LATO:
$t = \frac{Q_{\text{str}/j}}{\dot{m} \bullet c_{p}}$ $\dot{m} = V \bullet \rho\ \lbrack\text{kg}/s\rbrack$
$x = \frac{{\dot{m}}_{w}}{\dot{m}} = \frac{\dot{w}}{\dot{m}}$
N: tN = tP +/– ∆t
XN = xP – ∆x
hN = cp ∙ tN + xN ∙ (cpp ∙ tN + ro)
Z’: tZ’ = tZ – Φ ∙ (tZ – tP)
xZ’ = xZ = tZ – Φx ∙ (xP’ – xZ)
hZ’ = cp ∙ tZ’ + xZ’ ∙ (cpp ∙ tZ’ + ro)
1: x1 = xN; tD = 13 oC (wykres)
Qch = m’ ∙ (hZ’ – h1)
QNII = m’ ∙ (hN – h1)
$$x = 0,622 \bullet \frac{p_{w}"(t)}{p_{a} - p_{w}"(t)}\ \lbrack\frac{\text{kg}}{\text{kg}}\rbrack$$
ZIMA
N: jak lato
Z’: tZ’ = tZ + Φ ∙ (tZ – tP)
xZ’ = xZ
hZ’ = cp ∙ tZ’ + xZ’ ∙ (cpp ∙ tZ’ + ro)
2: x2 = xZ’; t2 = tN; h2
QNI = m’ ∙ (h2 – h Z’)
m'NP = m’ ∙ (xN – x2)
QNP = m’ ∙ (hN – h2)
M: tM = αM ∙ tZ + (1 – αM) ∙ tP
xM; hM. αM = Vo / (Vo + V2)
1: α1 = (Vo + V2) / V
tN = α1 ∙ t1 + (1 – α1) ∙ tP =>
t1 = tP – (tP – tN) / α1
x1 = xN = xP
Obiegi chłodnicze:
1’: x1’ = 1; h1’; s1’; υ1’
1: h1 = h1’ + qr
2: s2 = s1; p2 = pk
3: x3 = 0; p3 = pk
t3 = tk; | h3 υ3 wykres
4: h4 = h3 – qr ∙ ηRWC
p4 = pk | t4 υ4 wykres
5: h5 = h4; p5 = po
2’: s2’ = s1’; p2’ = pk
4’: h4’ = h3
∆tr = t3 – t4
∆tp = t1 – t1’
qo = h1’ – h5
ws = h2 – h1
qk = h2 – h3
qv = qo / υ1
mc = Qo / qo
Qk = mc ∙ qk
Ns = mc ∙ ws
ε = qo / ws
V1 = mc ∙ υ1
qo’ = h1’ – h4’
ws’ = h2’ – h1’
Wymiarowanie:
λ = 0,3164 / Re0,25 (Re<106)
λ = 0,0032 + (0,221 / Re0,237) (Re > 106)
Re = (w ∙ d) / ν
ν = μ/ρ = μ ∙ υ
m’ = Qk / (h2 – h3)
m’ = (Qo + Ns) / (h2 – h3)
m’ = Qo / (h1 – h4 )
tłoczny: 100% Qo:
V2 = m’ ∙ υ2 [m3/s]
100%: w = wmax = 18 m/s;
F = V2 / wśr = V2 / 18
Dw = √(4 ∙ F) / π| typoszer, obl. wrz
25%: w(25) = V2 (25) / F
F1 = V2 (25) / w(25)
Σ F = V2 (100) / wmax
F2 = Σ F – F1
Dw2 = √(4 ∙ F2) / π| typoszereg, obl. F2rz
∆p1 = λ ∙ [(L+Lz) / dw1] ∙ (w12/2) ∙ (1/υ2)
∆p1 = ∆p2 ; V = V1 + V2
w2 = (4 ∙ V) / [π ∙ (dw12 ∙ √(dw1/dw2)| + dw22)]
w1 = w2 ∙ √(dw1/dw2)|
Chłodnica: Vch = 4000 m3/h; t1/φ1 = 28oC/55%; x1 = 12,4 g/kg; h1 = 60 kJ/kg
