Section IV – First Law Applied to Open SystemsPage 1

SECTION IV – FIRST LAW APPLIED TO OPEN SYSTEMS

Refrigerant-12 vapour enters a steady-flow compressor as a saturated vapour at 12C. The outlet conditions are 0.6 MPa and 50C, and the process is assumed to be adiabatic. Calculate the power in Kilowatts, required if the refrigerant flow rate is 20 kg/min. Determine the diameter of the inlet tubing to the compressor if the inlet velocity is not to exceed 3 m/s.

Schematic of compressor

Inlet conditions TI = 12C, saturated vapour

Exit Conditions

Pe = 0.6 MPa, Te = 50C

Pe < Osat = 1.2193 MPa corresponding to

Te = 50C  Refrigerant-12 is superheated

at the exit.

Process is adiabetic 

Flow through compressor is steady 

Assumptions

- Properties are uniformly distributed at the inlet and exit.

- epot and ekin are negligible compared to h.

Steady-state, steady-flow form of the conservation of energy for the compressor as an open system is then:

-

Steam enters a turbine at 600C and 6 MPa with a velocity of 300 m/s. The mass flow rate entering the turbine is 400 kg/min. Assume the turbine is well-insulated and the exhaust steam leaves the turbine at a low velocity. The steam exits the turbine at 200 KPa and 260C. Calculate the power developed by the turbine and the inlet-duct area.

Schematic of turbine

Inlet Conditions: TI = 600C, PI = 6 MPa,

Exit Conditions: Te = 260C, Pe = 200 KPa

Turbine is well-insulated 

Assumptions

- flow through turbine is steady i.e. no mass build-up in turbine

- flow properties are uniformly distributed at the inlet and exit to turbine

- changes in potential energy of working fluid are negligible compared to change in enthalpy

- neglect kinetic energy at the exit since the velocity will be low

Steady-state, steady-flow form the conservation of energy for the turbine as an open system is then:

  • steam is superheated at the inlet since T > Tsat = 275.64C corresponding to P = 6 MPa hI = 3658.4
  • steam is also superheated at the outlet since P < Psat = 4.688 Mpa corresponding to P = 200 KPa


4-64 Boles & Cengel

Refrigerant –12 at 1MPa and 80C is cooled to 1MPa and 30C in a condenser by air. The air enters at 100KPa and 27C with a volume flow rate of 800 m3/min and leaves at 95KPa and 60C. Determine the mass flow rate of the refrigerant.

Can air be treated as an ideal gas under the given conditions?

Z3 = Z4 1 (See p. 807 of Text)

Assumptions:

  • Flow of Refrigerant –12 is steady
  • Flow of air is steady
  • Heat lost by R-12 is gained by air  Qie air = -Qie R-12
  • Properties are uniformly distributed at 1 & 2 and at 3 & 4.
  • Conservation of energy applied to the flow of air:

ie air ie air = (h4 – h3)

see p. 788 of text

  • Conservation of energy applied to the flow of Refrigerant –12:

(no work done on or by the Refrigerant –12)

T1 = 80C  Tsat = 41.64C corresponding to P1 = 1MPa  R-12 is a superheated vapor at 1

h1 = 232.91 KJ/Kg (see p.784 of Text)

T2 = 30C  Tsat = 41.64C corresponding to P2 = 1MPa  R-12 is a subcooled liquid at 2

(see pp. 780-781 of text)




The operating data from the simple, steady-flow, steam-power plant cycle shown in the accompanying figure are summarized in the table below. Determine the power output of the turbine, the heat-transfer rate in the steam generator, the power input to the pump, and the thermal efficiency of the cycle.

LocationPressure, MpaQuality or Temperature
10.010.0
248°C
35450° C
40.95
Other data:
Mass flow rate of steam = 20 kg/s
Negligible pressure drops through steam generator and condenser


Location / Pressure, Mpa / Quality or Temperature
1 / 0.01 / 0.0
2 / 5 / 48C
3 / 5 / 450
4 / 0.01 / 0.95

Assumptions:

  • ep < h, ek < h for all units
  • Properties are uniformly distributed at the inlet and exit of each unit of the cycle

-Pressure drop through steam generator is negligible  P2  P3 = 5Mpa

-Pressure drop through condenser is negligible  P4 P1 = 0.01 Mpa = 10KPa

  • Flow through the steam generator:

First law:

T2 = 48C < Tsat = 263.99C

Corresponding to P2 = 5MPa

 water is a compressed liquid at state (2)

T3 = 450C > Tsat = 263.99C corresponding to P3 = 5Mpa  water is a superheated vapor at state(3).

h3 = 3316.2 KJ/Kg


Flow through the turbine:

First Law:


  • Flow through the pump:

First law:


 Power input

Flow through the condenser:

First law:

  • Thermal Efficiency:

Check:

(17024 – 283.8) = (52203.6 – 45463.4)

16740.216740.2Check!