The difference in water surface level in two tanks, which are connected by three pipes in series of length 300 m, 170 m and 210 m and of diameter 300 mm, 200 mm and 400 mm respectively is 12 m. Determine the rate of flow of water if coefficient of friction are 0.005,0.0052 and 0.0048 respectively. Considering (i) minor losses also (ii) neglecting minor losses

 The difference in water surface level in two tanks, which are connected by three pipes in series of length 300 m, 170 m and 210 m and of diameter 300 mm, 200 mm and 400 mm respectively is 12 m. Determine the rate of flow of water if coefficient of friction are 0.005,0.0052 and 0.0048 respectively.

Considering

(i) minor losses also

(ii) neglecting minor losses

Solution:- 

Given:

Difference of water level, H=12m

Length of pipe 1, L1=300m$L_1=300m$

Diameter, d1=300mm=0.3m$d_1=300mm=0.3m$

Length of pipe 2, L2=170m$L_2=170m$

Diamter, d2=200mm=0.2m$d_2=200mm=0.2m$

Length of pipe 3, L3=210m$L_3=210m$

Diamter, d3=400mm=0.4m$d_3=400mm=0.4m$

Case-1: Considering minor losses

From continuity, we have

A1V1=A2V2=A3V3$A_1{V}_1=A_2{V}_2=A_3{V}_3$

∴V2=A1V1A2=π4(d1)2π4(d2)2V1$\therefore V_2={\displaystyle \frac{A_1{V}_1}{A_2}}={\displaystyle \frac{{\displaystyle \frac{\pi }{4}}(d_1{)}^2}{{\displaystyle \frac{\pi }{4}}(d_2{)}^2}}{V}_1$

V2=d21d22V1$V_2={\displaystyle \frac{d_1^2}{d_2^2}}{V}_1$

V2=(0.30.2)2×V1$V_2={\left({\displaystyle \frac{0.3}{0.2}}\right)}^2\times V_1$

V2=2.25V1..........................(1)$V_2=2.25V_1..........................(1)$

Now,

V3=A1V1A3=π4(d1)2π4(d3)2V1$V_3={\displaystyle \frac{A_1{V}_1}{A_3}}={\displaystyle \frac{{\displaystyle \frac{\pi }{4}}(d_1{)}^2}{{\displaystyle \frac{\pi }{4}}(d_3{)}^2}}{V}_1$

V3=d21d23V1$V_3={\displaystyle \frac{d_1^2}{d_3^2}}{V}_1$

V3=(0.30.4)2×V1$V_3={\left({\displaystyle \frac{0.3}{0.4}}\right)}^2\times V_1$

V3=0.5625V1..........................(2)$V_3=0.5625V_1..........................(2)$

Using the equation of flow through pipes, we have

∴H=0.5V212g+4f1L1V21d1×2g+0.5V222g+4f2L2V22d2×2g+(V2−V3)22g+4f3L3V23d3×2g+V232g$\therefore H={\displaystyle \frac{0.5V_1^2}{2g}}+{\displaystyle \frac{4f_1{L}_1{V}_1^2}{d_1\times 2g}}+{\displaystyle \frac{0.5V_2^2}{2g}}+{\displaystyle \frac{4f_2{L}_2{V}_2^2}{d_2\times 2g}}+{\displaystyle \frac{(V_2-V_3{)}^2}{2g}}+{\displaystyle \frac{4f_3{L}_3{V}_3^2}{d_3\times 2g}}+{\displaystyle \frac{V_3^2}{2g}}$

Substituting the values of V2$V_2$ and V3$V_3$,

12.0=0.5V212g+4×0.005×300×V210.3×2g+0.5(2.25×V1)22g+4×0.0052×170×(2.25×V1)20.2×2g+(2.25V1−0.562V1)22g+4×0.0048×210×(0.5625×V1)20.4×2g+(0.5625V1)22g$12.0={\displaystyle \frac{0.5V_1^2}{2g}}+{\displaystyle \frac{4\times 0.005\times 300\times V_1^2}{0.3\times 2g}}+{\displaystyle \frac{0.5(2.25\times V_1{)}^2}{2g}}+{\displaystyle \frac{4\times 0.0052\times 170\times (2.25\times V_1{)}^2}{0.2\times 2g}}+{\displaystyle \frac{(2.25V_1-0.562V_1{)}^2}{2g}}+{\displaystyle \frac{4\times 0.0048\times 210\times (0.5625\times V_1{)}^2}{0.4\times 2g}}+{\displaystyle \frac{(0.5625V_1{)}^2}{2g}}$

12.0=V212g[0.5+20.0+2.53+89.505+2.847+3.189+0.316]$12.0={\displaystyle \frac{V_1^2}{2g}}[0.5+20.0+2.53+89.505+2.847+3.189+0.316]$

V1=12×2×9.81118.887−−−−−−−−−−−√$V_1=\sqrt{{\displaystyle \frac{12\times 2\times 9.81}{118.887}}}$

Therefore, V1=1.407m/s$V_1=1.407m/s$

Therefore, rate of flow,

Q=A×V=A1×V1$Q=A\times V=A_1\times V_1$

Q=π4×(d1)2×V1$Q={\displaystyle \frac{\pi }{4}}\times (d_1{)}^2\times V_1$

Q=π4×(0.3)2×1.407$Q={\displaystyle \frac{\pi }{4}}\times (0.3{)}^2\times 1.407$

Q=0.09945m3/s$Q=0.09945m^3/s$

Case-2: neglecting minor losses:-

H=4f1L1V21d1×2g+4f2L2V22d2×2g+4f3L3V23d3×2g$H={\displaystyle \frac{4f_1{L}_1{V}_1^2}{d_1\times 2g}}+{\displaystyle \frac{4f_2{L}_2{V}_2^2}{d_2\times 2g}}+{\displaystyle \frac{4f_3{L}_3{V}_3^2}{d_3\times 2g}}$

12.0=V212g[4×0.005×3000.3+4×0.0052×170×(2.25)20.2+4×0.0048×210×(0.5625)20.4]$12.0={\displaystyle \frac{V_1^2}{2g}}\left[{\displaystyle \frac{4\times 0.005\times 300}{0.3}}+{\displaystyle \frac{4\times 0.0052\times 170\times (2.25{)}^2}{0.2}}+{\displaystyle \frac{4\times 0.0048\times 210\times (0.5625{)}^2}{0.4}}\right]$

12.0=V212g[20+89.505+3.189]$12.0={\displaystyle \frac{V_1^2}{2g}}[20+89.505+3.189]$

V1=12×2×9.81112.694−−−−−−−−−−−√$V_1=\sqrt{{\displaystyle \frac{12\times 2\times 9.81}{112.694}}}$

Therefore, V1=1.445m/s$V_1=1.445m/s$

Therefore, rate of flow,

Q=A×V=A1×V1$Q=A\times V=A_1\times V_1$

Q=π4×(d1)2×V1$Q={\displaystyle \frac{\pi }{4}}\times (d_1{)}^2\times V_1$

Q=π4×(0.3)2×1.445$Q={\displaystyle \frac{\pi }{4}}\times (0.3{)}^2\times 1.445$

Q=0.1021m3/s$Q=0.1021m^3/s$

or Q=102.1 lit/s