Unit Four Quiz Solutions and Unit Five Goals

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Unit Four Quiz Solutions and Unit Five Goals

• Mechanical Engineering 370
• Thermodynamics
• Larry Caretto
• March 4, 2003

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Outline

• Quiz three and four solutions are similar
• Finding work (area under path) and u
• Q m (ufinal uinitial) W
• Unit five open systems
• View first law as a rate equation
• Have mass crossing system boundaries
• Flows across boundaries have several energy forms
  including internal (u) , kinetic and potential
  energy plus flow work (Pv)

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Quiz Three Results

• 21 students max 30 mean 18.6
• 10 10 14 14 15 15 15 15 15 18 18
• 20 20 21 22 24 24 24 24 24 28
• Q m(ufinal uinitial) W
• To compute u h Pv
• Use specific volume in m3/kg
• convert P to kPa for h and u in kJ/kg

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Quiz Three Path

• Quiz three gave neon data near and in mixed
  region.
• T1, P1, V1, P2, V2
• T3, V4
• The quiz three diagram is shown here.
• This was not an ideal gas so we had to use
  property tables

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Quiz Four Path

• Quiz four has the same path with the same data
  items as quiz three.
• T1, P1, V1, P2, V2
• T3, V4
• The quiz four path diagram is shown here.
• This was an ideal gas so we use PV mRT and du
  ?cvdT

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Quiz Four Solution

• Given Water in three-step process
• T1 300oC, V1 1 m3, P1 100 kPa
• 1-2 is a linear path to P2 300 kPa, V2 0.8 m3
• 2-3 is constant volume with T3 400oC
• 3-4 is constant pressure with V4 0.4 m3
• Find the heat transfer, Q, using ideal gas
• Find Q from first law (ideal gas with cv const)
• Q DU W m(u4 u1) W mcv(T4 T1) W
• Work is (directional) area under path

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Path for This Process

• Work area under path trapezoid area plus
  rectangle area
• W (P1 P2)(V2 V1)/2 P3-4 (V4 V3)
• DV lt 0 means work will be negative

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P
3
1
4
V
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Finding the Answer

• Use properties at the initial state to find the
  mass

• Find P3-4 P3 P4 from state 3 defined by T3
  400oC and v3 v2 V2/m

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Finding the Answer (contd)

• Find T4 from m, P4 P3 and V4 0.4 m3

• Now find heat and work

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Finding the Answer (concluded)
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Future Quizzes

• Can use equation summary
• Download from course web page (follow course
  notes link)
• May have unannounced open book exams to allow use
  of tables
• If you are late for a quiz you can
• Come to class after quiz is over or
• Start quiz and receive grade

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Unit Five Goals

• Topic is first law for open systems, i.e.,
  systems in which mass flows across the boundary
• Will look at general results and focus on
  steady-state systems.
• As a result of studying this unit you should be
  able to
• understand all the terms (and dimensions) in the
  first law for open systems

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Open System Concepts
the useful work rate or mechanical power (ML2T-3)
the mass flow rate (MT-1)
the kinetic energy per unit mass (L2T-2)
gz the potential energy per unit mass (L2T-2)
total energy (ML2T-2)
heat transfer rate (ML2T-3)
rate of energy change (ML2T-3)
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Unit Five Goals Continued

• use the equation relating velocity, mass flow
  rate, flow area, A, and specific volume

• use the mass balance equation

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Flow Work

• For open systems work is done on (or by) mass
  entering and leaving the system
• Flow work is Pv times mass flow rate
• Add this flow work to internal energy (times mass
  flow rate)
• First law for mass flows has h u Pv (sum of
  internal energy plus flow work)

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Unit Five Goals Continued

• use the first law for open systems

• use the steady-state assumptions and resulting
  equations

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Steady-state equations

• Steady-state first law for open systems

• Steady-state mass balance for open systems

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Unit Five Goals Continued

• recognize that kinetic and potential energies are
  usually negligible
• A 1oC temperature change in air (ideal gas with
  cp 1.005 kJ/(kgK) has Dh 1005 J/kg
• A similar kinetic energy change requires a
  velocity increase from zero to 45 m/s (100 mph)
• A similar potential energy change requires an
  elevation change of 102 m (336 ft)

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Unit Five Goals Concluded

• handle simplest case steady-state, one inlet,
  one outlet (one mass flow rate), negligible
  changes in kinetic and potential energies

• work with ratios q and w in simplest case

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Example Calculation

• Given 10 kg/s of H2O at 10 MPa and 700oC renters
  a steam turbine the outlet is at 500 kPa and
  300oC. There is a heat loss of 400 kW.
• Find Useful work rate (power output)
• Assumptions Steady-state, negligible changes in
  kinetic and potential energies
• Configuration one inlet and one outlet

First law
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Getting the Answer

• At Tin 700oC and Pin 10 MPa, hin 3870.5
  kJ/kg (p. 836)
• At Tout 300oC and Pout 500 kPa, hout
  3061.6 kJ/kg
• Heat loss is negative Q Qin - Qout

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Review Ideal Gases

• For ideal gases du cvdT dh cpdT
• Ideal gas DH DU RDT
• May have molar DH data
• Last week we looked at problem with H2O as an
  ideal gas, where T1 200oC and T2 400oC
• How do we handle cv(T)?

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Ideal Gas with cv(T)

• Find a, b, c and d from Table A-2(c), p 827
• Use T1 473.15 K and T2 673.15 K
• Molar enthalpy change 7229.3 kJ/kmol

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Getting Du from molar Dh

• Use molar Dh just found, data on R and M, and DT
  200oC 200 K

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Ideal Gas Tables
 

• Find molar u(T) for H2O in Table A-23 on page 860
• Have to interpolate to find u1 u(473.15 K)
  11,953 kJ/kmol and u2 u(673.15 K) 17,490
  kJ/kmol
• Du (17,490 kJ/kmol - 11,953 kJ/kmol) / (18.015
  kg / kmol) 307.4 kJ/kg