Chapter 6, Problem 96P

Water flows at mass flow rate m through a 90° vertically oriented elbow of elbow radius R (to the centerline) and lime, pipe diameter D as sketched. The outlet s exposed to the atmospheric. (Hint: This means that the pressure at the outlet is atmospheric pressure.) The pressure at the inlet must obviously be higher than atmospheric in order to push the water through the elbow and to raise the elevation of the water. The irreversible head loss through the elbow is hL. Assume that the kinetic energy flux correction factor a is not unity, but is the same at the inlet and outlet of the elbow $\left(a_1=a_2\right)$ , Assume that the same thing applies to the momentum flux correction factor $\beta \left(i.e.{\beta }_1={\beta }_2\right)$

(a) Using the head form of the energy equation, derive an expression for the gage pressure $P_{gauge}$ at the center of the inlet as a function of the other variables as needed

(b) Plug in these numbers and solve for $P_{gauge};\rho =998.0kg/m^3,D=10.0cm,R=35.0cm,h_L=0.259m,$ (of equivalent water column height), $a_1=a_2=1.05,{\beta }_1={\beta }_2=1.03$ , and $m=25.0kg/s$ . Use $g=9.807m/s^2$ for consistency. Your answer should lie between $ and 6 kPa.

(c) Neglecting the weight of the elbow itself and the weight of the water in the elbow, calculator the x and z components of the anchoring force required to hold the elbow in place. Your final answer for the anchoring force should be given as a vector. $F\to Fxt\to +Fzk$ . Your answer for $F_x$ should lie between -120 and -1.40N, and your answer for $F_z$ should lie between 80 and 90N.

(d) Repeat Part (c) without neglecting the weight of the water in the elbow. Is it reasonable to neglect the weight of the water in this problem?

To determine

(a)

The expression of gage pressure using the head form of energy equation.

Answer to Problem 96P

The expression of gage pressure is $\left(h_L+R\right)\rho g$.

Explanation of Solution

Given information:

The elbow is $90°$ vertically oriented, the mass flow rate is $\dot{m}$, the radius of the elbow is $R$, the inner pipe diameter is $D$, the pressure at outlet is atmospheric pressure, the irreversible head loss through the elbow is $h_L$, the kinetic energy correction factor at inlet and outlet is same i.e. $\left({\alpha }_1={\alpha }_2=\alpha \right)$ and momentum flux correction factor at inlet and outlet is same i.e. $\left({\beta }_1={\beta }_2=\beta \right)$.

Write the expression of Bernoullis equation at inlet and outlet of elbow pipe.

  $\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1{}^2}{2g}+z_1=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2{}^2}{2g}+z_2+h_L$.....(I)

Here, the pressure at inlet is $P_1$, the pressure at outlet is $P_2$, the velocity at inlet is $V_1$, the velocity at outlet is $V_2$, the datum head at inlet is $z_1$, the datum head at outlet is $z_2$, the kinetic energy flux correction factor at inlet is ${\alpha }_1$, the kinetic energy flux correction factor at outlet is ${\alpha }_2$, the head loss is $h_L$, the density of the fluid is $\rho$ and the acceleration due to gravity is $g$.

Write the expression of gage pressure.

  $P_{\text{gage;1}}=P_1-P_{\text{atm}}$.....(II)

Here, the inlet pressure is $P_1$ and the atmospheric pressure is $P_{\text{atm}}$.

Consider, the velocity of flow at inlet and outlet is same i.e. $\left(V_1=V_2=V\right)$ and datum head at inlet is $0$.

Substitute, $P_{\text{atm}}$ for $P_2$

  $V$ for $V_1$ and $V_2$, $\alpha$ for ${\alpha }_1$ and ${\alpha }_2$, $R$ for $z_2$ and $0$ for $z_1$ in Equation (I).

  $\begin{array}{l}\frac{P_1}{\rho g}+\alpha \frac{V^2}{2g}+0=\frac{P_{atm}}{\rho g}+\alpha \frac{V^2}{2g}+R+h_L\\ \frac{P_1}{\rho g}-\frac{P_{atm}}{\rho g}=\alpha \frac{V^2}{2g}-\alpha \frac{V^2}{2g}+h_L+R\\ \frac{1}{\rho g}\left(P_1-P_{atm}\right)=\left(h_L+R\right)\\ \left(P_1-P_{atm}\right)=\left(h_L+R\right)\rho g\end{array}$....... (III)

Substitute $P_{\text{gage;1}}$ for $\left(P_1+P_{atm}\right)$ in Equation (III).

