Question 10.2.1: The diameter and stroke length of a single-acting reciprocat......

We learnt that separation is due to inertia head at the beginning of suction stroke or end of discharge stroke, since at these two points vacuum may be enough within the pump to cause separation of the flow. Inertia head at these points is obtained from Equation 10.2.8 and 10.2.9, respectively, by substituting θ = 0 or 2π in these equations. We are given all the data to evaluate these terms.

=-H_s-H_{as}=-H_s-(L_s/g)\times (A_p/A_s)\omega ^2  r  \cos  \theta            0\leq \theta \leq \pi          (10.2.8)

=H_d-H_{ad}=H_d-(L_d/g)\times (A_p/A_d)\omega ^2  r  \cos  \theta            0\leq \theta \leq 2\pi          (10.2.9)

Given:

\mathrm{Pump  D = 0.075  m → A_p = π/4  D^2 = 0.0044  m^2}

L = 0.15  m → r = L/2 = 0.075 m

Suction line H_s = 3  m,  L_s = 5  m,  d_s = 0.04  m → A_s = 0.001256  m^2

Discharge line H_d = 12  m,  L_d = 15  m,  d_d = 0.03  m → A_d = 0.0007068  m^2

Separation head, H_{sep} = 75  kN/m^2 = 75 × 1000/(9.81 × 1000)

= 7.645 m of water below atmosphere

∴ H_{as} at beginning of suction stroke = (L_s/g) × (A_p/A_s)  ω^2r = 0.134  ω^2, and

H_{ad} at end of discharge stroke = –(L_d/g) × (A_p/A_d)  ω^2r = – 0.7168  ω^2

To avoid separation, gauge pressure should be less than or equal to separation pressure. This gives: During suction stroke, (H_s + H_{as}) ≤ H_{sep} = H_{sep} = 7.645 in limit

→

3 + 0.134  ω² = 7.645 ⇒ ω = 5.88 rad/s = 2π N/60 → N = 56.15 rpm

During delivery stroke, (H_d + H_{ad}) ≤ H_{sep} = H_{sep} = 7.645 in limit

→

12 – 0.7168 ω² = 7.645 ⇒ ω = 5.235 rad/s = 2π N/60 → N = 50.0 rpm

The pump therefore can operate at maximum speed of 50 rpm and separation would occur at the end of discharge stroke. Power of motor will be chosen on the basis of pump rpm = 50. Eq 10.2.12 gives area of the indicator diagram, which is the work done per cycle. Therefore

Average work done =wAN /60 \times area of indicater diagram

=wAN/60\times (H_s+H_d+2/3  h_{fs}+2/3  h_{fd})L

=W(H_s+H_d+2/3  h_{fs}+2/3  h_{fd}),             (10.2.12)

H = total head to be developed by pump = (H_s + H_d + 2/3  h_{fs} + 2/3  h_{fd}),

where

h_{fs} = A_p  L_s/d_s  and  h_{fd} = A_p  L_d/d_d. Substituting the given values, we get

h_{fs} = A_p  L_s/d_s = 0.0044 × 5/0.04 = 0.55  m  and  h_{fd} = 0.0044 × 15/0.03 = 2.2  m

→

H = 3 + 12 + 2/3 × 0.0044 + 2/3 × 2.2 = 20.5 m

Q = \text{discharge}/s = A_p × L × N/60 = 0.00055  m^3/s

∴

P = wQH/1000 kW = 9.81 × 0.00055 × 20.5 = 0.11 kW