Transmission of Power

A Nozzle discharges 175 galls per min. under a head of 200 ft. The diameter of the nozzle is 1 in. and the diameter of the jet is 0.9 in.

Find:

1. The coefficient of velocity for the jet.
2. The head lost in the nozzle.
3. The horse power available in the jet.

• a) The Coefficient of Contraction, $C_D$ is given by:

$C_D = \left (\frac{0.9}{1.0} \right )^2 = 0.81$

The theoretical discharge $Q$ = The nozzle area $X$. From equation (5) velocity (neglecting losses) is $\sqrt{2\;g\;H} = \sqrt{2\times 32.2 \times 200}$ Therefore: $Q = \left (\frac{\pi }{4}\times \frac{1}{12^2} \right )\times \sqrt{2\times 32.2\times 200} = 0.619\;ft.^3\;sec.^{-1}$

Since 1 gallon of water weighs 10 lb. and 1 cubic foot of water weighs 62.4 lb, then $Q = \frac{175\times 10}{60\times 62.4} = 0.467\;ft^3\;sec^{-1}$ And the coefficient of discharge $= \frac{0.467}{0.619} = 0.755$

From equation (7), the coefficient of velocity: $=\frac{C_D}{C_C} = \frac{0.755}{0.81} = 0.932$

• b) Using Equation (5)

$H = H_l + \frac{V^2}{2\;g}$ where $H_l$ is the head lost in nozzle i.e. $H = H_L + {C_{V}}^{2} H$ Therefore: $H_L = H\left (1 - {C_{V}}^{2} \right ) = 200(1 - 0.932^2)$

Thus: $H_L=26.28\;ft.$

• c) The horse power of the jet is dependent upon the weight of water per second and the head of water in the jet

Thus:

Horse power will be: $\frac{W \times H_J}{550}$ $=\frac{175 \times 10}{60}\times (200 - 26.28)\times \frac{1}{550} = 9.21\;h.p.$

• a) The coefficient of velocity is $0.932$
• b) The head lost in nozzle is $H_L=26.28\;ft.$
• c) The horse power available in the jet is $9.21\;h.p.$

The water available for a Pelton wheel is $150\;ft^3/sec.$ and the total head from the reservoir level to the nozzles is 900 ft. The turbine has two runners with two jets per runner. All four jets have the same diameter. The pipeline is 10,000 ft. long. The efficiency of power transmission through the pipe line is 91 % and the efficiency of each runner is 90%. The velocity coefficient of each nozzle is 0.975 and the coefficient of friction for the pipe-line is 0.0045.

Determine:

1. The horse power developed by the turbine.
2. The diameter of the jets.
3. The diameter of the pipe-line.

Drawing from notes.

• a) To find the Horse-power of the turbine

The efficiency of the pipe and nozzles is 0.91% and the runners 90%. Hence the efficiency of the whole system is given by:

$\eta =0.91\times 0.9$

The total power available is: $\frac{62.4\times 150\times 900}{550}=15.316\;h.p.$

Therefore the power output is the total power available $X$ the efficiency of the system

i.e. $15316\times 0.91\times 0.9=12.544\;h.p.$

• b) To find the diameter of the jets

Let the velocity of water from the jets be $v$

i.e. $15316\times 0.91\times 0.9=12.544\;h.p.$ The velocity head of the jets is: $\frac{v^2}{2g}=0.91\times 900=819\;ft.$ The jet velocity is: $v=\sqrt{819\times 2\times 32.2}=229.7\;ft/sec.$

This is the actual velocity through the jet. The theoretical velocity, based on the diameter of the jet will be greater.

i.e. $v_t=\frac{229.5}{0.975}=235.4\;ft./sec.$ The flow of water is $150\;ft^3./sec.$ which is $\displaystyle\frac{150}{4}\;ft^3./sec$

$\therefore \;\;\;\;\;\;\frac{\pi d^2}{4}\times 235.5=\frac{150}{4}$ $\therefore \;\;\;\;\;d=\sqrt{\frac{150\times 4\times 12^2}{4\times\pi 235.4}}=5.40 \;in.$

• c) The Diameter of the Pipe

If $H_1$ is the Total Head behind the nozzle:

$v=C_v\sqrt{2gH_1}$ $\therefore \;\;\;\;\;H_1=\frac{v^2}{2g\;{C_{v}}^{2}}=\frac{819}{0.975^2}=861.5\;ft$