  $P_{\text{gage;1}}=\left(h_L+R\right)\rho g$.....(IV)

Conclusion:

The expression of gage pressure is $\left(h_L+R\right)\rho g$.

To determine

(b)

The gage pressure.

Answer to Problem 96P

The gage pressure is $5.96\text{kPa}$.

Explanation of Solution

Given information:

The density of the fluid is $998\text{kg}/{\text{m}}^{\text{3}}$, the diameter of the pipe is $10\text{cm}$, the radius of the elbow is $35\text{cm}$, the head loss is $0.259\text{m}$, the kinetic energy flux correction factor is $1.05$, the momentum flux correction factor is $1.03$, the mass flow rate is $25\text{kg}/\text{s}$ and the accelaretion due to gravity is $9.807\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.807\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.259\text{m}$ for $h_L$ and $35\text{cm}$ for $R$ in Equation (IV).

  $\begin{array}{c}{P}_{\text{gage;1}}=\left(0.259\text{m}+35\text{cm}\right)\times \left(998\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.807\text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(0.259\text{m}+\left(35\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right)\times \left(998\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.807\text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(0.609\text{m}\right)\times \left(9787.386\text{kg}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\right)\times \left(\frac{1\text{Pa}}{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}\right)\times \left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\\ =5.96\text{kPa}\end{array}$

Conclusion:

The gage pressure is $5.96\text{kPa}$.

To determine

(c)

The $x$ and $z$ component of anchoring force required to hold elbow in its place.

Answer to Problem 96P

The $x$ component of anchoring force is $128.671\text{N}$ in opposite direction.

The $z$ component of anchoring force is $81.88\text{N}$.

Explanation of Solution

Given information:

Neglect the weight of elbow and weight of the water.

Write the expression of $x$ component of anchoring force.

  $F_{x,\text{anchoring}}=\beta \dot{m}\left(V_2\cos \theta -V_1\right)-P_{\text{gage};1}A$.....(V)

Here, the momentum flux correction factor is $\beta$, the mass flow rate is $\dot{m}$, the velocity at outlet is $V_2$, the velocity at inlet is $V_1$, the gage pressure is $P_{\text{gage;1}}$, the area of flow is $A$ and the angle of orientation of elbow is $\theta$.

Substitute, $V$ for $V_1$ and $V_2$ in Equation (V).

  $\begin{array}{c}{F}_{x,\text{anchoring}}=\beta \dot{m}\left(V\cos \theta -V\right)-P_{\text{gage};1}A\\ =\beta \dot{m}V\left(\cos \theta -1\right)-P_{\text{gage};1}A\end{array}$.....(VI)

Write the expression of $z$ component of anchoring force.

  $F_{z,\text{anchoring}}=\beta \dot{m}{V}_2\sin \theta$.....(VII)

Here, the momentum flux correction factor is $\beta$, the mass flow rate is $\dot{m}$, the velocity at outlet is $V_2$ and the angle of orientation of elbow is $\theta$.

Substitute $V$ for $V_2$ in Equation (VII)

  $F_{z,\text{anchoring}}=\beta \dot{m}V\sin \theta$.....(VIII)

Write the expression of mass flow rate of fluid.

  $\begin{array}{l}\dot{m}=\rho AV\\ V=\frac{\dot{m}}{\rho A}\end{array}$.....(IX)

Here, the density of the fluid is $\rho$, area of flow of fluid is $A$ and the velocity of the fluid is $V$.

Write the expression of area of flow.

  $A=\frac{\pi }{4}{D}^2$.....(X)

Here, the diameter of the pipe is $D$.

Calculation:

Substitute $10\text{cm}$ for $D$ in Equation (X).

  $\begin{array}{c}A=\frac{\pi }{4}\times {\left(10\text{cm}\right)}^2\\ =\frac{\pi }{4}\times {\left(\left(10\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =\frac{\pi }{4}\times 0.01{\text{m}}^2\\ =0.00785{\text{m}}^2\end{array}$

Substitute $25\text{kg}/\text{s}$ for $\dot{m}$, $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $0.00785{\text{m}}^2$ for $A$ in Equation (IX).

