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Transcription:

基礎からの冷凍空調 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/067311 このサンプルページの内容は, 初版 1 刷発行当時のものです.

http://www.morikita.co.jp/support. 03-3817-5670FAX 03-3815-8199

i () () Q&A

ii 2007 2

iii 1 1 2 1 3 2 4 4 5 5 1.1 8 1.2 10 1.2.1 10 1.2.2 15 1.2.3 19 1.3 26 1.3.1 26 1.3.2 26 1.3.3 29 1.3.4 29 1.4 32 1.4.1 32 1.4.2 34 1.4.3 39 1.4.4 41 42 2.1 45 2.1.1 45 2.1.2 56 2.1.3 64 2.1.4 69 2.1.5 78 2.2 82 2.2.1 83 2.2.2 87 2.2.3 92 2.2.4 93 2.3 94

iv 2.3.1 94 2.3.2 96 2.4 99 2.4.1 99 2.4.2 99 2.4.3 101 2.4.4 102 103 3.1 104 3.1.1 104 3.1.2 105 3.1.3 111 3.2 117 3.2.1 117 3.2.2 122 3.2.3 125 3.2.4 127 3.3 128 3.3.1 128 3.3.2 131 3.3.3 133 3.4 136 3.4.1 136 3.4.2 140 144 4.1 145 4.1.1 145 4.1.2 147 4.1.3 151 4.2 154 4.2.1 154 4.2.2 156 4.2.3 160 4.2.4 161 4.2.5 162 4.3 164 4.3.1 164 4.3.2 165 4.3.3 168 4.3.4 172 4.4 175 4.4.1 175 4.4.2 176

v 179 180 181 185 188

8 1.1 (a) 1.1 (b) (refrigerating machine) (heat pump) (refrigeration cycle) (refrigerant) (vapor compression) 1.2 1.3

1.1 9 W [J/s] H 2 =H 1 +mgh=h 1 +W [J/s] H 1 [J/s] h [m] (a) W [J/s] Q 2 =Q 1 +W [J/s] Q 1 [J/s] DT ( b ) 1.1 1.4 (Stirling cycle) 1.1 (b) (coefficient of performance) COP 1.1 (b) COPε R = Q 1 W COPε H = Q 2 W (1.1) (1.2) Q 2 = Q 1 + W (1.3) COP

10 1 ε H = Q 1 + W W =1+ε R (1.4) ε R 34 ε H 5 (Rtrefrigeration ton) 24 0 C 1 0 C 1 333.6 kj/kg 1 = 333.6 1000 24 3600 =3.861 kw (1.5) lb () BTU () (1 =3.517 kw) 1.2 (compressor) (condenser) (expansion valve) (evaporator) p h 1.2

1.2 11 (Mollier diagram) 1.3 1.2 q H p H p p L w h 3 =h 4 q L h 1 h 2 h 1.3 1 2 p L ( 1) p H (>p L ) ( 2) 2 3 p H ( 3) 3 4 p L ( 4) - (Joule-Thomson effect) 4 1 p L ( 1) -

12 1 w w = h 2 h 1 (1.6) q H q L q H = h 2 h 3 (1.7) q L = h 1 h 4 (1.8) q L [J/kg] COP ε R = q L w = h 1 h 4 h 2 h 1 (1.9) ε H = q H w = h 2 h 3 h 2 h 1 (1.10) COP T H T L (1.6)(1.8) G [kg/s] (W = w G) (Q L = q L G) R134a 55 C 10 C 0.05 kg/s COP 1.4 10 C h 1 = 393 kj/kg 55 C 1.49 MPa 1 1.49 MPa h 2 = 435 kj/kg 1.49 MPa h 3 = 279 kj/kg Q L [kj/s] W [kj/s]cop Q L = G(h 1 h 3)=5.7 kj/s W = G(h 2 h 1)=2.1 kj/s ε R = QL W =2.71

1.2 13 R 134a 55 MPa] -10 J/ kg] 1.4 R134a [71] (180 ) µ (Joule-Thomson coefficient) ( ) dt µ (1.11) dp h µ =0 1.5 µ>0 (subcool) COP (superheat)

14 1 1.6 COP (1.6)(1.10) h 3 h 3, h 1 h 1 (1.12) 700 600 T [K] 500 400 300 200 100 0 1.5 20 40 60 80 p [MPa] p p H ' p L ' h 3 ' =h4 h h 1 ' h 2 1.6

