题目
100 如图 1-44 所示,料液由敞口高位槽流入精馏塔中。塔内进料处的压力为-|||-30kPa(表压),输送管路为 times 2.5mm 的无缝钢管,直管长为10m。管路中装有-|||-180°回弯头一个,90°标准弯头一个,标准截止阀(全开)一个。若维持进料量为 (m)^3/h,-|||-问高位槽中的液面至少高出进料口多少米(已知:操作条件下料液的物性 rho =890kg/(m)^3,-|||-mu =1.3times (10)^-3Pacdot S ?-|||-Pa-|||-1 1`-|||-二-|||-=-|||-2-|||-2`-|||-77-|||-图 1-44 [例 https:/img.cdnjtzy.com/zyb_88ce7062e81a6151c92539724679ddb2.jpg-16] 附图
题目解答
答案
解析
步骤 1：确定能量守恒方程
根据伯努利方程，能量守恒方程为：
${z}_{1}g+\dfrac {1}{2}{{u}_{1}}^{2}+\dfrac {{p}_{1}}{p}={z}_{2}g+\dfrac {1}{2}{{u}_{2}}^{2}+\dfrac {{p}_{2}}{p}+\sum _{i}$
步骤 2：确定各参数值
${z}_{1}=H\quad {u}_{1}\approx 0\quad {p}_{1}=0$ (表压)
${z}_{2}=0\quad {p}_{2}=30kPa$ (表压)
${u}_{2}=\dfrac {qv}{\dfrac {\pi }{4}{d}^{2}}=\dfrac {5/3600}{0.785\times {0.04}^{2}}=1.1m/s$
步骤 3：计算管路总阻力
管路总阻力 $\sum _{i=1}^{n}{w}_{i}={w}_{i}=(\lambda \dfrac {l}{d}+\sum _{i}s)\dfrac {{u}^{2}}{2}$
$Re=\dfrac {deu}{u}=\dfrac {0.04\times 890\times 1.1}{1.3\times {10}^{-3}}=3.01\times {10}^{4}$
取管壁绝对粗糙度 $e=0.3mm$ ,则 $\dfrac {e}{d}=\dfrac {0.3}{40}=0.0075$
从图 1-36 中查得摩擦系数 $\lambda =0.036$ 。由表 1-3 查出局部阻力系数ξ如下:
进口突然缩小 $S=0.5$ 90°标准弯头 5=0.75 180°回弯头 5=1.5 标准截止阀(全开) 5=6.0
${S}_{y}=0.5+1.5+0.75+6.0=8.75$
$2.{m}_{1}=(2\dfrac {1}{4}+2.5)\dfrac {{n}^{2}}{2}=(0.036\times \dfrac {10}{0.04}+8.75)\times \dfrac {1.12}{2}=10.745188$
步骤 4：计算位差
所求位差 $H=(\dfrac {{m}_{2}}{e}+\dfrac {{{m}_{2}}^{2}}{2}+{{I}_{2}{N}_{3}}^{2})''=(\dfrac {30\times {10}^{3}}{890}+\dfrac {{1.1}^{2}}{2}+10$
根据伯努利方程，能量守恒方程为：
${z}_{1}g+\dfrac {1}{2}{{u}_{1}}^{2}+\dfrac {{p}_{1}}{p}={z}_{2}g+\dfrac {1}{2}{{u}_{2}}^{2}+\dfrac {{p}_{2}}{p}+\sum _{i}$
步骤 2：确定各参数值
${z}_{1}=H\quad {u}_{1}\approx 0\quad {p}_{1}=0$ (表压)
${z}_{2}=0\quad {p}_{2}=30kPa$ (表压)
${u}_{2}=\dfrac {qv}{\dfrac {\pi }{4}{d}^{2}}=\dfrac {5/3600}{0.785\times {0.04}^{2}}=1.1m/s$
步骤 3：计算管路总阻力
管路总阻力 $\sum _{i=1}^{n}{w}_{i}={w}_{i}=(\lambda \dfrac {l}{d}+\sum _{i}s)\dfrac {{u}^{2}}{2}$
$Re=\dfrac {deu}{u}=\dfrac {0.04\times 890\times 1.1}{1.3\times {10}^{-3}}=3.01\times {10}^{4}$
取管壁绝对粗糙度 $e=0.3mm$ ,则 $\dfrac {e}{d}=\dfrac {0.3}{40}=0.0075$
从图 1-36 中查得摩擦系数 $\lambda =0.036$ 。由表 1-3 查出局部阻力系数ξ如下:
进口突然缩小 $S=0.5$ 90°标准弯头 5=0.75 180°回弯头 5=1.5 标准截止阀(全开) 5=6.0
${S}_{y}=0.5+1.5+0.75+6.0=8.75$
$2.{m}_{1}=(2\dfrac {1}{4}+2.5)\dfrac {{n}^{2}}{2}=(0.036\times \dfrac {10}{0.04}+8.75)\times \dfrac {1.12}{2}=10.745188$
步骤 4：计算位差
所求位差 $H=(\dfrac {{m}_{2}}{e}+\dfrac {{{m}_{2}}^{2}}{2}+{{I}_{2}{N}_{3}}^{2})''=(\dfrac {30\times {10}^{3}}{890}+\dfrac {{1.1}^{2}}{2}+10$