Question

The figure illustrates flow through a pipe with diameters of 1.0 mm and 2.0 mm and with different elevations. Px is the pressure in the pipe, and Vx is the speed of a non-viscous incompressible fluid at locations x = Q,R,S,T, or U. Options are: Greater than, Less than, Equal to

PU is ... PQ

VU is ... 2VT

PR is ... PU.

VR is ... VS

VQ is ... VU

PR is ... PS

Answer 1

By Bernoulli's equation as well as the Equation of continuity the relation between pressure and velocities at different points are,

1. $P_u > P_q$
2. $V_u > 2V_t$
3. $P_u > P_r$
4. $V_r=V_s$
5. $V_q=V_u$
6. $P_r > P_s$

To find the answer, we need to know about the Bernoulli's equation as well as the Equation of continuity.

How to find the solution?

1) We have to find the relation between pressure at U and Q.

•      We have the Bernoulli's equation,

                  $P+\frac{1}{2}dV^2+dgh=constant.$

where, P is the pressure, V is the velocity, d is density, g is acceleration due to gravity and h is the height of the flow.

• By using the equation, we can find the pressure at U and Q.

                       $P_u+\frac{1}{2}dV_u^2+dg*0= P_q+\frac{1}{2}dV_q^2+dgh\\where,\\V_u=V_q, \\since, D_p=D_q\\Thus,\\P_u=P_q+dgh$

                       $P_u > P_q$

2) We have to find the relation between velocity at U and T.

• For this, we have the equation of continuity as,

                           $AV=constant\\A_1V_1=A_2V_2$

• From the diagram, we have,

                          $A_u=\pi r^2=\pi *(0.5)^2=0.25\pi *mm^2\\A_t=\pi *1=\pi mm^2\\V_u=V\\V_t=?$

• Thus, the relation between velocity at U and T is,

                      $V_t=\frac{A_uV_u}{A_t}=\frac{V}{4}$

                      $2V_t=2*\frac{V}{4} =\frac{V}{2}\\V_u=V$

                       $V_u > 2V_t$

3) We have to find the relation between pressure at R and U

                 $P_u+\frac{1}{2}dV_u^2+dg*0= P_r+\frac{1}{2}dV_r^2+dg*0\\\\V_u=V , then\\V_r=V_t=\frac{V}{4}=\frac{V_u}{4} \\\\P_u=P_r+\frac{1}{16}$

                   $P_u > P_r$

4)  We have to find the relation between velocity at R and S

• Both points R and S, have same area, thus same velocity.

                                     $V_r=V_s$

5)  We have to find the relation between velocity at Q and U

• Both points Q and U, have same area, thus same velocity.               

                                    $V_q=V_u$

6)  We have to find the relation between pressure at R and S

• Both points R and S, have same area and thus, same velocities.

                          $P_r=P_s+dgh$

                           $P_r > P_s$

Thus, we can conclude that, By Bernoulli's equation as well as the Equation of continuity the relation between pressure and velocities at different points are,

1. $P_u > P_q$
2. $V_u > 2V_t$
3. $P_u > P_r$
4. $V_r=V_s$
5. $V_q=V_u$
6. $P_r > P_s$

Learn more about the Bernoulli's equation here:

https://brainly.com/question/9506577

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