The pipe friction is the difference between the Total head and the head behind the nozzle

i.e. $H-H_1=900-861.5=38.5\;ft.$

Using the Darcy Equation:

$\frac{4\times 0.0045\times 10,000\times v^2}{2gD}=38.5$

$\therefore \;\;\;\;\;\frac{v^2}{D}=13.74$

By continuity:

$\frac{\pi\;D^2}{4}\times v=150$

Combining the above two equations to eliminate $v$ gives:

$D^5=2655$\ And thus: $D=4.84\;ft.$

• The horse power developed by the turbine is $12,544\;h.p.$

• The diameter of the jets is $d=5.40 \;in.$

• The diameter of the pipe-line is $D=4.84\;ft.$

During 9 months of each year there is an ample supply of water for a Power station and for 5 hours each day water is pumped to a high level storage basin situated at 300 ft. above the turbines. The basin supplies the turbines for the remaining 3 months of the year through two equal pipe-lines which are arranged in parallel and are 1,000 ft. long. The turbines develop 1500 h.p. for 10 hours per day during these 3 months with an efficiency of 0,86 as reckoned on the conditions at the turbine. Taking each month as 30 days, $f$ = 0.005 and assuming that the friction is limited to 15 ft.

Find:

• a) The minimum size of the basin required

• b) The rate of discharge of the pumps.

• c)The diameter of the two pipes.

Let $d$ be the diameter of the pipe and $v$ the velocity of flow.

The head of water at the turbines $=H-1-15=285\;ft.$

Therefore the Power supplied th the turbines is given by:

$\frac{WH_1}{550}=\frac{wQH_1}{550}=\frac{62.4\times Q\times 285}{550}=\frac{150}{0.86}$ $\therefore \;\;\;\;\;Q=54\;ft.^3/sec.$

The minimum size of the basin is given by $Q$ and the number of seconds that the turbines operate over the three month period

i.e. the Minimum Basin size is : $54\times 3\times 30\times 10\times 60^2=175\times 10^6\;ft.^3$

The required Pump output is :

Volume of Basin/Pump operating Time $=\frac{175\times 10^6}{9\times 30\times 5\times 60^2}=36\;ft.^3/sec.$

The head lost due to friction in the pipes is given as 15 ft. So using Darcy's equation:

$h_f=15=\frac{4\times 0.005\times 1000}{2g}\times \frac{v^2}{d}$

From which:

$v^2=48.3\;d$

As there are two pipes:

$Q=2av=2\times \frac{\pi d^2}{4}\times v$

Substituting in values:

$v=\frac{54\times 4}{2\pi d^2}=\frac{34.4}{d^2}$ $\therefore \;\;\;\;\;48.3d=\frac{34.4^2}{d^4}$

From which $d = 1.89 ft.$

• The minimum size of the basin required is $175\times 10^6\;ft.^3$

• The rate of discharge of the pumps is $36\;ft.^3/sec.$

• The diameter of the two pipes is $d = 1.89 ft.$

In a Power Scheme using a Pelton Wheel, water is to be supplied from a reservoir,which will be 1200 ft. above the turbine, through a pipe-line 5400 ft. long. The Turbine is to produce 7000 b.h.p. at 500 r.p.m. Pipe friction is limited to 10 % of the gross head and the ratio of wheel diameter to jet diameter must not be less than 11:1

Calculate the number and diameter of the jets, the mean diameter of the wheel and the supply-pipe diameter. Assume$C_v$ for the jets to be 0.98, bucket speed 0.45 jet speed and the efficiency of the turbine 87 %, $f$ for the pipe friction is 0.0048.

Let

• $v_1=$ the velocity of water in the pipe.

• $d_1=$ the diameter of the supply pipe.

• $d^2=$ the diameter of the jets.

• $v_2=$ the jet velocity.

• $n=$ the number of jets..

• $D=$ the diameter of the wheel.

