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hw02 Solution

This document contains solutions to homework problems involving refrigeration cycles. It includes calculations of various refrigeration cycle parameters such as COP, mass flow rate, power input, and refrigeration capacity. An additional problem estimates the daily electricity cost of operating 7 chillers on campus based on their refrigeration capacity and an assumed COP value and electricity pricing structure.

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shum kenneth
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111 views2 pages

hw02 Solution

This document contains solutions to homework problems involving refrigeration cycles. It includes calculations of various refrigeration cycle parameters such as COP, mass flow rate, power input, and refrigeration capacity. An additional problem estimates the daily electricity cost of operating 7 chillers on campus based on their refrigeration capacity and an assumed COP value and electricity pricing structure.

Uploaded by

shum kenneth
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOC, PDF, TXT or read online on Scribd
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Solutions for Home Work 2

15 – 1
i1 = 42.5 Btu / lbm = 98.86 KJ / Kg R22

i2 = i1
60 °C ( 140°F)
i3 = 108.2 Btu / lbm = 251.7 KJ / Kg
1 4
43 °C ( 110°F)
i4 = 119.5 Btu / lbm = 278.0 KJ / Kg

P
3
COP = qe /w; Use P–I diagram 2 4.5 °C ( 40°F)

-w = i4 – i3 = 119.5 – 108.2

=11.3 Btu/lbm

qc = i1 – i4 = 42.5 – 119.5
i
= –77Btu/lbm

qe = – qc + w = 77 – 11.3

= 65.7 Btu/lbm

(a) COP = 65.7 / 11.3 = 5.814

(b) COP carnot = Te / Tc – Te = (4.5+273.15)/(43–4.5) =7.21

ηr = COP / COP carnot = 5.814 / 7.21 = 0.806

(c) HP / ton = 4.716 / COP = 4.716 / 5.814 = 0.811

KW / KW = 1/ COP = 1 / 5.814 = 0.172

q e 10  12000
(d) m    1826.5 lbm / hr  0.2301 kg / s
qe 65.7

q 10  12000
(e) W  e   20640 Btu / hr  8.112 HP  6.049 kW
COP 5.814

(f) V3 = 0.04099 m3/kg = 0.6566 ft3/lb

PD  mV
 3  0.2301  0.04099  0.009432 m 3 / s  19.99 ft 3 / min
15 – 2

W  2.5kW ; m  0.05kg / s

q c  i1  i4 ;  w  i4  i3
i4  i3  W / m  398.4  2.5 / 0.05  448.4 kJ / kg
q c  i1  i4  260.3  448.4  188.1 kJ / kg
(a)
w  i3  i4  398.4  448.4  50 kJ / kg
q e  q c  w  188.1  50  138.1 kJ / kg
Qr  q c  m  188.1  0.05  9.405 kW

COP  138.1/ 50  2.76


(b) 255
COPcarnot   3.86
321  255

(c) i4  448.4 kJ / kg from part (a )

(d) r  2.76 / 3.86  0.72 or 72%

Additional Problem

Estimate the daily electricity cost ( in HK$) of 7 chillers at HKUST.  Refrigeration


capacity of each chiller is 3517 kW and assume COP=3.7.

Solution:

The power input for each chiller


         = Refrigeration capacity/ COP=3517/3.7=950.54 [kW]

Assume the chillers operated 12 a day and the unit electricity cost is as follows:
                First 5000 units (kWh) is  HK$0.974/unit
                Over 5000 units (kWh) is HK$0.964/unit

Then, the electricity cost of 7 chillers per day is:


       5000 x 0.974 + ( 950.54 x 7 x 12 - 5000) x 0.964
     = 77021 [HK$]
The daily electricity cost of 7 chillers at HKUST is about HK$77021.

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