URGENT QUESTION Q1 Oil with specific gravity 108 is to be tr

URGENT QUESTION!!!

Q1. Oil with specific gravity 1.08 is to be transferred a tank at elevation 0 m to another tank at elevation 5 m through a piping system. The pipe diameter is 400mm, overall length of pipe is 80m and frictional factor is 0.032, and minor loss coefficient of the piping system is K-4.0 There are two pumps available. The characteristics of the two pumps are given in the tables below, where Hp is the pump head in meter and Q is flow rate in m/s. The efficiency for both pumps is 80% Characteristics of Pump A 0.15 0.2 0.25 0.3 0.340.38 12.8 m /S 0.05 Hp (m)15 14.7 14 8.8 6 3.43 0.56 Characteristics of Pump B 0.05 15.7 Q(m/s)0 Hp (m)16 0.15 13.8 0.2 0.25 0.3 0.34 0.38 12 9.8 7 4.41.53 1) When only pump A is used, determine the system characteristics, operating pump head, discharge and power required. (Ans: Q0.274 m/s) 2) If the elevation between the two tanks is increased to 20m, write down the new systenm characteristics equation. Can pump A work? How would you overcome this deficiency? (Ans: if operated in series, Q-0.217 m\'/s) 3) If pump A and pump B are to operate in parallel, determine the new operating pump head and discharge when the elevation difference is 5m. (Note oPog SGx Pwaterxg) (Ans: Q 0.423 m\'/s) 5m D. L-80 d- 6.4

Solution

Given:

Sp.gravity of oil = 1.08 =1.08x1000 = 1080 kg/3³

d= 0.4m

L = 80m

f =0.032

k=0.4

h_f =H/3 = 15/3 = 5m

h_f = 4fLV²

       2gd

5 = 4x0.032x80xV²

            2x9.81x0.4

V= 1.957m/s

Discharge = Q= VxA = 1.957x /4 x 0.4²

Q= 0.274m³/s

Power required P = pg Q hf

                                       1000

                                  = 1080x9.81x0.274x5

                                              1000

                                 P = 14.581KN

2) Pipes connected in series

Elevation = 20m = 4fLV²

                             2gd

20= 4x0.032x80xV²

           2x9.81x0.4

V= 3.915m/s

A ₁ V ₁= A₂V₂

V₂ = A₁ V_{1               =} d₁²      V₁

         A₂                       d₂²

             3.915=0.4² x 1.957

                     d₂²

d₂ = 0.28m

Q=AV =/4 x 0.28² x3.915

Q=0.241m/s

hence pump A works

3) Pipes connected parallel

Pump 1

V₁ =1.957m/s

Q₁ = 0.274m³/s

Pump 2

h_f = H/3 =16/3 = 5.33m

               h_f = 4fLV²       

                        2gd

           5 = 4x0.032x801.957         + 4x0.032x80xV₂²

                        2x9.81x 0.4              2x9.81x 0.4

5=2.553+ 1.304 V₂²

             

V₂ = 1.36m/s

Q = VA = 1.36x /4 x 0.4²

Q= 0.427m³/s