The head lost in the supply pipe is 10 % of the total head so:

$h_f=1200\times \frac{10}{100}=120\;ft.$

Using Darcy's equation:

$120=\frac{4\times 0.0048\times 5400}{2g}\times \frac{{v_{1}}^{2}}{d_1}=1.61\frac{{v_{1}}^{2}}{d_1}$

Applying Bernoulli:

$1200=hf\;\text{for pipe}+\text{nozzle loss}+\frac{{v_{2}}^{2}}{2g}$ $=120+\left ( \frac{1}{0.98^2} -1\right )\frac{{v_{2}}^{2}}{2g}+\frac{{v_{2}}^{2}}{2g}$

$\therefore \;\;\;\;\;\frac{{v_{2}}^{2}}{2g}=\frac{1084}{1.04}$ And: $v_2=258\;ft./sec.$

The jet Horse-power is given by:

$\frac{nwa_2v_2}{550}\times \frac{{v_{2}}^{2}}{2g}$

Substituting in values from above: $\frac{7000}{0.87}=\frac{n\times 62.4\times a_2\times 258}{550}\times \frac{1080}{1.04}$

i.e.

But:

From equations (12) and (13)

$\frac{\pi}{4}{d_{1}}^{2}v_1=0.264\times 258$ $\therefore \;\;\;\;\;{v_{1}}^{2}=\frac{7530}{{d_{1}}^{4}}$

Substituting into equation (11)

$74.7d_1=\frac{7530}{{d_{1}}^{4}}$ $\therefore \;\;\;\;\;d_1=2.516\;ft.$

Substituting the above into equation (11)

${v_{1}}^{2}=74.7\times 2.516$ $\therefore \;\;\;\;\;v_1=13.7\;ft./sec.$

And: $Q=a_1v_1=\frac{\pi \times 2.516\\62}{4}\times 13.7=68\;ft.^2/sec.$

Using equation (13)

$68=na_2v_2=na_2\times 258$

From the question:

The bucket speed $= 0.45\times \text{Jet speed}$

i.e. $\frac{500}{60}\times 2\pi\times \frac{D}{2}=0.45\times 258$ $\therefore \;\;\;\;\;D=4.43\;ft.$

The following conditions were imposed by the question:

• $\displaystyle\frac{D}{d_2}$ must not be >11

• $d_2$ must not be >$\displaystyle\frac{4.43}{11}$

• ${d_{2}}^{2}$ must not be >0.1625

From equation (14) and the condition required for $\displaystyle {d_{2}}^{2}$

• $\displaystyle\frac{0.336}{n}$ must not be >0.1625

i.e. $n=3$

From equation (14) ${d_{2}}^{2}=\frac{0.336}{3}$ $\therefore \;\;\;\;\;d_2 =0.335\;ft$ which is The diameter of the jets.

• The number of jets is $n=3$
• The diameter of the jets is $d_2 =0.335\;ft$
• The mean diameter of the wheel is $D=4.43\;ft.$
• The supply-pipe diameter is $d_1=2.516\;ft.$

A twin jet Pelton wheel is supplied by 3 equal pipes in parallel each having a length $L_1$, diameter $d_1$ and friction coefficient $f$ connected through a short common pipe to the nozzles. Ignoring losses other than pipe friction, find the nozzle diameters for maximum kinetic energy of the jet.

Calculate the Horse-power developed by the machine if for each pipe $d_1=6\;in.$, $L_1=7430\;ft.$ and $f=0.005$ and given that the efficiency of the machine is $81.5\%$. The permissible loss of head due to pipe friction is 5% of the gross head and $C_v=0.985$

Let

• $v_1$ be the water velocity in the pipes.

• $d_2$ be the jet velocity.

• $v_2$ be the jet velocity.

• $H$ be the gross head,

Applying Bernoulli and neglecting all losses except pipe friction:

$H=\frac{4fl{v_{1}}^{2}}{2gd_1}+\frac{{v_{2}}^{2}}{2g}=\frac{{v_{2}}^{2}}{2g}\left ( \frac{4fL_1}{d_1}\times \frac{{v_{1}}^{2}}{{v_{2}}^{2}}+1\right )$

But: $Q=3a_1v_1=2a_2v_2$ $\therefore \;\;\;\;\;3{d_{1}}^{2}v_1=2{d_{2}}^{2}v_2$

From the above equations:

Where: $k=\frac{16fL_1}{9{d_{1}}^{5}}$

The Kinetic Energy of the jets per second is :

From equations (17) and (18)

The kinetic energy per second $=\frac{\pi w{d_{2}}^{2}}{4g}\times \frac{\left ( 2gH \right )^{\frac{3}{2}}}{(1+k{d_{2}}^{4})^{\frac{3}{2}}}$