  $\begin{array}{c}V=\frac{\left(25\text{kg}/\text{s}\right)}{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(0.00785{\text{m}}^2\right)}\\ =\frac{\left(25\text{kg}/\text{s}\right)}{\left(7.83\text{kg}/\text{m}\right)}\\ =3.18\text{m}/\text{s}\end{array}$

Substitute $1.03$ for $\beta$, $25\text{kg}/\text{s}$ for $\dot{m}$, $3.18\text{m}/\text{s}$ for $V$, $0.00785{\text{m}}^2$ for $A$, $90°$ for $\theta$ and $5.96\text{kPa}$ for $P_{\text{gage};1}$ in Equation (VI).

  $\begin{array}{c}{F}_{x,\text{anchoring}}=\left(1.03\right)\times \left(25\text{kg}/\text{s}\right)\times \left(3.18\text{m}/\text{s}\right)\times \left(\cos 90°-1\right)-\left(5.96\text{kPa}\right)\times \left(0.00785{\text{m}}^2\right)\\ =\left[\begin{array}{l}\left(81.885\text{kg}\cdot \text{m}/{\text{s}}^2\right)\times \left(-1\right)-\\ \left(5.96\text{kPa}\right)\times \left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\times \left(\frac{1\text{kg}/\text{m}\cdot {\text{s}}^2}{1\text{Pa}}\right)\times \left(0.00785{\text{m}}^2\right)\end{array}\right]\\ =\left\{\left(-81.885\text{kg}\cdot \text{m}/{\text{s}}^2\right)-\left(46.786\text{kg}\cdot \text{m}/{\text{s}}^2\right)\right\}\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\\ =-128.671\text{N}\end{array}$

So, the force is acting on opposite direction.

Substitute $1.03$ for $\beta$, $25\text{kg}/\text{s}$ for $\dot{m}$, $3.18\text{m}/\text{s}$ for $V$, and $90°$ for $\theta$. in Equation (VIII).

  $\begin{array}{c}{F}_{z,\text{anchoring}}=\left(1.03\right)\times \left(25\text{kg}/\text{s}\right)\times \left(3.18\text{m}/\text{s}\right)\times \left(\sin 90°\right)\\ =\left(1.03\right)\times \left(25\text{kg}/\text{s}\right)\times \left(3.18\text{m}/\text{s}\right)\times 1\\ =\left(81.88\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =81.88\text{N}\end{array}$

Conclusion:

The $x$ component of anchoring force is $128.671\text{N}$ in opposite direction.

The $z$ component of anchoring force is $81.88\text{N}$.

To determine

(d)

The $x$ and $z$ component of anchoring force required to hold elbow in its place without neglecting weight of the water in the elbow.

Answer to Problem 96P

The $x$ component of anchoring force is $128.671\text{N}$ in opposite direction.

The $z$ component of anchoring force is $39.71\text{N}$.

Explanation of Solution

Given information:

The weight of water in the elbow is considered.

Write the expression of $x$ component of anchoring force by considering the weight of water.

  $F_{x,\text{anchoring}}=\beta \dot{m}\left(V_2\cos \theta -V_1\right)-P_{\text{gage};1}A$.....(XI)

Here, the momentum flux correction factor is $\beta$, the mass flow rate is $\dot{m}$, the velocity at outlet is $V_2$, the velocity at inlet is $V_1$, the gage pressure is $P_{\text{gage;1}}$, the area of flow is $A$ and the angle of orientation of elbow is $\theta$.

Substitute, $V$ for $V_1$ and

  $V_2$ in Equation (XI)

  $\begin{array}{c}{F}_{x,\text{anchoring}}=\beta \dot{m}\left(V\cos \theta -V\right)-P_{\text{gage};1}A\\ =\beta \dot{m}V\left(\cos \theta -1\right)-P_{\text{gage};1}A\end{array}$.....(XII)

Write the expression of $z$ component of anchoring force.

  $F_{z,\text{anchoring}}=\beta \dot{m}{V}_2\sin \theta -W$.....(XIII)

Here, the momentum flux correction factor is $\beta$, the mass flow rate is $\dot{m}$, the velocity at outlet is $V_2$, the weight of water is $W$ and the angle of orientation of elbow is $\theta$.

