FE Civil Fluid Mechanics practice problems

The bulk modulus relates pressure change to fractional volume change: $E_v = -\Delta P/(\Delta V/V)$, so $|\Delta V/V| = \Delta P/E_v = (5 \times 10^6)/(2.2 \times 10^9) = 0.002273 = 0.227\%$. The 0.0227% choice is off by a factor of 10 (a power-of-ten slip in the exponents). The 2.27% choice is off by a factor of 10 the other way. The 0.114% choice uses $2E_v$ in the denominator, an erroneous factor of 2 that does not belong in the bulk-modulus definition.

1. Definition of bulk modulus:
   $E_v = -\frac{\Delta P}{\Delta V / V}$
2. Solve for the fractional volume change:
   $\left|\frac{\Delta V}{V}\right| = \frac{\Delta P}{E_v} = \frac{5 \times 10^6}{2.2 \times 10^9}$
3. Compute:
   $\left|\frac{\Delta V}{V}\right| = 0.002273 = 0.227\%$

FE Reference Handbook 10.6, p. 181

Liquids and gases respond to temperature in opposite ways. In a liquid, viscosity comes mainly from intermolecular cohesion; heating weakens those bonds, so liquid viscosity DROPS as temperature rises (hot oil flows more easily). In a gas, viscosity comes from molecular momentum exchange between layers; heating speeds the molecules up, so gas viscosity RISES with temperature. The two "both fluids behave the same" choices incorrectly assume oil and air respond identically. The choice with the oil becoming more viscous and the air becoming less viscous reverses the two behaviors.

1. In liquids, viscosity arises from intermolecular cohesion, which weakens as temperature rises.
2. Therefore liquid (oil) viscosity decreases with increasing temperature.
3. In gases, viscosity arises from momentum transfer between molecules, which increases as the molecules move faster.
4. Therefore gas (air) viscosity increases with increasing temperature.

FE Reference Handbook 10.6, p. 181

For a differential manometer between two points at the same elevation with water above the mercury on both sides, the pressure difference is $P_A - P_B = (\gamma_{Hg} - \gamma_w)h$. The water columns partially cancel, so you subtract water from mercury. $\gamma_{Hg} = 13.6 \times 9{,}810 = 133{,}416$ N/m³. $P_A - P_B = (133{,}416 - 9{,}810)(0.3) = 123{,}606 \times 0.3 = 37{,}082$ Pa $= 37.1$ kPa. The 40.0 kPa choice uses $\gamma_{Hg}h$ alone and forgets to subtract the water column. The 2.94 kPa choice uses $\gamma_w h$ (treats the gauge fluid as water). The 37,100 kPa choice forgets to convert Pa to kPa.

1. Specific weight of mercury:
   $\gamma_{Hg} = 13.6 \times 9{,}810 = 133{,}416\,\text{N/m}^3$
2. Differential manometer relation (water above mercury on both legs):
   $P_A - P_B = (\gamma_{Hg} - \gamma_w)h = (133{,}416 - 9{,}810)(0.3)$
3. Compute:
   $P_A - P_B = 123{,}606 \times 0.3 = 37{,}082\,\text{Pa} \approx 37.1\,\text{kPa}$

FE Reference Handbook 10.6, p. 183

Depth comes from the GAUGE pressure, not the absolute pressure. Gauge $= 250 - 101.3 = 148.7$ kPa. Seawater specific weight $= 1.03 \times 9{,}810 = 10{,}104$ N/m³. Depth $= P_{\text{gauge}}/\gamma_{sw} = 148{,}700/10{,}104 = 14.72$ m. The 15.16 m option uses freshwater $\gamma = 9{,}810$ instead of seawater. The 24.74 m option divides the full absolute pressure by the seawater specific weight (forgets to subtract atmospheric). The 25.49 m option uses the absolute pressure AND freshwater (two errors).

1. Gauge pressure:
   $P_{\text{gauge}} = P_{\text{abs}} - P_{\text{atm}} = 250 - 101.3 = 148.7\,\text{kPa}$
2. Seawater specific weight:
   $\gamma_{sw} = 1.03 \times 9{,}810 = 10{,}104\,\text{N/m}^3$
3. Depth:
   $h = \frac{P_{\text{gauge}}}{\gamma_{sw}} = \frac{148{,}700}{10{,}104} = 14.72\,\text{m}$

FE Reference Handbook 10.6, p. 182

A floating body displaces its own weight. Setting weight equal to buoyancy: $\gamma_{\text{wood}} V = \gamma_{\text{oil}} V_{\text{sub}}$, so the submerged fraction is $V_{\text{sub}}/V = \gamma_{\text{wood}}/\gamma_{\text{oil}} = SG_{\text{wood}}/SG_{\text{oil}} = 0.80/0.90 = 0.889$. The choice 0.80 uses the block’s SG directly, which only works when the fluid is water ($SG = 1$). The choice 0.111 gives the fraction ABOVE the surface (1 - 0.889). The choice 1.125 inverts the ratio (0.90/0.80), which is greater than 1 and physically impossible for a floating body.

1. Equilibrium (weight = buoyant force):
   $\gamma_{\text{wood}} V = \gamma_{\text{oil}} V_{\text{sub}}$
2. Submerged fraction:
   $\frac{V_{\text{sub}}}{V} = \frac{\gamma_{\text{wood}}}{\gamma_{\text{oil}}} = \frac{SG_{\text{wood}}}{SG_{\text{oil}}}$
3. Compute:
   $\frac{V_{\text{sub}}}{V} = \frac{0.80}{0.90} = 0.889$

FE Reference Handbook 10.6, p. 184

The resultant force uses the VERTICAL depth to the centroid: $F_R = \gamma h_C A$. The centroid lies at the midpoint of the slope, a slant distance $y_C = 2.5/2 = 1.25$ m from the surface. The vertical depth is $h_C = y_C \sin 60\degree = 1.25 \times 0.866 = 1.0825$ m. Area $A = 2 \times 2.5 = 5$ m². $F_R = 9{,}810 \times 1.0825 \times 5 = 53{,}095$ N $\approx 53.1$ kN. The 61.3 kN choice uses the slant distance 1.25 m as the depth (forgets the $\sin 60\degree$). The 30.7 kN choice uses $\cos 60\degree$ instead of $\sin 60\degree$. The 106.2 kN choice uses the vertical depth to the bottom edge ($2.5\sin 60\degree = 2.165$ m) instead of to the centroid.

1. Slant distance to the centroid:
   $y_C = \frac{2.5}{2} = 1.25\,\text{m}$
2. