混凝土温度计算

胶凝材料用量水化热总量混凝土比热混凝土质量绝热温升（kJ/kg）(kJ/kg.℃)密度(kg/m3)(℃)(kg/m3)3203203203203203203203203203203203203203203203203203203203203203203203203203203203203003003003003003003003003003003003003003003003003003003003003003003003003003003003000.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.960.96240024002400240024002400240024002400240024002400240024002400240024002400240024002400240024002400240024002400240027.4131.7034.7036.7938.2639.2840.0040.5040.8541.1041.2741.3941.4741.5341.5741.6041.6241.6341.6441.6541.6641.6641.6641.6641.6641.6641.6741.67降温系数0.7400.7370.7320.7300.7260.7230.7200.6900.6700.6500.6100.5800.5500.5200.4900.4600.4400.4100.3700.3400.3200.3000.2800.2700.2500.2480.2430.240TH*降温系数20.2923.3625.4026.8627.7728.4028.8027.9527.3726.7125.1724.0022.8121.6020.3719.1418.3117.0715.4114.1613.3312.5011.6711.2510.4210.3310.1210.00混凝土内部计浇筑温度算温度T1（T）Tj（℃）（℃）43.2923.0046.3623.0048.4023.0049.8623.0050.7723.0051.4023.0051.8023.0050.9523.0050.3723.0049.7123.0048.1723.0047.0023.0045.8123.0044.6023.0043.3723.0042.1423.0041.3123.0040.0723.0038.4123.0037.1623.0036.3323.0035.5023.0034.6723.0034.2523.0033.4223.0033.3323.0033.1223.0033.0023.00龄期(天)3456789101112131415161718192021222324252627282930m0.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.360.36说明：最终绝热温升=水泥用量×水化热量/0.96/2400×（1-e-mt）；混凝土内部温度（TH*降温系数）=绝热温升×降温系数；混凝土内部实际温度=混凝土内部温度+浇筑温度。保温材料所需厚度计算公式：

0.5hλi(TbTq）λ0(TmaxTb）Kb式中 i----保温材料所需厚度(m)； h----结构厚度(m)；

λi----结构材料导热系数(W/m.K)； Tmax---混凝土中心最高温度(℃)； Tb---混凝土表面温度(℃)； K---透风系数。一、计算参数 混凝土的导热系数 保温材料的导热系数 大体积混凝土结构厚度 混凝土浇注体表面温度

混凝土达到最高温度(浇筑后3d-5d)的大气平均温度 透风系数

混凝土浇筑体内的最高温度二、保温材料所需厚度

λ0=λi=h=Tb=Tq=Kb=Tmax=δ=

2.30.14425.00201.351.8029.5

℃mmW/m.kW/m.Km℃℃

Ta---空气平均温度(℃)；

三、龄期t时混凝土表层温度Tb(t)=Tq+4h’(H- h’)[T1(t)-Tq]/H2λ—混凝土导热系数取2.33W/mKh—混凝土实际厚度h’—混凝土结构的虚厚度(m)H—混凝土计算厚度h’=Kλ/β=0.67*2.33/β

H=h’=

4.540.27

mm

h=

4

m

β—模板及保温层传热系数(W/m2•K)：1/( ∑δi/λi+1/βg)材料一模板δi—模板及各种保温材料厚度(m)0.0300λi—模板及各种保温材料的导热系数(W/m•K)βg—空气层传热系数可取β=1/( ∑δi/λi+1/βg) 计算Tq —大气平均温度

H混凝土计算厚度（H=h+2h’）T1(t)—混凝土中心温度龄期t时混凝土表面温度Tb(t)四、混凝土内部与混凝土表层温差△t

正确与否自行判断

0.230.1723W/m2.K5.8188Tq=20H=

4.54T1(t)=51.80Tb(t)=27.08△t=

24.72

℃m℃℃℃