High viscosity is obviously an undesirable property of good refrigerant as it increases the losses of energy converts the useful energy in to non-conservative frictional energy.
A solid metallic block weighing 5 kg has an initial temperature of 500°C. 40 kg of water initially at 25°C is contained in a perfectly insulated tank. The metallic block is brought into contact with water. Both of them come to equilibrium. Specific heat of block material is 0.4 kJ.kg⁻¹. K⁻¹. Ignoring the effect of expansion and contraction and also the heat capacity to tank, the total entropy change in kJ.kg⁻¹ , K⁻¹ is
By energy balance mcp(Ti−Tf) of block=mcp(Tf−Ti) of water So, we will get Tf=303.16k We know, ΔStotal=ΔSBLOCK+ΔSWATER⇒Δs=mcplnTfTI+mcplnTfTi ⇒ on substitution w will get Δs=−1.869+3.118=1.26
If we see the typical p−T plot any substance below the triple point we have only sublimation zone that means no liquefaction is not possible if we still decrease the pressure hence always the pressure should be greater than triple point pressure and the value of temperature depends on the pressure at which the substance is present.
For a spontaneous process of a system the change in entropy should be always greater than zero and for an equilibrium (more or less reversible) process the entropy must be equal to zero.
The change in enthalpy s given as enthalpy of the products – enthalpy of the reactants since, the given reaction is opposite of given reaction only the sign changes.
We know COP of a refrigerator =Cooling effectWork taken=QLWnet=Q2Q1−Q2 For a reversible refrigerator COP =T2T1−T2 Since T1−T2<T2 So COP is always greater than zero.
. Compressor the process equipment is usually not used during the vapour compression refrigeration system and absorption compression refrigeration system.
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