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Determine for the ideal cycle:$\ \ \ \$ i. C.O.P. $\ \$ ii. Mass of air circulated per min $\ \$ iii. Theoretical piston displacement of compressor and expander $\ \$ iv. Net power per TR

### Show the cycle on P-V and T-S diagrams. C for air, take $\gamma$=1.4, Cp=1 kJ/kgK.

Mumbai University > Mechanical Engineering > Sem8 > Refrigeration and Air Conditioning

Marks: 12M

Year: Dec 2013

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Given: Q=10 TR; p2 = p3 = 4.2 bar;p1=p4=1.4 bar;T3=50℃=323K

T1= -20℃ = 253K

Process 1-2 = Isentropic compression in compressor

Process 2-3 = Constant pressure heat rejection in heat exchanger

Process 3-4 = Isentropic expansion in expander

Process 4-1 = Constant pressure heat absorption in refrigerator

i. C.O.P.

Assume compression and expansion to be isentropic.

$\frac{T2}{T1}=\big(\frac{p2}{p1}\big)^{\frac{\gamma - 1}{\gamma }}=\big(\frac{4.2}{1.4}\big)^{\frac{1.4-1}{1.4}}=1.369 \\ ∴ T2=253×1.369=346K \\ \frac{T3}{T4}=\big(\frac{p3}{p4}\big)^{\frac{\gamma - 1}{\gamma }}=\big(\frac{4.2}{1.4}\big)^{\frac{1.4-1}{1.4}}=1.369 \\ ∴ T4=323/1.369=236K \\ C.O.P.=\frac{(T1-T4)}{(T2-T3)-(T1-T4)}=2.83$ Alternatively,

COP for ideal Reverse Brayton Cycle is

$$C.O.P.= \frac1{(r_p )^{\frac{(γ-1)}{γ}}-1}=2.71 \\ where \ \ \ r_p=\frac{p2}{p1}=\frac{p3}{p4}$$

ii) Mass of air circulated per min

Let Heat absorbed per sec=$Q ̇_a$=3.5×capacity

Capacity = 10TR

∴ $Q ̇_a$=3.5×10=35 kJ/sec

Also $Q ̇_a=m ̇_a×C_p×(T1-T4)$

35= $m _a$×1×(253-236)

∴ $m _a$=2.0588 kg/sec=123.53 kg/min

iii. Theoretical piston displacement of compressor and expander

$$η_vol=\frac{V ̇_a}{V ̇_s}$$

Assuming 100% volumetric efficiency ,hence V ̇a=V ̇s for both the cases

PV=mRT

Let V1= Theoretical piston displacement of compressor per min

V1=$\frac{(m_a RT1)}{P1}=\frac{(123.53×287×253)}{(1.4×10^5 )}=64.06 m^3/min$

Let V4= Theoretical piston displacement of expander per min

$\frac{V4}{T4}=\frac{V1}{T1} \\ V4=\frac{(64×236)}{253}=59.7 m^3/min$

Note:

a) The size of compressor is described by volume of air entering it.

b) The size of expander is described by volume of air leaving it.

iv) Net power per TR

Net power consumed by plant $=W ̇_C-W ̇_E \\ = m ̇_a × C_p × [(T2-T1)-(T3-T4)] \\ =12.3528 kW$

Hence net power per TR =$\frac{12.3528}{10}=1.235 \frac{kW}{TR}$