1.2 15 Q1.1 5 C COP 5 C 50 C 1.4 1 5 C h 1 = 397 kj/kg Q1.1 h 2 = 440 kj/kg 3 50 C h 3 = 272 kj/kg COP ε R = h 1 h 3 397 272 = h 2 h 1 440 397 =2.91 1.1 1.1 [MPa] [ C] R22 R134a R407C R410A 30 1.192 0.770 1.356 1.885 1.166 40 1.534 1.017 1.744 2.419 1.553 50 1.943 1.318 2.210 3.061 2.031 1.7 (a) COP 1.7 (b) (cascade) (two-stage cascade refrigeration system)

16 1 (a) (b) 1.7 1.8 1.9 COP 1.8 1.9 w = w 12 + w 56 =(h 2 h 1 )+(h 6 h 5 ) [kg/s] G H G L [J/s] W

1.2 17 1.8 W = G H w 12 + G L w 56 = G H (h 2 h 1 )+G L (h 6 h 5 ) (1.13) 1.9 q L [J/kg] Q L [J/s] Q L = G L (h 5 h 8 ) (1.14) Q H [J/s] Q m [J/s] Q H = G H (h 2 h 3 ) (1.15) Q m = G L (h 6 h 7 )=G H (h 1 h 4 ) (1.16) COP ε R = Q L W = Q L G H w 12 + G L w 56 = G L (h 5 h 8 ) G H (h 2 h 1 )+G L (h 6 h 5 ) (1.17) COP ε RH ε RL ε RH = Q m G H w 12 (1.18)

18 1 q H p 2 p G H p 1 w 12 h 3 =h 4 h 1 h 2 h p 6 q m p G L p 5 w 56 q L h 7 =h 8 h 5 h 6 h 1.9 ε RL = Q L G L w 56 (1.19) (1.18)(1.19) (1.17) ε R = Q L ε RH ε RL = (1.20) Q m /ε RH + Q L /ε RL ε RH +(Q m /Q L )ε RL

1.2 19 Q m Q L = Q L + G L w 56 Q L =1+ 1 ε RL (1.21) (1.21) (1.20) COP COP ε R = ε RH ε RL ε RH + ε RL +1 (1.22) (1.16) G H = h 6 h 7 (1.23) G L h 1 h 4 80 C 1995 R22 R13 R22/R23 [1] (mixing chamber) (two-stage compression refrigeration system) 1.10 ( 4) (flash chamber) ( 5) ( 6) ( 7) ( 8) ( 9) ( 5) ( 1) p V 1.11 p m [Pa] 1 2 w [J/kg]

104 (room air conditioner) 1 2 (air conditioning) 3.1 3.1

3.1 105 21% 3.1 (humid air) (dry air) 3.1 N 2 O 2 Ar CO 2 [%] 78.09 20.95 0.93 0.03 [%] 75.53 23.14 1.28 0.05

106 3 ( ) p p a p v (Dalton) p = p a + p v (3.1) 3.2 (saturated air) p s 3.2 p v p s (unsaturated air) p p p s p v p a p v p s 3.2 (relative humidity) φ p v p s % φ = p v p s (3.2) p v p s p v = p s φ =1 100% 3.3 (absolute humidity) 1kg

