Mechanical Heat Pump

    7 Votes

Objective of this paper presentation is to study the performance characteristics of Hilton Mechanical Heat Pump. Experimental set up is mounted on a single platform. The Hilton Heat Pump is designed solely for educational purpose and data in qualitative form can be obtained for analysis of the heat pump. It consists of

  • Two cylindrical water tanks (Condenser tank – red & Evaporator tank- blue)
  • Material—Stainless steel 
  • Hermetically sealed reciprocating type single cylinder compressor
  • Power—0.5 H.P.
  • Operating--220V 50Hz
  • Manually variable expansion valve
  • Silica gel drier
  • Refrigerant flow meter
  • Pressure cut-outs(LP cut-out: 2-6 bar, HP cut-out: 6-32 bar)
  • Two water flow meters
  • Watt hour meter(c = 150rev/kWh )
  • Two pressure gauges(Condenser: 0-2000kN/m2 & Evaporator: 0-600kN/m2)
  • Pipe for refrigerant flow with insulation
  • Nickel plated copper

Heat Pump

Principle: It works on the vapor compression refrigeration cycle. Heat from the atmosphere (i.e. water or air) at low temperature is extracted at the evaporator and this heat with input of compressor work is delivered to the condenser. The heat rejected (useful heat) at the condenser can be used for various heating purposes.

The working medium is R-12 (Dichlorodifluoromethane CF2Cl2) and the heat source and heat sink is cold running water. The refrigerant is superheated in the evaporator tank and then compressed and pumped through insulated copper pipes to the nickel plated copper condensing coil in the condensing tank. Liquid refrigerant coming from the condenser is passed through silica gel drier to remove any traces of moisture present and then it is expanded through manually operated expansion valve. Now low pressure and low temperature refrigerant is sent to evaporator tank and the cycle repeats.

The actual cycle differs from ideal cycle. The schematic diagram of ideal cycle is shown.

Heat Pump Line Diagram

Processes

A. Vapor Compression Cycle: There are two constant pressure processes, one reversible adiabatic process and one is enthalpic throttling process.

  • Process 4—1: Constant pressure heat addition process in evaporator. The refrigerant takes latent heat from water around evaporator coils and changes into saturated vapour.
  • Process 1—2: Reversible adiabatic (isentropic) compression with superheated vapour at compressor outlet.
  • Process 2—3: Constant pressure heat rejection process. The refrigerant rejects heat to water around the condenser tubes and gets condensed.
  • Process 3—4: Throttling (not isentropic but isenthalpic) process. The refrigerant’s temperature and pressure falls.

Observation Table

Sr. NoEvaporator gauge pressure kN/m2Condenser gauge pressure kN/m2Condenser temperature InletCondenser temperature OutletEvaporator temperature InletEvaporator temperature OutletEnergy meter,time for 5 rotations T, in sRotameter, g/s
1 135 870 54 35 -17 27 572 7
2 135 870 58 36.5 -20 26.5 430 7

Calculations

Observations of specific enthalpy from p-h or t-s charts at different state point:

For the 1st observation reading:

Sr. No.State PointPressure, barTemperature (o C)Specific enthalpy (kJ/kg)
1 1 1.35 -25 342
2 2 8.7 54 380
3 3 8.7 35 234
4 4 1.35 -25 234


For reverse Carnot cycle (ideal cycle)

By the means of Carnot cycle and Temperature-Entropy diagram: (refer fig: 3)

Ideal C.O.P. = T3/(T3-T1) = 308/(308-248) = 5.13

For theoretical simple saturation cycle

By the use of pressure-enthalpy diagram (see fig- 4)

  • Theoretical heating effect = h2-h3 = (380- 234) kJ/kg = 146 kJ/kg
  • Work done by compressor = h2-h1 = (380-342) kJ/kg = 38 kJ/kg
  • Theoretical C.O.P. = (heating effect/Work done by compressor)=(h2-h3)/(h2-h1) = 146/38= 3.84
  • Mass flow rate of refrigerant = m· = 7 g/s
  • Cooling capacity = m·(h1-h4) = 7 x (342- 234)= 108 W
  • Heating capacity = m·(h2-h3) = 7 x (380-234) = 146 W
  • Total work consumed = m·(h2-h1) = 7 x (380-342) = 266W

For actual cycle

  • Assume mass flow rate of water = mw(cond) = 5 g/s
  • Actual heating effect = mw(cond)Cp(?Tc) = 5 x 4.183 (54 - 35) = 0.39kW
  • (where ?Tc = difference between outlet and inlet water temperature in condenser from the table)
  • Actual work done obtained from watt hour meter =(5 x 60 x 60) ÷(t x c) kW = (5 x 60 x 60)/ (572 x 150) = 0.2 kW
  • (where c = 150 rev/kWh; t = time required for 5 rotations)
  • Actual C.O.P.= (actual heating effect) ÷ (actual work done) = 0.39/0.2 =1.85

Results

CycleCOP
Ideal(Carnot) 5.13
Theoretical 3.84
Actual 1.85

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