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# Solved Problems: Refrigeration and Air Conditioning

Mechanical and Electrical - Thermal Engineering - Refrigeration and Air Conditioning

SOLVED PROBLEMS

1.    A sling psychrometer gives reading of 250c dry bulb temperature 1 50c wet bulb temperature. The barome ter indicates 760 mm of hg assuming partial pressure of the vapour as 10 mm of Hg. Determine 1. Specific humidity 2. Saturation ratio.

Given Data:

Dry bulb  temper ature  td  =250c

Wet bulb tempe rature tw=150c Barometer pressure pb=760mm of Hg

Partial pressure pv= 10mm of Hg

To Find:

Specific humidit y

Saturation ratio.

Solution:

Specific humidity:

We know that Specific humidity =                                                             0.41

Result:

1.                                                             Specific humidity     0.0083 kg/kg of dry air

2.                                                             Saturation ratio . =  0.41

2. A two stages, single acting air compressor compresses air to 20bar. The air enters the L.P cylinder at 1bar a nd 27oc and leaves it at 4.7bar. the air en ters the H.P. cylinder at 4.5bar and 27oc. the size of the L.P cylinder is 400mm diameter and 5 00mm stroke. The clearance volume In both cylinder is 4% of the respective stroke volume. The compressor runs at 200rpm, taking index of compression and expansi on in the two cylinders as 1.3, estimate 1. The indicated power required to run the co mpressor; and 2. The heat rejected in the intercooler per minute.

Given data:

Pressure (P4)= 20bar

Pressure (P1) = 1 bar = 1× 105 N/m2

Temperature (T1 ) = 27oC = 27+273 = 300K Pressure (P 2) = 4.7bar

Pressure (P3) = 4.5bar

Temperature (T3 ) = 27oC = 27+273 = 300K Diameter ( D1) = 400mm 0.4m Stroke (L1) = 500 mm = 0.5m

K = 0.04

N = 200rpm ; n = 1.3

To Find:

Indicated power required to run the compressor

Solution :  = 2 043.5× 103 J/min = 2034.5 KJ/min

Total indicated work done by the compressor,

W = WL + WH = 2123.3 + 2034.5

= 4157.8 KJ/min

Indicated power required to run the compressor = 4157.8 / 60

= 69.3KW

3. In an oil gas turbine installation , air is taken as 1 bar and 30oC . The air is compressed

to 4bar and then heated by b urning the oil to a temperature of 500oC . If the air flows at the rate of 90Kg/min . Find the power developed by the plant take γ for air as 1.4 Cp as 1KJ/KgK . If 2.4Kg of oil having calorific value of 40,000 KJ/Kg if b urned in the combustion chamber per min ute. Find the overall efficiency of the plant.

Given Data:

Pressure (P4 = P3) = 1bar Pressure (P1 = P2) = 4bar

Temperature (T2) = 500oC  = 500+273 = 773K

Mass flow rate of air(ma) = 90Kg/min = 1.5Kg/sec Mass flow rate of fuel (mf) = 2.4Kg/min = 0.04Kg/sec

Temperature (T4) = 30oC = 30+273 = 303K γ = 1.4 ; Cp = 1KJ/KgK ; Cv= 40,000 KJ/Kg

To Find:

Power developed   by   the   plant

Performance                            of  the  gas  turbine

Overall efficiency of the plant

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Mechanical and Electrical - Thermal Engineering - Refrigeration and Air Conditioning : Solved Problems: Refrigeration and Air Conditioning |

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