System Components of a Cooling System
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Questions and Answers

What is the assumption made about the air?

  • Constant specific heats at 20 K
  • Constant specific heats at 300 K (correct)
  • Variable specific heats at 300 K
  • Constant specific heats at 25 K
  • What is the value of Cp, w for water?

  • 1.005 kJ/kg·K
  • 2.50 kJ/kg·K
  • 4.18 kJ/kg·K (correct)
  • 3.33 kJ/kg·K
  • What is the equation for the energy balance in the system?

  • E in - E out = ΔE system
  • E in = E out (correct)
  • E in × E out = ΔE system
  • E in + E out = ΔE system
  • What is the ratio of the mass flow rates of air to water?

    <p>1 kg air / 3.33 kg water</p> Signup and view all the answers

    What is the temperature difference between the initial and final states?

    <p>25 K</p> Signup and view all the answers

    What is the value of Cp, air for air?

    <p>1.005 kJ/kg·K</p> Signup and view all the answers

    What is the unit of the energy transfer rate?

    <p>kW</p> Signup and view all the answers

    What is the assumption made about the water?

    <p>Incompressible fluid with constant specific heat</p> Signup and view all the answers

    What is the ratio of the heat capacities of water and air?

    <p>Cp, w / Cp, air = 4.18 / 1.005</p> Signup and view all the answers

    What is the value of mw in the problem?

    <p>3.33 kg/s</p> Signup and view all the answers

    Study Notes

    Exercise 1-1: Work Done by Air Flowing Through a Turbine

    • High-pressure air at 1300 K flows into an aircraft gas turbine and undergoes a steady-state, steady-flow, adiabatic process to the turbine exit at 660 K.
    • The work done per unit mass of air flowing through the turbine is calculated using the conservation of energy equation: Wout = ṁ (h1 - h2).
    • The work done by the air per unit mass flow is equal to the enthalpy decrease of the air.
    • Using the air tables, the work done is calculated as wout = 103.9 kJ/kg.

    Exercise 1-3: Steam and Cold Water Flow Rates

    • Steam at 0.2 MPa, 300°C, enters a mixing chamber and is mixed with cold water at 20°C, 0.2 MPa, to produce 20 kg/s of saturated liquid water at 0.2 MPa.
    • The conservation of mass equation is applied: ṁ1 + ṁ2 = ṁ3.
    • The conservation of energy equation is applied: ṁ1h1 + ṁ2h2 = ṁ3h3.
    • The required steam and cold water flow rates are calculated as ṁ1 = 2.82 kg/s and ṁ2 = 17.18 kg/s.

    Exercise 1-4: Air and Water Flow Rates in a Heat Exchanger

    • Air is heated in a heat exchanger by hot water, with a 20°C drop in temperature.
    • The conservation of mass equation is applied: ṁair,1 + ṁw,1 = ṁair,2 + ṁw,2.
    • The conservation of energy equation is applied: ṁair(hair,1 - hair,2) = ṁw(hw,2 - hw,1).
    • The ratio of mass flow rate of air to mass flow rate of water is calculated as ṁair / ṁw = 3.33.

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    Description

    This quiz covers the main components of a cooling system, including air and water inlets, control surfaces, and exit points.

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