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

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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

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

Answer from: Quest

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