t2/φ2 = 15oC/85%; x2 = 9,0 g/kg; h2 = 38 kJ/kg; tz/tp = 7/12oC
Klimak: Q1KK = 3,0 kW; n = 10 QKK = 30 kW; tz/tp = 7/12oC; ∆p = 5 kPa = 0,05 bar
Belki: QB = 40 kW; tz/tp = 14/17oC
R407C: to = 5oC; tK = 45oC; tD = 40oC; ∆tp = 15oC
Moc chłodnicza
Qch = m ∙ (h1 – h2) = 1,33 ∙ (60 – 38) = 29 kW
m = Vch ∙ ρ = 4000/3600 ∙ 1,2 = 1,33 kg/s
QA = Qch + QKK + QB = 29 + 30 + 40 = 99 kW
Parametry obiegu chłodniczego:
1: p1 = po = 0,52 MPa t1 = to + ∆t = 5+15= 20oC; h1= 428 kJ/kg;
s1= 1,84 kJ/kgK; υ1 = 0,050 m3/kg
2: s2 = s1= 1,84; p2 = pk = 2,0 MPa; h2= 464; υ2 = 0,014
3: p3 = pk= 2; t3 = tD = 40; h3= 260; υ3 = 0,00094; s3= 1,21
4: h4 = h3= 260 p4 = po= 0,52;
qo = h1 – h4 = 428 – 260 = 168 kJ/kg
ws = h2 – h1 = 464 – 260 = 204 kJ/kg
qk = h2 – h3 = 464 – 428 = 36 kJ/kg
mC = QA/qo = 99/168 = 0,59 kg/s
qv = qo / υ1 = 168/0,050 = 3360 kJ/m3
QK = mC ∙ qk = 0,59 ∙ 204 = 120 kW
Ns = mC ∙ ws = 0,59 ∙ 36 = 21 kW
Średnice rurociągów:
Cieczowy: zał: w = 0,4÷0,8 m/s
Vci = mC ∙ υ3 = 0,59 ∙ 0,00094 = 0,000555 m3/s
Vci = A ∙ w A = Vci/w;
A = (π∙d2)/4 d = √((4 ∙ A)/π)|
A = 0,000688 d= 0,03 35x1,5
Arz = 0,000804 wrz = Vci/Arz = 0,69 m/s
Tłoczny: zał: w = 12÷18 m/s (min 7)
100%: V100 = mC ∙ υ2 = 0,59 ∙ 0,014 = 0,00826 m3/s
66%: V66 = 0,00545 m3/s; 33%: V33 = 0,00273 m3/s
A = V100/wmax = 0,00826/18= 0,000459 m2;
d = 0,024 28x1,5 Arz = 0,000491m2;
Wrz = V100/Arz =16,82 m/s
33%: w = V33/Arz100 = 0,00273/0,000491 = 5,56
A33 = V33/wmax = 0,00273/18 = 0,000152 m2;
d33 = 0,013 18x1 A33rz = 0,000201 m2;
ΣA = V100/wmax = 0,000459 m2;
A2 = ΣA – A33rz = 0,000459 – 0,000201= 0,000258
d = 0,018 22x1 A2rz = 0,000314
ΣA = A33rz + A2rz = 0,000201 + 0,000314= 0,000515
Wrz = V100/ΣA = 17,7 m/s
Pompy: chłodnica: Qch = mch ∙ cw ∙ (tp – tz)
mch = Qch / cw ∙ (tp – tz) = 29 / 4,19 ∙ (12-7)=1,38 kg/s
Vch = mch / ρ = 1,38 / 1000 = 0,00138 m3/s = 4,97 m3/h
Kv: Kv = V / √∆pc| ∆pc = ∆p + ∆pv100
∆p = 5 kPa = 0,05 bar; ∆pv100 = [a/ (1 – a)] ∙ ∆ p
Kvch = 4,97 / √0,1 = 15,72