Chapter 2, Problem 127P

The rotating parts of a hydroelectric power plant having power capacity W have a rotational synchronous speed it. The weight of the rotating parts (the hydroturbine and its electric generator) is supported in a thrust bearing having amulet form between D and d diameters as sketched the thrust hearing is operated with a very thin oil film of thickness e and dynamic viscosity. It is armed that the oil is a Newtonian fluid and the velocity is approximated as linear in the hearing. Calculate the ratio of lost power in the thrust heating to the produced power in the hydraulic power plant. Use
   $\begin{array}{l}W=48.6MW,\mu =0.035Pa\cdot s,n=500\text{rpm},e=0.25\text{mm,}D=3.2m,\\ d=2.4\text{}m\end{array}$

To determine

The ratio of lost power in the thrust bearing.

Answer to Problem 127P

The ratio of lost power in the thrust bearing is $4.46\times {10}^{-3}$.

Explanation of Solution

Given information:

The larger diameter of the bearing is $3.2\text{m}$, the smaller diameter of the bearing is $2.4\text{m}$, the power capacity of power plant is $48.6\text{MW}$, the speed of the plant is $500\text{rpm}$, the dynamic viscosity of the fluid is $0.035\text{Pa}\cdot \text{s}$ and the thickness of the film is $0.25\text{mm}$.

Write the expression for the area in differential form.

  $dA=2\pi rdr$   ...... (I)

Here, the area of the plant is $A$ and the radius of the bearing is $r$.

Write the expression for the radius of the smaller bearing.

  $R_1=\frac{d}{2}$   ...... (II)

Here, the radius of the smaller bearing is $R_1$ and the diameter of the smaller bearing is $d$.

Write the expression for the radius of the larger bearing.

  $R_2=\frac{D}{2}$   ...... (III)

Here, the radius of the larger bearing is $R_2$ and the diameter of the larger bearing is $D$.

Write the expression for the volume of the plant in differential form.

  $dV=\omega dr$   ...... (IV).

Here, the volume of the plant is $V$ and the angular velocity is $\omega$.

Write the expression for the angular velocity.

  $\omega =\frac{2\pi \dot{n}}{60}$   ...... (V)

Here, the speed of the plant is $\dot{n}$.

Write the expression for the shear stress due to viscosity.

  $d\tau =\mu \frac{dV}{dh}$   ...... (VI)

Here, the shear stress due to viscosity is $\tau$, the thickness of film is $h$ and the dynamic viscosity is $\mu$.

Write the expression for the force due to shear stress.

  $dF=\tau dA$   ...... (VII)

Here, the force due to shear stress is $F$.

Write the expression for the power loss due to viscosity of oil.

  $dP=FdV$   ...... (VIII)

Here, the power loss is $P$.

Write the expression for the net power produced.

  $P_{net}=\dot{W}-P$   ...... (IX)

Here, the net power produced is $P_{net}$ and the power capacity of the plant is $\dot{W}$.

Write the expression for the ratio of power loss.

  $R=\frac{P}{P_{net}}$   ...... (X)

Here, the ratio of power loss is $R$.

Calculation:

Substitute $2.4\text{m}$ for $d$ in Equation (II).

  $\begin{array}{c}{R}_1=\frac{2.4\text{m}}{2}\\ =1.2\text{m}\end{array}$

Substitute $3.2\text{m}$ for $D$ in Equation (III).

  $\begin{array}{c}{R}_2=\frac{3.2\text{m}}{2}\\ =1.6\text{m}\end{array}$

Substitute $500\text{rpm}$ for $\dot{n}$ in Equation (V).

  $\begin{array}{c}\omega =\frac{2\pi \times 500\text{rpm}}{60}\\ =\frac{3141.6}{60}\text{rad}/\text{s}\\ =52.36\text{rad}/\text{s}\end{array}$

Substitute $0.035\text{Pa}\cdot \text{s}$ for $\mu$ and $0.25\text{mm}$ for $h$ in Equation (VI).

  $\begin{array}{c}d\tau =\frac{0.035\text{Pa}\cdot \text{s}}{0.25\text{mm}}\times dV\\ =\frac{\left(0.035\text{Pa}\cdot \text{s}\times \frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\right)}{\left(0.25\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}\times dV\\ =140\text{N}\cdot \text{s}/{\text{m}}^3\times dV\end{array}$   ...... (XI)

Substitute $\omega dr$ for $dV$ in Equation (XI).