3.1 107 3.2 (1 mbar = 100 Pa) t [ C] p s [mmhg] p s [mbar] h [kj/kg ] c p [kj/(kg K)] c v [kj/kg K] x [g/kg ] 30 0.38 0.509 29.386 1.0056 0.7722 0.312 28 0.46 0.613 27.218 1.0057 0.7723 0.377 26 0.55 0.737 25.020 1.0058 0.7724 0.453 24 0.66 0.882 22.789 1.0060 0.7725 0.542 22 0.79 1.053 20.518 1.0062 0.7727 0.647 20 0.94 1.254 18.202 1.0064 0.7729 0.771 18 1.12 1.487 15.834 1.0067 0.7731 0.914 16 1.32 1.759 13.407 1.0070 0.7733 1.082 14 1.56 2.075 10.912 1.0074 0.7736 1.276 12 1.83 2.440 8.339 1.0078 0.7739 1.502 10 2.15 2.862 5.678 1.0083 0.7742 1.762 8 2.51 3.347 2.916 1.0088 0.7746 2.062 6 2.93 3.905 0.040 1.0094 0.7751 2.407 4 3.41 4.544 2.964 1.0102 0.7757 2.802 2 3.96 5.274 6.115 1.0110 0.7763 3.255 0 4.58 6.107 9.428 1.0120 0.7770 3.771 2 5.29 7.053 12.927 1.0130 0.7778 4.360 4 6.10 8.128 16.632 1.0143 0.7787 5.030 6 7.01 9.345 20.569 1.0157 0.7798 5.790 8 8.04 10.720 24.766 1.0173 0.7810 6.651 10 9.20 12.270 29.252 1.0191 0.7823 7.625 12 10.51 14.015 34.063 1.0211 0.7839 8.724 14 11.98 15.974 39.236 1.0234 0.7856 9.963 16 13.63 18.170 44.810 1.0260 0.7875 11.358 18 15.47 20.627 50.833 1.0289 0.7897 12.925 20 17.53 23.370 57.354 1.0321 0.7921 14.685 22 19.82 26.427 64.429 1.0357 0.7948 16.657 24 22.37 29.828 72.120 1.0398 0.7979 18.866 26 25.21 33.605 80.495 1.0444 0.8013 21.336 28 28.35 37.792 89.630 1.0495 0.8051 24.098 30 31.82 42.426 99.610 1.0552 0.8093 27.182 32 35.66 47.546 110.529 1.0615 0.8141 30.624 34 39.90 53.195 122.493 1.0686 0.8194 34.464 36 44.57 59.417 135.620 1.0765 0.8253 38.746 38 49.70 66.259 150.042 1.0853 0.8319 43.520 40 55.33 73.771 165.911 1.0952 0.8393 48.842 42 61.51 82.009 183.396 1.1061 0.8475 54.776 44 68.28 91.027 202.691 1.1183 0.8566 61.394 46 75.67 100.887 224.018 1.1320 0.8668 68.779 48 83.75 111.651 247.632 1.1472 0.8782 77.027 50 92.55 123.387 273.825 1.1642 0.8909 86.246

108 3 (1+x) kg x kg kg m v m a m p p v p a 3.3 x [kg/kg ] m a v a m v v v x x = m v = v a (3.3) m a v v ( ) (dry-bulb temperature) t [ C] (wet-bulb temperature) t [ C] t t ( ) (dew point temperature) (perfect gas ideal gas) pv = mrt (Dalton) p = p a + p v

3.1 109 ( ) (a) T p V 3.3 p a p v m a m v p a V = m a R a T (3.4) p v V = m v R v T (3.5) R a R v R a = 287.2 J/(kg K)R v = 461.6 J/(kg K) (3.3) (3.4)(3.5) x x = m v m a = p vr a p a R v =0.622 p v p a (3.6) (3.2) (3.1) x φ x =0.622 p v =0.622 φp s (3.7) p p v p φp s 3.2 (b) V R (3.8) (3.9) (3.10) pv = mrt (3.8) m = m a + m v (3.9) mr = m a R a + m v R v (3.10) v 1kg (3.8) (3.10) (3.3) v (3.11) v = V m a = (m a + m v )RT m a p = (R a + xr v )T p = 461.6(0.622 + x)t p (3.11)

110 3 (c) 0 C 1kg (1 + x) [kg] h [kj/kg ] h = c pa t +(h L + c pv t)x =1.005t + (2500 + 1.846t)x (3.12) c pa (= 1.005 kj/(kg K))h L 0 C (= 2500 kj/(kg K))c pv (= 1.846 kj/(kg K)) 1kg (3.11)(3.12) 1kg (1 + x) [kg] 1kg ( (1 + x) [kg] ) (3.11) 1kg (1 + x) [kg] 1kg (3.11) (1 + x) [kg] 1kg v n (3.11) (1 + x) v n = 461.6(0.622 + x)t p(1 + x) (3.13) (3.13) [m 3 /kg] (3.11) (1 + x) [kg] (1 + x) [kg] [m 3 /kg ] (3.12) (1 + x) [kg] [kj/kg ] 28 C 60% 100% 28 C 3.2 p s =37.792 mbar 1 bar=10 5 Pa p s =37.792 10 5 /1000 = 3779.2 Pa 1.01325 10 5 Pa φ = 60% = 0.6 φ = 100% = 1.0 (3.7) φ = 60% x =0.622 (0.6 3779.2/(1.01325 10 5 0.6 3779.2))