This will be a maximum when :

$(1+k{d_{2}}^{4})^{\frac{3}{2}}\times 2d_2={d_{2}}^{2}\times \frac{3}{2}(1+k{d_{2}}^{4})^{\frac{1}{2}}\times 4k{d_{2}}^{3}$ i.e. When: $2k{d_{2}}^{4}=1$

Substituting the above in equation (16)

$d_2=\left ( \frac{9{d_{1}}^{5}}{32fL_1} \right )^{\frac{1}{4}}$

From Bernoulli and equation (6)

$1500=h_f\;\text{For pipes}+\text{Nozzle losses}+\frac{{v_{2}}^{2}}{2g}$ $=\frac{5}{100}\times 1500+\left ( \frac{1}{0.985-1} \right )\frac{{v_{2}}^{2}}{2g}+\frac{{v_{2}}^{2}}{2g}$

$\therefore \;\;\;\;\;\frac{{v_{2}}^{2}}{2g}=1385$ And: $v_2=298\;ft./sec.$

The frictional head lost is 5% of the total head and thus from the Darcy equation:

$\frac{5}{100}\times 1500=\frac{4\times 0.005\times 7430{v_{1}}^{2}}{2g\times \frac{6}{12}}=4.63{v_{1}}^{2}$ Hence: $v_1=4.02\;ft./sec.$

Substituting into equation (15)

$\frac{4.02}{298}=\frac{{d_{2}}^{2}}{\displaystyle\frac{1}{4}}\times \frac{2}{3}$ $\therefore \;\;\;\;\;a_2=\frac{\pi {d_{2}}^{2}}{4}=\frac{3.96}{10^3}\;ft.^2$

From equation (9)

The jet horse-power $=\frac{2wa_2v_2}{550}\times \frac{{v_{2}}^{2}}{2g}$

The machine h.p.= The jet h.p. multiplied with Efficiency $=\frac{2\times 62.4}{550}\times\frac{3.96}{10^3}\times298\times 1385\times \frac{81.5}{100}=303 \;h.p.$

The Horse-power is $303 \;h.p.$

The output of a multi-cylinder hydraulic motor is required to be 180 h.p. when its efficiency is 73 %. A hydraulic power station developing a pressure of 1200 $lb./in^2$ supplies the motor through four 3 in. pipes, 2 miles long.

Determine the pressure at the motor, the velocity of flow in the pipes and the efficiency of transmission $f$=0.008

Let

• $p_1$ be the supply pressure in lb./sq.in.

• $p_2$ be the pressure at the motor

• $v$ be the velocity of water in the pipes

Note: 1 mile is 1769 yards or 5280 ft.

The head lost in a pipe is the difference in pressure between the ends.

$(p_1-p_2)\times \frac{12^2}{62.4}=h_f$ (The pressures were converted to ft. of water) $\therefore \;(p_i-p_2)\times \frac{144}{62.4}=\frac{4\times 0.008\times 2\times 5280}{\frac{3}{12}}\times\frac{v^2}{2g} =1350\frac{v^2}{2g}$

Water h.p.at the motor: $\left ( \frac{180}{0.73} \right )=\frac{WH}{550}=\frac{4wav}{550}\times \frac{p_2\times 12^2}{62.4}$ $\frac{180}{0.73} =4\times \frac{\pi\;(\frac{3}{12})^2}{4}\times \frac{12^2}{550}\times vp_2$ $\therefore \;\;\;\;\;p_2v=4800$ Or:

From Equations (19) and (20)

$0.1105(1200-p_2)=\frac{(23\times 10^6)}{{p_{2}}^{2}}$ $\therefore \;\;\;\;\;\frac{208\times 10^6}{{p_{2}}^{2}+p_2}=1200$

Trial and error gives a value for $p_2$ of 990 $lb./in^2$

From equation (19)

$v^2=0.1105\times (1200-990)$ $\therefore \;\;\;\;\;v=4.85\;ft./sec.$ Transmission efficincy $=\frac{p_2}{p_1}=\frac{990}{1200}=82.5\;\%$

• The pressure at the motor is $p_2=990 lb./in^2$
• The velocity of flow in the pipes is $v=4.85\;ft./sec.$
• The efficiency of transmission is $82.5\;\%$