  $d\tau =140\text{N}\cdot \text{s}/{\text{m}}^3\times \left(\omega dr\right)$   ...... (XII)

Substitute $52.36\text{rad}/\text{s}$ for $\omega$ in Equation (XII).

  $\begin{array}{c}d\tau =\left(140\text{N}\cdot \text{s}/{\text{m}}^3\right)\times \left(52.36\text{rad}/\text{s}\right)\times dr\\ =\left(140\times 52.36\right)\text{N}/{\text{m}}^3\times dr\\ =\left(7330.4\text{N}/{\text{m}}^3\right)\times dr\end{array}$   ...... (XIII)

Integrate Equation (XIII) under the lower limit $1.2\text{m}$ and upper limit $1.6\text{m}$ for $r$.

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{\tau }{\int }}d\tau }=\left(7330.4\text{N}/{\text{m}}^3\right)\times {\displaystyle \underset{1.2}{\overset{1.6}{\int }}dr}\\ \tau =\left(7330.4\text{N}/{\text{m}}^3\right)\times \left[1.6\text{m}-1.2\text{m}\right]\\ \tau =2932.16\text{N}/{\text{m}}^2\end{array}$

Substitute $2932.16\text{N}/{\text{m}}^2$ for $\tau$ and $2\pi rdr$ for $dA$ in Equation (VII).

  $\begin{array}{c}dF=\left(2932.16\text{N}/{\text{m}}^2\right)\times \left(2\pi rdr\right)\\ =\left(18423.3\text{N}/{\text{m}}^2\right)\times \left(rdr\right)\end{array}$   ...... (XIV)

Integrate Equation (XIV) under the lower limit $1.2\text{m}$ and upper limit $1.6\text{m}$ for $r$.

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{F}{\int }}dF}=\left(18423.3\text{N}/{\text{m}}^2\right)\times {\displaystyle \underset{1.2}{\overset{1.6}{\int }}rdr}\\ F=\left(18423.3\text{N}/{\text{m}}^2\right)\times \frac{1}{2}\left[{\left(1.6\text{m}\right)}^2-{\left(1.2\text{m}\right)}^2\right]\\ F=10317.04\text{N}\end{array}$

Substitute $10317.04\text{N}$ for $F$ and $\omega dr$ for $dV$ in Equation (VIII).

  $dP=\left(10317.04\text{N}\right)\times \left(\omega dr\right)$   ...... (XV)

Substitute $52.36\text{rad}/\text{s}$ for $\omega$ in Equation (XV).

  $\begin{array}{c}dP=\left(10317.4\text{N}\right)\times \left(52.36\text{rad}/\text{s}\right)\times dr\\ =\left(10317.4\times 52.36\right)\text{N}/\text{s}\times \left(dr\right)\\ =\left(540219.06\text{N}/\text{s}\right)\times dr\end{array}$   ...... (XVI)

Integrate Equation (XVI) under the lower limit $1.2\text{m}$ and upper limit $1.6\text{m}$ for $r$.

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{P}{\int }}dP}=\left(540219.06\text{N}/\text{s}\right)\times {\displaystyle \underset{1.2}{\overset{1.6}{\int }}dr}\\ P=\left(540219.06\text{N}/\text{s}\right)\times \left[1.6\text{m}-1.2\text{m}\right]\\ P=\left(\left(216087.6\text{N}\cdot \text{m}/\text{s}\right)\times \frac{1\text{MW}}{{10}^6\text{N}\cdot \text{m}/\text{s}}\right)\\ P=0.216\text{MW}\end{array}$

Substitute $0.216\text{MW}$ for $P$ and $48.6\text{MW}$ for $\dot{W}$ in Equation (IX).

  $\begin{array}{c}{P}_{net}=\left(48.6\text{MW}\right)-\left(0.216\text{MW}\right)\\ =\left(48.6-0.216\right)\text{MW}\\ =\text{48}\text{.384}\text{MW}\end{array}$

Substitute $0.216\text{MW}$ for $P$ and $48.384\text{MW}$ for $P_{net}$ in Equation (X).

  $\begin{array}{c}R=\frac{0.216\text{MW}}{48.384\text{MW}}\\ =\frac{0.216}{48.384}\\ =4.46\times {10}^{-3}\end{array}$

Conclusion:

The ratio of lost power in the thrust bearing is $4.46\times {10}^{-3}$.