145 2 3 ) ) 4.1 (a) T 1 [K, T 1 (a) (b) A (c) (d) T 2 Q x=0 x=l 4.1

146 4 C] L [m] (A [m 2 ]) (b) T 2 [K, C] T 1 T 2 (b) (c) (d) (heat conduction) x Q [W] x dt/dx [K/m] Q = Aλ dt dx (4.1) A () (4.1) (Fourier s law) dt/dx () () λ [W/(m K)] (thermal conductivity) 4.1 4.1 (4.2) (x, y, z) cρ T t = ( λ T ) + ( λ T ) + ( λ T ) (4.2) x x y y z z () λ

4.1 147 4.1 (300K) [66] λ [W/m K] 398 237 35.2 (273 K) 2.210 1.38 0.61 0.144 0.0261 (4.2) cρ T ( 2 ) t = λ T x 2 + 2 T y 2 + 2 T z 2 (r, φ, z) cρ T [ ( 1 t = λ r T ) + 1 2 ] T r r r r 2 φ 2 + 2 T z 2 (4.3) (4.4) (4.3) (4.4) 4.1 () 2 ( ) 4.1 (d) (4.3) (4.3)

148 4 d 2 T =0 (4.5) dx2 (4.5) T = C 1 x + C 2 (4.6) C 1 C 2 4.1 x =0T = T 1 x = L T = T 2 (4.6) T = T 1 + x L (T 2 T 1 ) (4.7) (4.1) (4.7) Q = Aλ T 2 T 1 L (4.8) (4.7) (4.8) (4.8) (T 1 T 2 ) 25 mm 0.65 m 2 (λ =0.21 W/(m K)) 34 C 2 C (4.8) T2 T1 Q = Aλ = 0.65 0.21 2 34 L 0.025 = 174.7 W (4.7) x =12.5 mm T = T 1 + x (T2 T1) =34+0.0125 L 0.025 (2 34) = 18 C 4.2 Q

4.1 149 Q = Aλ 1 T 1 T 2 L 1 = Aλ 2 T 2 T 3 L 2 = Aλ 3 T 3 T 4 L 3 = = Aλ n T n T n+1 L n (4.9) L k /λ k (k =1, 2, n) 4.2 A(T 1 T n+1 ) Q = L 1 + L 2 + L 3 + + L (4.10) n λ 1 λ 2 λ 3 λ n L 1 L 2 L 3 l 1 l 2 l 3 L n l n T 1 T L k l k T 3 T l l l l n T n L 1 L 2 L 3 L n T n+1 4.2 ( ) (4.4) (4.4) ( 1 d r dt ) = 0 (4.11) r dr dr

150 4 T = C 1 ln r + C 2 (4.12) (4.12) (4.7) 4.3 r = r 1 T = T 1 r = r 2 T = T 2 C 1 C 2 T T 1 = ln(r/r 1) T 2 T 1 ln(r 2 /r 1 ) (4.13) L Q Q = 2πrLλ dt dr =2πLλ T 1 T 2 (4.14) ln(r 2 /r 1 ) T 1 T 2 Q =2πLλ 1 ln(r 2 /r 1 ) =2πLλ T 2 T 3 2 ln(r 3 /r 2 ) = =2πLλ T n T n+1 n ln(r n+1 /r n ) (4.15) Q = 2πL(T 1 T n+1 ) ln(r 2 /r 1 ) + ln(r 3/r 2 ) + + ln(r n+1/r n ) λ 1 λ 2 λ n (4.10) (4.16) T 1 T 2 r 1 r L l T 1 T 2 r2 a 4.3

4.1 151 30 mm 8.0 mm 0.15 W/(m K) 15 mm 0.03 W/(m K) 25 C 200 C 2 (4.16) L =1mT 1 = 200 CT 3 =25 C Q = ln(r 2/r 1) λ 1 =86.8 W 2πL(T 1 T 3) + ln(r3/r2) λ 2 = ln(0.038/0.030) 0.15 2π(200 25) + ln(0.053/0.038) 0.03 86.8 W (4.15) T 2 = T 1 Q ln(r 2/r 1) 2πL = 200 86.8 λ 1 2π T 2 = T 3 + Q ln(r 3/r 2) 2πL =25+ 86.8 λ 2 2π ln(0.038/0.030) 0.15 ln(0.053/0.038) 0.03 = 178.2 C = 178.2 C T T w A [m 2 ] Q T T w T >T w Q = Aα(T T w ) (4.17) α (heat transfer coefficient) [W/(m 2 K)] 4.4 T in >T 1 >T 2 >T out

