传热学(习题2-3)

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tw1 tw 2
tw1 tw 2
0.24 0.02 δ 3 0.24 0.02 + + + ( ) × 0.2 = , 0.7 0.58 0.06 0.7 0.58 δ 3 = 0.0905m = 90.5mm
西安建筑科技大学 Xi’an University of Architecture and Technology
ηf =
th(ml ) ηf = = 0.831 ml
也可查图 lc
32
2h 1 2 ( ) , f = lcδ , lc = l + δ 2 λf
西安建筑科技大学 Xi’an University of Architecture and Technology
高等传热学
2-28 解：每米共有1000 / s =
θ = 0.01θ0
= 0.01
τ = 14.43s
t = t f + 0.01(25 120) = 119.05
西安建筑科技大学 Xi’an University of Architecture and Technology
高等传热学
3-8
39 × 1.5 × 103 Bi = = = 1.21 × 103 < 0.1 48.5 λ
高等传热学
2-10
t w1 + t w2 解:t = = 250 2 λ = 0.094 + 0.000125 × 250 = 0.12525W /(m K ) (t w1 t w2 ) × λ (450 50) × 0.12525 δ= = = 0.147m q 340
dt 另解 : λ = q = const ,即 : λ dt = q dx dx
高等传热学
第三章非稳态导热 3-7 解：判别式 瞬态导热，首先判断能否用集总参数法 瞬态导热，
95 × 0.25 × 103 Bi = = = 5.97 × 105 < 0.1 λ 398 可用集总参数法。 可用集总参数法。 圆球定型尺寸 V R L= = ρ cV A 3 τ hR
hA
θ BiFo =e =e θ0
e ml + e ml e6.266 + e 6.266 ch ( ml ) = = = 263.2, 2 2 1 40 θl = θ0 = = 0.152, 得tl = 100 0.152°C = 99.8°C , ch( ml ) 263.2 t = tl t f = 0.152°C
西安建筑科技大学 Xi’an University of Architecture and Technology
高等传热学
∑ Rλ = Rλ
1
+ Rλ 2 + Rλ 3 = 0.797 (m K)/W
(tw1 tw 2 ) (2) : ql = = 313.8W / m; ∑ Rλ (3) : tw 2 = tw1 ql Rλ 1 = 299.9°C ≈ 300°C tw 3 = tw 4 + ql Rλ 3 = 137.6°C
θ 80 20 = =e θ0 34 20
ρ cV
hA
τ
=e
ρ cd
4h
τ
h = 83.2W/(m K)
2
1 2 V 4πd L 1 = = d A π dL 4
验证
Bi =
hR
λ
= 2.16 × 103 < 0.1
假设正确
西安建筑科技大学 Xi’an University of Architecture and Technology
西安建筑科技大学 Xi’an University of Architecture and Technology
高等传热学
2-18
解 : Rλ = ln d2 d1 = ln (1 + 2 × 0.5) 1 = 0.7358(m K/W),
2π × 0.15 25 (tmax ql = = = 33.97W/m = I 2 R R 0.7358 33.9 ql 2 I = = = 15300.43, I = 123.7A 3 R 2.22 × 10
d3
2πλ2 ln d4 2πλ3
=
170 = 0.518( m K / W ), 2π × 0.093 ln 310 230 = 0.279(m K / W ). 2π × 0.17
ln 230
d3
=
与另两个热阻 比较极小，可 忽略
西安建筑科技大学 Xi’an University of Architecture and Technology
Rλ 1 = 1 1 1 = + = 1.5559,即R = 0.643K / W ′ ′′ R Rλ 1 Rλ 2 × 2 + Rλ 2
λ2 = 0.48, 查得 = 0.96 λ1
修正
Rq = R × A × = 0.643 × (0.31 + 0.02) × 0.96 = 0.204( m 2 K ) / W
λ1
λ3
λ1
λ2
λ3
tw1 tw 2 tw1 tw 2 = 0.12 δ ′ 3 0.12 0.05 0.25 + + + 0.8 0.6 0.8 0.12 0.6 δ ′ 3 = 0.500m = 500mm
西安建筑科技大学 Xi’an University of Architecture and Technology
高等传热学
F肋 = 2π ( r2 c 2 r12 ) = 2.32 × 102 m 2
Φ 0 = hF肋 (t0 t f ) = 60 × 2.32 × 102 × (320 75) = 341.04W Φ 肋 = η f Φ 0 = 259.2W
(2) Φ 光 A光 = π d1 ( s δ ) = π × 0.077 × (0.025 0.004) = 0.00508m 2 Φ 光 = hF光 (t0 t f ) = 60 × 0.00508 × (320 75) = 71.64W
西安建筑科技大学 Xi’an University of Architecture and Technology
高等传热学
2-16.
解 : d 3 = d 2 + 2δ 2 = 170 + 60 = 230mm; d 4 = d 3 + 2δ 3 = 230 + 80 = 310mm (1) : Rλ 1 = Rλ 2 = Rλ 3 = ln ln d2 2πλ d2 d1 = ln 170 160 = 1.66 × 104 (m K / W ), 2π × 58
高等传热学
3-11
分析，瞬态非稳态导热，第三类边界条件 θ x ＝f ( Bi , Fo, ) θ0 δ
h → ∞, BiA = BiB → ∞
θ θ x x ( ) A = ( ) B， ( ) A = ( ) B＝0 θ0 θ0 δ δ
如何改进？ 如何改进？强化措施
西安建筑科技大学 Xi’an University of Architecture and Technology
高等传热学
2-26
解:m = hU 20 × π d 20 = = = 39.16, λA 16.3 × π d δ 16.3 × 0.0008 ml = 39.16 × 0.16 = 6.266,
高等传热学
2-15
Φ
1. 热流方向 2. 单元体 3. 单位面积
Φ1
Φ2
Φ
Φ q= A
Φ1 Φ2
R q = RΦ A
西安建筑科技大学 Xi’an University of Architecture and Technology
高等传热学
δ 0.2 = = 6.536 K / W λ1 A1 1.53 × 0.021 0.035 δ′ ′2 = Rλ = = 0.0738 K / W λ1 A2 1.53 × 0.311 0.13 δ ′′ ′′ Rλ 2 = = = 0.5652 K / W λ2 A2 0.742 × 0.311
2πλ tmin )
西安建筑科技大学 Xi’an University of Architecture and Technology
高等传热学
2-25
hU 20 × π d 解:m = = = 16.58, 48.5 × π d δ λA ml = 16.58 × 0.12 = 1.99, ch( ml ) = 3.76, θ0 1 θl = θ0 = ch( ml ) 3.76 t f = 99.94°C = 100°C , t = tl t f = 15.94°C
高等传热学
传热学
Heat Transfer
郭亚军
环境与市政工程学院热工流力教研室 境与市政工程学院热
Xi'an 710049 China
西安建筑科技大学 Xi’an University of Architecture and Technology
高等传热学
第二章 稳态导热 2-9
tw1 tw 2 解: q = = δ1 δ 2 0.24 0.02 + + λ1 λ2 0.7 0.58 tw1 tw 2 q = = δ1 δ 2 δ 3 0.24 0.02 δ 3 + + + + λ1 λ2 λ3 0.7 0.58 0.06
50
∫ (0.094 + 0.000125t )dt = ∫
450
δ
0
340dx
δ = 0.147m
西安建筑科技大学 Xi’an University of Architecture and Technology