Complete Question
The complete question is shown on the first uploaded image
Answer:
The power loss between section (1) and section (2) is [tex]p_{(loss)} = 185626 \ ft \cdot lb/s[/tex]
Explanation:
The velocity at the first section is mathematically represented as
[tex]v_1 = \frac{Q}{a_1}[/tex]
Where Q is the flow rate given as [tex]150ft^3/s[/tex] and [tex]a_1[/tex] is the area at first section which is evaluated as
[tex]a_1 = \pi d_1 ^2[/tex]
The value of diameter [tex]d_1[/tex] is = 3 ft
So
[tex]a_1 = \pi 3^2 = 28.278 ft^2[/tex]
Now [tex]v_1 = \frac{150}{28,278}[/tex]
[tex]= 21.2 \ ft/s[/tex]
The continuity equation for first and second section is mathematically represented as
[tex]a_1v_1 = a_1 v_2[/tex]
[tex]\frac{\pi}{4}d_1^2 v_1 = \frac{\pi}{4} d_2^2 v_2[/tex]
[tex]v_2 = [\frac{d_1}{d_2} ]^2 v_1[/tex]
From the question [tex]d_2 = 4 \ ft[/tex]
Therefore [tex]v_2 = [\frac{3}{4}]^2 * 21.23[/tex]
[tex]= 11.94 \ ft/s[/tex]
Generally balanced energy equation between first and second section is mathematically represented as
[tex]-[\frac{dp}{\rho} + d[\frac{v^2}{2} ]+ gdz ] = \delta w_{sn} - \delta w_{(loss)}[/tex]
Now this to obtain the energy equation for the between the start (point 1)and end(point 2) points
[tex]-[\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)}} \,[/tex]
[tex]- [\frac{p_2 -p_1}{\rho} + \frac{v^2_2 -v^2_1}{2} + g(z_2 -z_1) ] = w_{(sn)} -w_{(loss)}[/tex]
[tex]\frac{p_2 -p_1}{\rho} + \frac{v^2_2 -v^2_1}{2} + g(z_2 -z_1) ] = w_{(sn)}-w_{(loss)} } ---(1)[/tex]
The negative sign is to show that energy is transferred from the turbine
Where
[tex]v_1 , z_1 , p_1[/tex] are the velocity , height , pressure of the datum at first section
[tex]v_2 , p_2 , z_2[/tex] are the velocity , height , pressure of the datum at second section
[tex]\rho[/tex] is the density of water with value of [tex]1.94\ slugs /ft^3[/tex]
[tex]\delta w_{(loss)}[/tex] is the loss of workdone
g is the acceleration due to gravity with a value of 32.3 ft/[tex]s^2[/tex]
Given that [tex]p_1 = 60psi = 60 \frac{lb}{in^2} = 60 \frac{(12 in)^2}{1 ft} * \frac{lb}{in^2} = 8640\ lb/ft^2[/tex]
[tex]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[/tex]
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
[tex]p_{(sn)} = -2500 \ hp * \frac{550 \ ft \cdot lb/s}{1 \ hp} = -1375000ft \cdot lb/s[/tex]
The negative sign show that power is transferred from the turbine
When we multiply the equation 1 above by the mass flow rate [tex]\r m[/tex] we get
[tex]\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)} ][/tex]
[tex]\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)[/tex]
The mass flow rate is mathematically represented as
[tex]\r m = \rho Q[/tex]
[tex]= 1.94 * 150[/tex]
[tex]= 291 \ slugs/s[/tex]
The height [tex]z_1 = 0ft \ and \ z_2= 10 ft[/tex]
Substituting values into equation 2
[tex][291 [\frac{-707.26-8640}{1.94} + \frac{11.94^2 -21.23^2}{2} + 32.2 * (0-10)]] = -1375000 - p_{(loss)}[/tex]
[tex]-1540626 = - 1375000 -p_{(loss)}[/tex]
[tex]p_{(loss)} = 185626 \ ft \cdot lb/s[/tex]
