he rate of flow of water pump into a pipe ABC, which is 200 m long and its diameter is 100 mm, while the length of the portion BC is also 100 m but its diameter is 200 mm. The change of diameter at B is sudden. The floor is taking place from A to C, where the pressure at A is 19.62N/cm219.62N/cm2 and C is connected to a tank. Find the pressure at C and draw the hydraulic gradient and Total energy line. Take f= 0.008.

 he rate of flow of water pump into a pipe ABC, which is 200 m long and its diameter is 100 mm, while the length of the portion BC is also 100 m but its diameter is 200 mm. The change of diameter at B is sudden. The floor is taking place from A to C, where the pressure at A is 

19.62N/cm2$19.62N/c{m}^2$ and C is connected to a tank. Find the pressure at C and draw the hydraulic gradient and Total energy line. Take f= 0.008.

Solution:

 Given:-

Length of Pipe, ABC=200m

Discharge, Q=20lit/s=0.02m3/s$Q=20lit/s=0.02m^3/s$

Slope of pipe, i=1 in 40=140${\displaystyle \frac{1}{40}}$

Length of Pipe, AB=100m

Diameter of pipe AB=100mm=0.1m

Length of pipe, BC=100m

Diameter of pipe BC=200mm=0.2m

Pressure of A, PA=19.62N/cm2=19.62×104N/m2$P_A=19.62N/c{m}^2=19.62\times {10}^{4}N/m^2$

f=0.008

Step 1: Velocity of water in pipe AB

V1=QArea of AB$V_1={\displaystyle \frac{Q}{\text{Area of AB}}}$

V1=0.02π4(0.1)2=2.54m/s$V_1={\displaystyle \frac{0.02}{{\displaystyle \frac{\pi }{4}}(0.1{)}^2}}=2.54m/s$

Step 2: Velocity of water in pipe BC

V2=QArea of BC$V_2={\displaystyle \frac{Q}{\text{Area of BC}}}$

V2=0.02π4(0.2)2=0.63m/s$V_2={\displaystyle \frac{0.02}{{\displaystyle \frac{\pi }{4}}(0.2{)}^2}}=0.63m/s$

Applying Bernoulli's equation to point A and C,

PAρg+V2A2g+zA=PCρg+V2C2g+zC+total losses from A to C...............(1)${\displaystyle \frac{P_A}{\rho g}}+{\displaystyle \frac{V_A^2}{2g}}+z_A={\displaystyle \frac{P_C}{\rho g}}+{\displaystyle \frac{V_C^2}{2g}}+z_C+\text{total losses from A to C}...............(1)$

Step 3: Total losses from A to C

• Loss of head due to friction in pipe AB,

hf1=4fLV2d×2g$h{f}_1={\displaystyle \frac{4fL{V}^2}{d\times 2g}}$

hf1=4×0.008×100×(2.54)20.1×2×9.81=10.52m$h{f}_1={\displaystyle \frac{4\times 0.008\times 100\times (2.54{)}^2}{0.1\times 2\times 9.81}}=10.52m$

• Loss of head due to friction in pipe BC,

hf2=4fLV2d×2g$h{f}_2={\displaystyle \frac{4fL{V}^2}{d\times 2g}}$

hf2=4×0.008×100×(0.63)20.2×2×9.81=0.323m$h{f}_2={\displaystyle \frac{4\times 0.008\times 100\times (0.63{)}^2}{0.2\times 2\times 9.81}}=0.323m$

• Loss of head due to enlargement at B

he=(V1−V2)22g=(2.54−0.63)22×9.81=0.186m$h_e={\displaystyle \frac{(V_1-V_2{)}^2}{2g}}={\displaystyle \frac{(2.54-0.63{)}^2}{2\times 9.81}}=0.186m$

∴$\therefore$ Total losses from A to C =hf1+he+hf2=10.52+0.186+0.323=11.03m$=h{f}_1+h_e+h{f}_2=10.52+0.186+0.323=11.03m$

Substituting the values in equation (1), we get

PAρg+V2A2g+zA=PCρg+V2C2g+zC+11.03..................................(2)${\displaystyle \frac{P_A}{\rho g}}+{\displaystyle \frac{V_A^2}{2g}}+z_A={\displaystyle \frac{P_C}{\rho g}}+{\displaystyle \frac{V_C^2}{2g}}+z_C+\mathrm{11.03..................................}(2)$

Taking datum line passing through 'A', we have

zA=0$z_A=0$

zC=140×total length of pipe$z_C={\displaystyle \frac{1}{40}}\times \text{total length of pipe}$

zC=140×200=5m$z_C={\displaystyle \frac{1}{40}}\times 200=5m$

Also, PA=19.62×104N/m2$P_A=19.62\times {10}^{4}N/m^2$

VA=V1=2.54m/s,VC=V2=0.63m/s$V_A=V_1=2.54m/s,V_C=V_2=0.63m/s$

Substituting the values in equation (2), we get

19.62×1041000×9.81+2.5422×9.81+0=PCρg+0.6322×9.81+5+11.03${\displaystyle \frac{19.62\times {10}^4}{1000\times 9.81}}+{\displaystyle \frac{{2.54}^2}{2\times 9.81}}+0={\displaystyle \frac{P_C}{\rho g}}+{\displaystyle \frac{{0.63}^2}{2\times 9.81}}+5+11.03$

20.328=PCρg+16.05$20.328={\displaystyle \frac{P_C}{\rho g}}+16.05$

∴PCρg=20.328−16.05=4.278m$\therefore {\displaystyle \frac{P_C}{\rho g}}=20.328-16.05=4.278m$

PC=4.278×ρ×g$P_C=4.278\times \rho \times g$

PC=4.278×1000×9.81N/m2$P_C=4.278\times 1000\times 9.81N/m^2$

PC=4.278×1000×9.81104N/cm2$P_C={\displaystyle \frac{4.278\times 1000\times 9.81}{{10}^4}}N/c{m}^2$

PC=4.196N/cm2$P_C=4.196N/c{m}^2$