Question 15.4: A single acting reciprocating pump, having cylinder diameter......

Given: D = 15 cm = 0.15 m, L = 30 cm = 0.3 m, r = L/2 = 0.3/2 = 0.15 m,
N= 50 rpm, H= 35 m, \omega={\frac{2\pi N}{60}}={\frac{2\times\pi\times50}{60}}= 5.236 rad/s , f= 0.01,

Now,              A={\frac{\pi}{4}}(0.15)^{2}=0.01767\,\mathrm{m^{2}\quad}

Again,              Q_{\mathrm{th}}={\frac{A L N}{60}}={\frac{0.01767\times0.3\times50}{60}}=0.0044177\,\mathrm{m^{3}/s}

or                     P_{\mathrm{th}}={\frac{1000\times9.81\times0.0044177\times35}{1000}}=1.5168\,\mathrm{kW}

∴    Percentage of slip = \left\lgroup{\frac{Q_{\mathrm{th}}-Q_{\mathrm{ac}}}{Q_{\mathrm{th}}}}\right\rgroup \times100={\frac{0.0044177-0.004}{0.0044177}}\times100=9.455\%

Again, the head loss due to acceleration in the delivery pipe is

or             h_{d}=\mathrm{\frac{~22~}{~9.81~}}\times\frac{0.001767~}{{0.007853}}\times0.15\times5.236\times\cos{\theta}

or               h_{d}=3.96274{\ \mathrm{cos}}\,\theta

At the beginning of stroke, θ = 0, cos θ = 1

Hence,     h_{dmax}=3.96274

Again, Maximum friction head loss   h_{fmax}=\frac{f l}{2g d}{\left\lgroup\frac{A}{a}r\omega\right\rgroup }^{2}

or              h_{fmax}={\frac{0.01\times35}{2g d}}\left\lgroup{\frac{0.01767}{0.0078539}}\times0.15\times5.236\right\rgroup ^{2}

or          h_{\mathrm{fmax}}=1.795\;\mathrm{m}