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Refrigerant 134a is compressed from 3 bar, saturated vapor, to 10 bar, 50°C in a compressor operating at steady state. The mass flow rate of refrigerant entering the compressor is 7 kg/min, and the power input is 10.85 kW. Kinetic and potential energy effects can be neglected. (a) Determine the rate of heat transfer for the compressor, in kW. Note that heat transfer is positive going into the compressor. (b) If the heat transfer occurs at an average surface temperature of 50°C, determine the rate of entropy production, in kW/K. (c) Determine the rate of entropy production, in kW/K, for an enlarged control volume that includes the compressor and its immediate surroundings such that the heat transfer occurs at 300 K.

Answer :

A) The rate of heat transfer for the compressor, in kW is; Q' = -7.123 KW

B)  The rate of entropy production, in kW/K is; 0.02455 kW/K

C) The rate of entropy production for an enlarged control volume is; 0.02624 kW/K

What is the Rate of Entropy?

A) From energy equation, if we neglect kinetic and potential energy effects, we will get;

Q' - W' = m'(h2 - h1)

From the first table attached,

At pressure of 3bar = 300 KPa, hg = h1 = 250.88 KJ/Kg

Similarly, from second table attached, at pressure of 10 bar = 1000 KPa and temperature of 50°C, h2 = 282.8 KJ/Kg

Mass flow rate; m' = 7 kg/min = 7/60 kg/s = 0.1167 Kg/s

Power input; W' = -10.85 KW

So, from Q' - W' = m'(h2 - h1)

Let's make Q' the subject and plug in relevant values;

Q' = W' + m'(h₂ - h₁)

Q' = -10.85 + 0.1167(282.8 - 250.88)

Q' = -7.123 KW

B) Entropy could be defined as;

ΔS = ∫δQ/T + σ

As constant average values are used, we have;

∫δQ/T = Q/T

From the first table attached, specific entropy at pressure of 300 KPa is; s1 = 0.9312 KJ/Kg.K

Similarly, from second table attached, at pressure of 10 bar = 1000 KPa and temperature of 50°C,

Entropy, s2 = 0.9526 KJ/Kg.K

Now, the rate of entropy production is given as;

σ = ΔS - Q/T_Q = m'•ΔS - Q/T_Q = m'(s₂ - s₁) - Q/T

Thus; σ = m'(s₂ - s₁) - Q/T

Where T is the temperature at which heat is transferred = 50°C = 273 + 50 = 323 K

Plugging in the relevant values, we have;

σ = 0.1167(0.9526 - 0.9312) - (-7.123/323)

σ = 0.02455 kW/K

C) The formula for the rate of entropy production is;

σ = ΔS - Q/T_surr = m'(s2 - s1) - Q/T

Where T_surr = T = 300K

Thus; σ = 0.1167(0.9526 - 0.9312) - (-7.123/300) = 0.02624 kW/K

Read more about rate of Entropy at; https://brainly.com/question/16014998

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