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14-1. Either graphically or by using the computer, for an ambient temperature of 30 C develop the performance characteristics of a condensing unit (of the form of Fig. 14-6 or Table 14-3) if the compressor has performance shown by Fig. 14-1 [ or Eq. (14-1) and (14-2)] and the condenser has characteristics shown by Fig. 14-3 [ or Eq. (14-4)].
Solution: Use mathematical computation: Use Fig. 14-3 or Eq. 14- qc = (9.39 kW/K)(tc - tamb) at 30 C qc = (9.39 kW/K)(tc -30)
Range of Evaporator Temperature, Fig. 14-1. -10 C, -5 C, 0 C, 5 C, and 10 C.
Eq. (14-1), constant at Table 14-1, Fig. 14-1. 2 c
2 9 e
2 c 8 ec
2 6 ec 7 e
2 4 c 5 c
2 q (^) e =c 1 +c 2 te+c 3 te +ct +ct +ct t +ct t +ct t +ct t
Eq. (14-2) constant at Table 14-1, Fig. 14-1. 2 c
2 9 e
2 c 8 ec
2 6 ec 7 e
2 4 c 5 c
2 P =d 1 +d 2 te+d 3 te +dt +dt +d tt +dt t +dt t +dt t
Eq. (14-3) qc = qe + P
Solving for tc at te = -10 C
q e =137.402+4.60437 ( − 10 ) +0.061652( − 10 ) 2 - 1.118157tc−0.001525tc^2
−0.0109119 ( − 10 )t c −0.00040148(− 10 ) 2 tc−0.00026682( − 10 )t c^2
+0.00000387 3 ( − 10 ) 2 tc^2
2 q (^) e =97.5235−1.049186tc+0.0015305tc
P =1.00618−0.893222 ( − 10 ) −0.01426( − 10 ) 2 +0.870024tc-0.0063397tc^2
+0.033889 ( - 10 ) tc −0.00023875( − 10 ) 2 tc−0.00014746( − 10 )t c^2
+0.00000679 62 ( − 10 ) 2 tc^2
2 P =8.5124+0.507259tc −0.00418548tc
qc = qe + P
2 q (^) c =106.0359-0.541927tc−0.00265498tc
Then, qc = (9.39 kW/K)(tc -30)
2 9.39t (^) c - 281.7=106.0359-0.541927tc−0.00265498tc
0.00265498t (^) c 2 +9.931927tc−387.7359= 0
tc = 38.64 C
q e =97.5235−1.049186 ( 38.64) +0.0015305( 38.64)^2
qe = 59.3 kW at te = -10 C.
Solving for tc at te = -5 C
q e = 137.402+4.60437 ( − 5 ) +0.061652(− 5 ) 2 - 1.118157tc−0.001525tc^2
−0.0109119 ( − 5 )t c −0.00040148( − 5 ) 2 tc−0.00026682( − 5 )t c^2
+0.00000387 3 ( − 5 ) 2 tc^2
2 q (^) e =115.92145−1.0736345tc+0.000094075tc
P = 1.00618−0.893222 ( − 5 ) −0.01426( − 5 ) 2 +0.870024tc-0.0063397tc^2
+0.033889 ( - 5 )t c −0.00023875(− 5 ) 2 tc−0.00014746( − 5 ) tc^2
+0.00000679 62 (− 5 ) 2 tc^2
2 P =5.11579+0.69461025tc −0.005432495tc
qc = qe + P
2 q (^) c =121.03724-0.37902425tc−0.00552657tc
2 9.39t (^) c −281.7=121.03724-0.37902425tc−0.00552657tc
0.00552657t (^) c 2 +9.76902425tc−402.73724= 0
tc = 40.31 C
q e =115.92145−1.0736345 ( 40.31) +0.00000940 75 (40.31 )^2
qe = 72.5 kW at te = -5 C.
