Physics, 13.04.2020 16:44 jessb9583

Water is supplied at 150 ft3/s and 60 psi to a hydraulic turbine through a 3-ft inside-diameter inlet pipe as indicated in Fig. 4. The turbine discharge pipe has a 4-ft inside diameter. The static pressure at section (2), 10 ft below the turbine inlet, is 10-in. Hg vacuum. If the turbine develops 2500 hp, determine the power lost between sections (1) and (2).

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Answer from: tatemcgee

Complete Question

The complete question is shown on the first uploaded image

The power loss between section (1) and section (2) is $p_{(loss)} = 185626 \ ft \cdot lb/s$

Explanation:

The velocity at the first section is mathematically represented as

$v_1 = \frac{Q}{a_1}$

Where Q is the flow rate given as 150ft^3/s and a_1 is the area at first section which is evaluated as

$a_1 = \pi d_1 ^2$

The value of diameter d_1 is = 3 ft

$a_1 = \pi 3^2 = 28.278 ft^2$

Now $v_1 = \frac{150}{28,278}$

$= 21.2 \ ft/s$

The continuity equation for first and second section is mathematically represented as

a_1v_1 = a_1 v_2

$\frac{\pi}{4}d_1^2 v_1 = \frac{\pi}{4} d_2^2 v_2$

$v_2 = [\frac{d_1}{d_2} ]^2 v_1$

From the question $d_2 = 4 \ ft$

Therefore $v_2 = [\frac{3}{4}]^2 * 21.23$

$= 11.94 \ ft/s$

Generally balanced energy equation between first and second section is mathematically represented as

$-[\frac{dp}{\rho} + d[\frac{v^2}{2} ]+ gdz ] = \delta w_{sn} - \delta w_{(loss)}$

Now this to obtain the energy equation for the between the start (point 1)and end(point 2) points

$-[\int\limits^2_1 {\frac{dp}{\rho} } \, + \int\limits^2_1 {d[\frac{v^2}{2} } \, ] + \int\limits^2_1 gdz \, ] = \int\limits^2_1 {\delta w_{sn}} \, -\int\limits^2_1 {\delta w _{(loss)}} \,$

$- [\frac{p_2 -p_1}{\rho} + \frac{v^2_2 -v^2_1}{2} + g(z_2 -z_1) ] = w_{(sn)} -w_{(loss)}$

$\frac{p_2 -p_1}{\rho} + \frac{v^2_2 -v^2_1}{2} + g(z_2 -z_1) ] = w_{(sn)}-w_{(loss)} } ---(1)$

The negative sign is to show that energy is transferred from the turbine

Where

v_1 , z_1 , p_1 are the velocity , height , pressure of the datum at first section

v_2 , p_2 , z_2 are the velocity , height , pressure of the datum at second section

$\rho$ is the density of water with value of $1.94\ slugs /ft^3$

$\delta w_{(loss)}$ is the loss of workdone

g is the acceleration due to gravity with a value of 32.3 ft/ s^2

Given that $p_1 = 60psi = 60 \frac{lb}{in^2} = 60 \frac{(12 in)^2}{1 ft} * \frac{lb}{in^2} = 8640\ lb/ft^2$

$p_2 = -10 \ in \ of \ Hg = - 10 * \frac{70.726 \ lb/ft^2}{1 \ in \ of \ Hg} * in \ of \ Hg = -707.26 lb/ft^2$

The negative sign is to show that the pressure is acting against the pressure of the system

We are also told that the power developed by the turbine is

$p_{(sn)} = -2500 \ hp * \frac{550 \ ft \cdot lb/s}{1 \ hp} = -1375000ft \cdot lb/s$

The negative sign show that power is transferred from the turbine

When we multiply the equation 1 above by the mass flow rate $\r m$ we get

$\r m[\frac{p_2 -p_1}{\rho} + \frac{v^2_2 -v^2_1}{2} + g(z_2 -z_1) ] = \r m [ w_{(sn)}-w_{(loss)} ]$

$\r m[\frac{p_2 -p_1}{\rho} + \frac{v^2_2 -v^2_1}{2} + g(z_2 -z_1) ] = \r m [ p_{(sn)}-p_{(loss)}] ----(2)$

The mass flow rate is mathematically represented as

$\r m = \rho Q$

= 1.94 * 150

$= 291 \ slugs/s$

The height $z_1 = 0ft \ and \ z_2= 10 ft$

Substituting values into equation 2

$[291 [\frac{-707.26-8640}{1.94} + \frac{11.94^2 -21.23^2}{2} + 32.2 * (0-10)]] = -1375000 - p_{(loss)}$

$-1540626 = - 1375000 -p_{(loss)}$

$p_{(loss)} = 185626 \ ft \cdot lb/s$

Water is supplied at 150 ft3/s and 60 psi to a hydraulic turbine through a 3-ft inside-diameter inle

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