第6章作业答案6-21现代储蓄热能的一种装置的示意图如附图所示。一根内径d=25mm的圆管被置于一正方形截面的石蜡体中心，热水流过管内使石蜡熔解，从而把热水的显热转化成石蜡的潜热而储蓄起来。热水的入口温度为60℃，流量为0.15kg/s。石蜡的物性参数为：熔点为27.4℃，熔化潜热L=244kJ/kg，固体石蜡的密度ρs=770kg/m3。假设圆管表面温度在加热过程中一直处于石蜡的熔点，试计算把该单元中的石蜡全部熔化热水需流过多长时间？b=0.25m，l=3m。解：已知管壁温度等于石蜡熔点℃4.27w=t一个单元中石蜡全部融化需要热量kJ96.34950244770)30125.014.3325.0(22=××××−×=Q已知热水入口温度℃60f=′t，出口温度ft′′未知，但要求出加热所需时间需先知道出口水温ft′′，故先假设出口水温℃40f=′′t定性温度℃5024060m=+=t，自教材563页附录9查取水的物性参数：K)W/(m648.0⋅=λ，/sm10556.026−×=ν，54.3Pr=，3kg/m1.988=ρ，K)J/(kg4174⋅=pcm/s31.00125.014.31.98815.02=××=u461085.1393810556.0025.031.0Re>=××==−νud属于旺盛湍流冷却流体用迪－贝公式计算47.6954.385.13938023.0PrRe023.03.08.03.08.0=××==NuK)W/(m7.1800025.0648.047.692⋅=×==dNuhλ①按热平衡关系)()(wfffmtthAttcqp−=′′−′其中℃502fff=′′+′=ttt℃5.4323025.014.37.1800417415.0)304.27(3025.014.37.180060417415.02)2(mfwfmf=×××+×−××××+××=+′−+′=′′hAcqtthAtcqtpp与假设值相差%5%75.8%10040405.43>=×−=∆重新设℃5.43f=′′t进行计算定性温度℃75.5125.4360m=+=t，查得：K)W/(m65.0⋅=λ/sm10542.026−×=ν，44.3Pr=，3kg/m2.987=ρ，K)J/(kg9.4174⋅=pcm/s31.00125.014.32.98715.02=××=u461089.1429810542.0025.031.0Re>=××==−νud3.7044.389.14298023.03.08.0=××=NuK)W/(m8.1827025.065.03.702⋅=×==dNuhλ℃3.4323025.014.38.18279.417415.0)304.27(3025.014.38.1827609.417415.02)2(mfwfmf=×××+×−××××+××=+′−+′=′′hAcqtthAtcqtpp与假设值相差%46.0%1005.435.433.43−=×−=∆满足计算要求最后热水出口温度℃4.4323.435.43f=+=′′t对流传热量W10396)4.4360(9.417415.0)(ffm=−××=′′−′=Φ′ttcqpW10460)4.2724.4360(3025.014.38.1827)(wf=−+××××=−=Φ′′tthAW104282=Φ′′+Φ′=Φ需时min86.55s33521042834950960==②按热平衡关系wfwfwfwfmffmln)()()(tttttttthAthAttcqp−′′−′−′′−−′=∆=′′−′4.274.2760ln)4.27()4.2760(3025.014.37.1800)60(417415.0fff−′′−−′′−−××××=′′−××ttt方程式为隐式格式，用试算法计算见下表ft′′)(ffmttcqp′′−′=Φ′wfwfwfwfln)()(tttttttthA−′′−′−′′−−′=Φ′′4111895.99216.2444211269.89502.3724310643.79781.13743.510330.659917.98643.810142.829999.33243.9610042.6410042.49重新设℃44f=′′t进行计算定性温度℃5224460m=+=t，查得：K)W/(m6502.0⋅=λ/sm105404.026−×=ν，43.3Pr=，3kg/m1.987=ρ，K)J/(kg4175⋅=pcm/s31.00125.014.31.98715.02=××=u461023.14341105404.0025.031.0Re>=××==−νud4.7043.323.14341023.03.08.0=××=NuK)W/(m4.1828025.06502.03.702⋅=×==dNuhλ而wfwfwfwfmffmln)()()(tttttttthAthAttcqp−′′−′−′′−−′=∆=′′−′W10020)4460(417515.0=−××=Φ′W102084.27444.2760ln)4.2744()4.2760(3025.014.34.1828=−−−−−××××=Φ′′二者相差%88.1%100100201020810020−=×−=∆满足计算要求最后℃44f=′′tW101142=Φ′′+Φ′=Φ需时min6.57s34561011434950960==6-26在一摩托车引擎的壳体上有一条高2cm、长12cm的散热片（长度方向系与车身平行）。散热片表面温度为150℃。如果车子在20℃的环境中逆风前进，车速为30km/h，而风速为2m/s，试计算此时肋片的散热量（风速与车速相平行）。解：此题属于空气外掠平板问题车速m/s33.83600100030km/h301≈×==u风速m/s22=u故空气相对于散热片表面的合成流速m/s33.1021=+=uuu定性温度℃85220150m=+=t，查取空气物性表得K)W/(m1009.32⋅×=−λ、/...
