbm h, BTU/lbm 0.18 (psat) 0.01602 18.06 18.06 500 0.01600 18.02 19.50 Difference (500 – psat) -0.00002 -0.04 1.44 For v: effect of p @ const. T is negligible For u: effect of p @ const. T is negligible For h, h = u + pv Δh = Δu + Δpv = Δu + pΔv + vΔp Δh ≈ vΔp @ T = constant ENSC 2213 THERMODYNAMICS 25 Example 2. Air at 800°R enters a nozzle operating at steady state with a very low velocity and exits the nozzle at 570°R. Heat transfer occurs from the air to the surroundings at a rate of 10 Btu/lbm of air. Assuming ideal gas behavior and neglecting potential energy effects, determine the velocity at the exit in ft/s. ENSC 2213 THERMODYNAMICS Example 3. A well-insulated turbine operating at steady state is sketched in Fig. P4.40. Steam enters at 3 MPa, 400°C, _ With a volumetric flow rate of 85 m?/min. Some steam is extracted from the turbine at a pressure of 0.5 MPa and a temperature of 180°C. The rest expands to a pressure of 6 kPa and exits with a mass flow rate of 40,000 kg/h and a quality of 90%. Kinetic and potential energy effects can be neglected. Determine {a) the diameter, in m, of the duct through which steam is extracted, if the velocity there is 20 m/s. (b) the power developed by the turbine, in kW and Btu/h. - Power —p> 4 out 3 Py = 3MPa V Ty = 400°C up (AV), = 85 m/min ay p3 = 6kPa Vp = 20 mis xy = 90% P2 = 0.5MPo thea = 40,000 kg/h Ty = 180°C 26