Substitute $V$ for $V_2$ in Equation (XIII)

  $F_{z,\text{anchoring}}=\beta \dot{m}V\sin \theta -W$.....(XIV)

Write the expression of weight of water in the elbow.

  $W=mg$.....(XV)

Here, the mass of water is $m$ and acceleration due to gravity is $g$.

Write the expression of mass of water.

  $m=\rho \dot{V}$.....(XVI)

Here, the density of the fluid is $\rho$ and the volume is $\dot{V}$.

Write the expression of volume of water.

  $\dot{V}=\frac{A\pi R}{2}$.....(XVII)

Here, the area of flow is $A$ and the radius of the elbow is $R$.

Calculation:

Substitute $1.03$ for $\beta$, $25\text{kg}/\text{s}$ for $\dot{m}$, $3.18\text{m}/\text{s}$ for $V$, $0.00785{\text{m}}^2$ for $A$, $90°$ for $\theta$ and $5.96\text{kPa}$ for $P_{\text{gage};1}$ in Equation (XII).

  $\begin{array}{c}{F}_{x,\text{anchoring}}=\left(1.03\right)\times \left(25\text{kg}/\text{s}\right)\times \left(3.18\text{m}/\text{s}\right)\times \left(\cos 90°-1\right)-\left(5.96\text{kPa}\right)\times \left(0.00785{\text{m}}^2\right)\\ =\left[\begin{array}{l}\left(81.885\text{kg}\cdot \text{m}/{\text{s}}^2\right)\times \left(-1\right)-\\ \left(5.96\text{kPa}\right)\times \left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\times \left(\frac{1\text{kg}/\text{m}\cdot {\text{s}}^2}{1\text{Pa}}\right)\times \left(0.00785{\text{m}}^2\right)\end{array}\right]\\ =\left\{\left(-81.885\text{kg}\cdot \text{m}/{\text{s}}^2\right)-\left(46.786\text{kg}\cdot \text{m}/{\text{s}}^2\right)\right\}\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\\ =-128.671\text{N}\end{array}$

So, the force is acting on opposite direction.

Substitute $0.00785{\text{m}}^2$ for $A$ and $35\text{cm}$ for $R$ in Equation (XVII).

  $\begin{array}{c}\dot{V}=\frac{\left(0.00785{\text{m}}^2\right)\times \pi \times \left(35\text{cm}\right)}{2}\\ =\left(0.00785{\text{m}}^2\right)\times \left(1.57\right)\times \left(\left(35\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right)\\ =\left(0.00785{\text{m}}^2\right)\times \left(1.57\right)\times \left(0.35\text{m}\right)\\ =0.00431{\text{m}}^{\text{3}}\end{array}$

Substitute $0.00431{\text{m}}^{\text{3}}$ for $\dot{V}$ and $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in Equation (XVI).

  $\begin{array}{c}m=\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(0.00431{\text{m}}^{\text{3}}\right)\\ =\left(4.3\frac{\text{kg}}{{\text{m}}^{\text{3}}}\cdot {\text{m}}^{\text{3}}\right)\\ =4.3\text{kg}\end{array}$

Substitute $4.3\text{kg}$ for $m$ and $9.807\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (XV).

  $\begin{array}{c}W=\left(4.3\text{kg}\right)\times \left(9.807\text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(4.3\times 9.807\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\\ =42.17\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $1.03$ for $\beta$, $25\text{kg}/\text{s}$ for $\dot{m}$, $3.18\text{m}/\text{s}$ for $V$, $90°$ for $\theta$ and $42.17\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}$ for $W$ in Equation (XIII).

  $\begin{array}{c}{F}_{z,\text{anchoring}}=\left(1.03\right)\times \left(25\text{kg}/\text{s}\right)\times \left(3.18\text{m}/\text{s}\right)\times \left(\sin 90°\right)-\left(42.17\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\\ =\left(1.03\right)\times \left(25\text{kg}/\text{s}\right)\times \left(3.18\text{m}/\text{s}\right)\times 1-\left(42.17\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\\ =\left\{\left(81.88\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)-\left(42.17\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\right\}\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =39.71\text{N}\end{array}$

Conclusion:

The $x$ component of anchoring force is $128.671\text{N}$ in opposite direction.

The $z$ component of anchoring force is $39.71\text{N}$.