Solving for tc at te = 0 C
q e = 137.402+4.60437 ( 0 ) +0.061652( 0 ) 2 - 1.118157tc−0.001525tc^2
−0.0109119 ( 0 )t c −0.00040148( 0 ) 2 tc−0.00026682( 0 )t c^2
+0.00000387 3 ( 0 ) 2 tc^2
2 q (^) e =137.402−1.118157tc+0.0001525tc
P =1.00618−0.893222 ( 0 ) −0.01426( 0 ) 2 +0.870024tc-0.0063397tc^2
+0.033889 ( 0 ) tc −0.00023875( 0 ) 2 tc−0.00014746( 0 ) tc^2
+0.00000679 62 ( 0 ) 2 tc^2
2 P =1.00618+0.870024tc −0.0063397tc
qc = qe + P
P =1.00618−0.893222 ( 10 ) −0.01426( 10 ) 2 +0.870024tc-0.0063397tc^2
+0.033889 ( 10 )t c −0.00023875( 10 ) 2 tc−0.00014746( 10 )t c^2
+0.00000679 62 ( 10 ) 2 tc^2
2 P =−3.81643+1.03350025tc −0.006907095tc
qc = qe + P
2 q (^) c =180.25886-0.082435tc−0.01094058tc 2 9.39t (^) c −281.7=180.25886-0.082435tc−0.01094058tc
0.01094058t (^) c 2 +9.472435tc−461.95886= 0
tc = 46.29 C
q e =189.6109−1.267424 ( 46.29) +0.0038059( 46.29)^2
qe = 122.8 kW at te = 10 C.
Ans.
qe, kw 122.8 104.4 87.6 72.5 59. te, C 10 5 0 -5 - tc, C 46.29 44.14 42.14 40.31 38.
14-2. Combine the condensing unit of Problem 14-1 (using answers provided) with the evaporator of Fig. 14-8 to form a complete system. The water flow rate to the evaporator is 2 kg/s, and the temperature of water to be chilled is 10 C. (a) What are the refrigerating capacity and power requirement of this system? (b) This system pumps heat between 10 C and an ambient temperature of 30 C, which is the same temperature difference as from 15 to 35 C, for which information is available in Table 14-4. Explain why the refrigerating capacity and power requirement are less at the lower temperature level.
Solution:
(a) Eq. 14-6.
q e =6.0 1 +0.046 (t wi−te)
twi = 10 C Expressing qe = f(te) from Problem 14-1. 2 q (^) e =87.5914+3.178te+0.03457te Then:
q e =6.0[ 1 +0.046 ( 10 −te)]( 10 −te)
q e = ( 60 −6te)(1.46 −0.046te)
2 q (^) e =87.6−11.52te+0.276te 2 e e
2 87.6 −11.52te +0.276te =87.5914+3.178t +0.03457t
0.24143t (^) e 2 −14.698te+0.0086= 0 t (^) e ≈ 0 C,tc=42.14 C Then, qe = 87.6 kw - - - Ans.
2 P =1.00618+0.870024tc −0.0063397tc
P =1.00618+0.870024 ( 42.14) −0.0063397( 42.14)^2
P = 26.4 kw - - - Ans. qc = qe + p = 87.6 kw + 26.4 kw = 114 kw
(b) At lower temperature level, if twi = 15 C and ambient temperature= 35 C. From Fig. 14-9. 15 C Entering Water Te.mperature 35 C Ambient Temperature
te = Evaporator Temp = 4.4 C qe = Refrigerating Capacity = 96 kw
Table 14-3. P = 30 kw tc = 48.4 C qc = 125.8 kw
Answer. All values above are higher than low temperature level. Therefore refrigerating capacity and power are less at low temperature level due to lower ambient temperature and lower entering water temperature to be chilled.
14-3. Section 14-11 suggests that the influences of the several components shown in Table 14-6 are dependent upon the relative sizes of the components at the base condition. If the base system is the same as that tabulated in Table 14-6 except that the condenser is twice as large [ F = 18.78 kW/K in Eq. (14-4)], what is the increase in system capacity of a 10 percent increase in condenser capacity above this new base condition? The ambient temperature is 35 C, and the entering temperature of the water to be chilled is 15 C.
Solution: 35 C ambient temperature, tamb. 15 C entering temperature of water, twi.
Eq. 14-4.
q c =F ( tc-tamb)
q c =18.78 ( tc- 35 )
Eq. 14-
q e =6.0 1 +0.046 (t wi−te)
q e =6.0[ 1 +0.046 ( 15 −te)]( 15 −te)
Eq. 14-1.
2 c
2 e
2 c ec
2 ec e
2 c c
2 e e e 0.0109119tt 0.00040148t t 0.00026682tt 0.000003873t t
q 137.402 4.60437t 0.061652t 1.118157t 0.001525t
− − − +
Eq. 14-2.
2 c
2 e
2 c ec
2 ec e
2 c c
2 e e 0.033889tt 0.00023875t t 0.00014746t t 0.0000067962t t
P 1.00618 0.893222t 0.01426t 0.870024t 0.0063397t
Eq. 14-3. qc = qe + P
0.00966937t (^) c 2 +18.92925325tc−815.44872= 0 tc = 42.17 C
Equation B.