152 4 T in Q a in T 1 T 2 l L Tout a out 4.4 α in Q 1 = Aα in (T in T 1 ) (4.18) (4.17) (4.8) α out Q 2 = Aα out (T 2 T out ) (4.19) 4.4 Q = Aα in (T in T 1 )= Aλ (T 2 T 1 ) = Aα out (T 2 T out ) (4.20) L (4.20) T 1 T 2 Q = A(T in T out ) 1 + L α in λ + 1 (4.21) α out T 1 T 2 (4.21) (4.20) 1/α L/λ (4.10) 1 K = 1 + L α in λ + 1 (4.22) α out (4.22) K [W/(m 2 K)] A

4.1 153 Q = AK(T in T out ) (4.23) (n ) (4.22) 1 K = 1 + L 1 + L 2 + + L n + 1 (4.24) α in λ 1 λ 2 λ n α out 4.4 150 mm (1.2 W/(m K)) (1 m 2 ) 25 C 6W/(m 2 K) 15 C 20 W/(m 2 K) (4.21) Q = A(Tin Tout) 1 + L α in λ + 1 = α out 25 ( 15) 1 6 + 0.15 1.2 + 1 20 (4.20) T 1 = T in Q =25 117.1 =5.48 C α in 6 T 2 = T out + Q = 15 + 117.1 = 9.15 C α out 20 = 117.1 W Q (4.14) T in >T out Q 2πL(T in T out ) Q = 1 + 1 α in r 1 λ ln r 2 + 1 (4.25) r 1 α out r 2 2πL(T in T out ) Q = 1 + 1 ln r 2 + 1 ln r 3 + + 1 ln r n+1 1 + α in r 1 λ 1 r 1 λ 2 r 2 λ n r n α out r n+1 (4.26)

154 4 (convection heat transfer) () 4.5 u T y u u u T T d u d T T x T w T w x c 4.5 (velocity boundary layer) δ u 99% x c x (0.5ρu 2 ) (µu/x) (Reynolds number) Re x = u x/ν Re xc = u x c ν =(3 5) 10 5 (4.27) x c T T w

4.2 155 (x, y, z ) 4.5 u x + v y = 0 (4.28) ρu u u + ρv x y = u µ 2 y 2 (4.29) ρc p u T x + ρc pv T y = λ 2 T y 2 (4.30) Q 1 x Q 1 (x) ( ) T Q 1 (x) = Aλ (4.31) y y=0 x α x [W/(m 2 K)]() Q 1 (x) =Aα x (T w T ) T w >T (4.32)

156 4 (4.31) (4.32) ( ) T 1 α x = λ (4.33) y y=0 T w T ( ) 4.6 x y u=ay 3 +by 2 +cy+d b u c d d T T T=ay 3 +by 2 +cy+d a d x T 4.6 y δ u y (y =0) (u =0) (y = δ) (u = u )

4.2 157 u = 3 y u 2 δ 1 ( y ) 3 (4.34) 2 δ (a-b-c-d) d δ u(u u)dy + du δ ( ) u (u u)dy = ν (4.35) dx dx y 0 0 (4.34) (4.35) x δ νx x δ =4.64 =4.64 (4.36) u Rex Re x = u x/ν (4.36) y T T w = 3 y 1 ( ) 3 y (4.37) T T w 2 δ T 2 δ T δ T d δt ( ) T (T T )udy = a (4.38) dx 0 y y=0 a (= λ/(ρc p )) [m 2 /s] (4.35) (4.38) (Pr = ν/a) (Prandtl number) 4.2 1 (4.34) (4.37) (4.38) δ =1.026Pr 1 3 (4.39) δ T y=0

188 COP 9 GWP 137 ODP 137 30 77 10 49 75 74 45 136 137 137 169 70 45 13 13 105 108 49 19 30 100 72 82 82 74 10 104 57 163 99 26 19 28 47 34 167 60 67 46 162 139 108 105 112 76 36 11, 99 8 1 74 10, 64 63 77 60 55 54 9 9 97 106

189 94 49 106 167 47 47 78 49 117 136 23 71 34 74 71 45 169 15 62 29 19 7 158 164 152, 166 7, 151 146 6, 146 40 26 26 41 34 46 2 2 47 8, 128 67 173 106 29 157 6, 146 62 67 27 94 1 10, 70 106 159 46 4, 11 73 1 8 12 8 10 10 154 8 137 117 76 51 108

2007 JCLS Printed in Japan ISBN978-4-627-67311-3