RAC_TYS Exp._923

2017 WORKBOOK Detailed Explanations of

Try Yourself Questions Mechanical Engineering Refrigeration and Air-conditioning

1

Heat Engine, Heat Pump, Refrigerator and Reversed Carnot Cycle

T1 : Solution (COP)RE = (COP)HP – 1 = 4 – 1 = 3 (COP)RE = ⇒

Required cooling effect Power input

Required cooling effect = 3 × 3 = 9 kW = 9 × 60 kJ/min = 540 kJ/min

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2

Vapour Compression Refrigeration System

T1 : Solution T

2

P 2

3

3 1

4

4 s

1 h

From table

h1 = 183.2 kJ/kg h2 = 222.6 kJ/kg h3 = h4 = 84.9 Vs =

[3 to 4 is an Isenthelpic process]

1.5 m3 1000

N = 1600 m3 ηvol. = 80% =

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80 = 0.8 100


Mechanical Engineering • Refrigeration and Air-conditioning

4

ηvol. =

80 = 100

∴

& 1 mV N Vs × 60 m& × 0.0767 1.5 1600 × 1000 60

80 1.5 1600 1 m& = 100 × 1000 × 60 × 0.0767 = 0.4172 kg/s

R.E. = m& [h1 − h2 ] = 0.4172 × [222.6 – 183.2] = 16.437 = 16.44 kW T2 : Solution

h4 = h3 = 134 kJ/kg h1 = 244 kJ/kg h2′ = 285 kJ/kg

T

2 2′ 3 1.6 MPa

h2 ′ − h1 h2 = h1 + η comp.

=

QR = = Winput = = COP =

4

285 − 244 244 + = 292.2 kJ/kg 0.85 (h2 – h3) = 292.2 – 134 158.2 kJ/kg h2 – h1 = 292.2 – 244 48.2 kJ/kg

240 kPa

1

s

QR 158.2 = = 3.2821 Winput 48.2 T

T3 : Solution

s1 = sf1 +

x1hfg1 T1

x1 × 1297.68 263 = 0.5443 + 4.934x1

2

3

= 0.5443 +

s 2 = sf2 +


x2hfg 2 T2

4

1

s

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= 1.2037 +

5

0.95 × 1145.8 303

= 4.796 0.5443 + 4.934 x1= 4.796 x1 = 0.86 h 1 = hf + x1hfg 1 1 = 135.37 + 0.86 × 1297.68 = 1251.4 kJ/kg h2 = hf + x2hf g 2 2 = 323.08 + 0.95 × 1145.8 = 1411.6 kJ/kg We know that theoretical C.O.P. =

h1 − hf3 h2 − h1

=

1251.4 − 323.08 = 5.8 1411.6 − 1251.4

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3

Vapour Absorption Refrigeration System

T1 : Solution

COP =

TG − T0 TR × TG T 0 − TR

TR = Evapourator Temperature TG = Generator Temperature T0 = Ambient Temperature (condenser temperature) COP =

360 − 310 260 × 360 310 − 260

= 0.72 0.72 =

TG − 310 250 × TG 310 − 250

TG = 374.9 K


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4

Refrigerants

T1 : Solution R-502 is an azeotrope. A azeotrope a mixture of two or more refrigerants in which the liquid and vapour have same composetion at equilibrium. Azeotropes behaves like pure component refrigerants because there is no change in boiling temperature or shift in composition during phase change, equipment operation or leakage.

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5

Refrigeration Equipment and Gas Refrigeration

T1 : Solution

TH = 40°C Initial COP – 40 + 273 = 40 – (– 40) =

230 = 2.9125 80

TL = – 40°C

Final COP = 3.5 Let the change in higher temperature and lower change be “ΔT ”. New higher temperature = Th – ΔT New lower temperature = TL + ΔT, [COP]final =

TL + ΔT (TH − ΔT ) − (TL + ΔT )

3.5 =

233 + ΔT (40 − ΔT ) − ( −40 + ΔT )

3.5 =

233 + ΔT 80 − 2ΔT

280 – 7ΔT = 233 + ΔT 280 – 233 = 7ΔT + ΔT


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ΔT = Hence new higher temperature = = New lower temperature = = = Checking

New COP =

=

9

47 = 5.875°C 8

40 – 5.875 = 34.125°C 307.125 K – 40 + 5.875 – 34.125°C 238.875°C

(TL )NEW (TH )NEW − (TL )NEW 238.875 = 3.5 307.125 − 238.875

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6

Air-conditioning

T1 : Solution

Specific humidity, ω = ?

Patm DBT φ Pvs

= = = =

1 bar = 100 kPa 30°C 70% = 0.7 4.25 kPa

Pv φ = PV s Pv 4.25 Pv = 2.975 kPa

0.07 = Specific humidity,

ω = 0.622 × = 00.0191

2.975 Pv = 0.622 × 100 − 2.975 P − Pv

kg water vapour kg dry air

T2 : Solution Wet bulb depression at the inlet =

(t db − t wb )inlet

= (38 – 18)inlet = 20°C

Wet bulb depression at the outlet = (24 – 18) = 6°C Percentage change =


(∵ twb inlet = twb exit)

20 − 6 = 70% 20

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11

T3 : Solution Inlet bulb depression at inlet = DBT1 – WBT1 Range = Tw 1 – Tw 2 Approach = Tw 2 – WBT1 Sum of range and approach = Tw 1 – Tw 2 + Tw 2 – WBT1 It is said wet bulb depression at inlet is equal to seem of approach and range ∴

DBT1 – WBT1 = Tw 1 – Tw 2 + Tw 2 – WBT1 DBT1 = Tw 1

∴

DBT1 – Tw 1 = 0

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RAC_TYS Exp._923

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