X 18.88437 5 0.214348 5 1.11815742.17 0.00152542.
2 2
2 2
2 2
Say te = 4 C
Equation A.
0.00001066 92 ( ) 4 t 0
0.0229771 4 t 0.00064023 4 t 0.00041428 4 t
795.70818 3.711148 4 0.047392 4 19.028133t 0.0078647t
2 c
2
2 c c
2 c
2 c c
2
0.0093511128t (^) c 2 +18.94646828tc−811.311044= 0 tc = 41.95 C
Equation B.
X 18.88437 4 0.214348 4 1.11815741.95 0.00152541.
2 2
2 2
2 2
Say te = 3.5 C
Equation A.
0.00001066 92 (3.5 ) t 0
0.02297713.5t 0.000640233.5 t 0.000414283.5t
795.70818 3.7111483.5 0.0473923.5 19.028133t 0.0078647t
2 c
2
2 c c
2 c
2 c c
2
0.0091839823t (^) c 2 +18.95555597tc−809.27775= 0 tc = 41.845 C
Equation B.
X 18.884373.5 0.2143483.5 1.11815741.845 0.00152541.
2 2 2
2
2 2
Say te = 3.75 C
Equation A.
0.00001066 92 ( 3.75) t 0
0.02297713.75t 0.000640233.75 t 0.000414283.75t
795.70818 3.7111483.75 0.0473923.75 19.028133t 0.0078647t
2 c
2
2 c c
2 c
2 c c
2
0.009268214375t (^) c 2 +18.95097211tc−810.291435= 0 tc = 41.90 C
Equation B.
X 18.884373.75 0.2143483.75 1.11815741.90 0.00152541.
2 2 2
2
2 2
Therefore: te = 3.75 C and tc = 41.90 C
( )( ) 2 ( ) (^2 )^2
2 2
2 e
q 137.402 4.604373.75 0.0616523.75 1.11815741.
qe = 102.4 kW
or
qe= 6.0[ 1 +0.046 ( 15 − 3. 75 )]( 15 − 3. 75 )
qe = 102.4 kW
( )( ) 2 ( ) (^2 )^2
2 2
2
P 1.00618 0.8932223.75 0.014263.75 0.87002441.
P = 27.3 kW
qc = qe + P = 102.4 kW + 27.3 kW = 129.7 kW
or qc = 18.78 (tc - 35) = 18.78 (41.90 - 35) = 129.6 kW
New Base Conditions: Compressor = 27.3 kw Condenser = 129.7 kw Evaporator = 102.4 kw
If condenser capacity is increased by 10 % F = 18.78 x 1.1 = 20.
Equation A:
or
qe =6.0[ 1 +0.046 ( 15 −3.70)]( 15 −3.70)
qe = 103.04 kW
( )( ) 2 ( ) (^2 )^2
2 2
2
P 1.00618 0.8932223.7 0.014263.7 0.87002441.
P = 27.0 kW
qc = qe + P = 103 kW + 27 kW = 130 kW
or qc = 20.658 (tc - 35) = 20.658 (41.29 - 35) = 130 kW
102.4kw
103.04kw 102.4kw Increase insystemcapacity ×
Increase in system capacity = 0.62 % - - - Ans.
14-4. For the components of the complete system described in Secs. 14-7, 14-8, and 14-11 the following costs (or savings) are applicable to a 1 percent change in component capacity. An optimization is now to proceed by increasing or decreasing sizes of components in order to reduce the first cost of the system. What relative changes in components sizes should be made in order to reduce the first cost of the system but maintain a fixed refrigerating capacity?
Increase (saving) in first cost Component for 1 % increase (decrease) in component capacity
Compressor $ 2. Condenser 0. Evaporator 1.
Solution:
Tabulation of increase and decrease.
Compressor Condenser Evaporator Total Increase/ Reduction -2.80 +0.67 +1.40 -0. -2.80 -0.67 +1.40 -2. -2.80 +0.67 -1.40 -3. -2.80 -0.67 -1.40 -4. +2.80 +0.67 +1.40 +4. +2.80 -0.67 +1.40 +3. +2.80 +0.67 -1.40 +2. +2.80 -0.67 -1.40 +0.
The compressor should be increased to avoid freezing of water. So try evaporation reduced by 3 % or 2 %.
Compressor Condenser Evaporator Total Increase/ Reduction
Answer: Therefore use 3% evaporator capacity decrease for every 1 % increase in compressor capacity
