Refrigerant-134a enters an adiabatic compressor as saturated vapor at 160 kPa at a rate of 2.4 m3/min and is compressed to a pressure of 900 kPa. Determine the minimum power that must be supplied to the compressor. Use the tables for R-134a

Explanation:

Step one:

given data

from the table  for R-134a

For the given pressure state the following data were extracted.

s=0.94202kJ/kg.K

h1=241.14kJ/kg

∝1=0.12355m^3/kg

By interpolation, the final enthalpy and pressure with data from the table of  R-134a

h2=277.12kJ/kg

The minimum power output is then determined from the energy balance

W=m(h2-h1)

W=V1/∝1(h2-h1)

W=2.4/60/0.12355(277.12-241.14)

W=0.04/0.12355(35.98)

W=0.3237(35.98)

W=11.65kW Approx.

The minimum power output of the compressor is 11.65kW

power generated = 307.84 Kw

Explanation:

The Given data

Tk = 86 f = 303.15k

To = 41 f = 278.15k

thermal efficiency ( n ) = 0.025

Q = 13300 gpm = 0.84 m^3/sec

C = 1.0 Btu/Ibmf

density of seawater ( d ) = 64 Ibm/ft3 = 1025.18 kg/m^3

rise in temp = 6 f

we have to make assumptions

first we calculate

mass flow rate of water ( Mw )

Mw = density of seawater * Q

      =  64 * 13300 * 1/7.4804

      = 113790 Ibm/min  = 1896.51 Ibm/sec

therefore the rejection of the cooling water will be

 = mass flow rate * C * rise in temperature

= 1896.51 * 1.0 * 6

= 11379.06 BTU/s (Qout)

Determine the amount of power generated using thermal efficiency formula

n =  

W = power generated

n = thermal efficiency

Qout = rejection of cooling water

0.025 =

W = 0.025 W + 284.48

W ( 1 - 0.025 ) = 284.48

therefore   W = 284.48 / 0.975

W = 291.77 BTU/s

note : 1 kw = 0.9478 BTU/s

            x   = 291.77 BTU/s

W in kw ( x ) = 291.77 / 0.9478 = 307.84 Kw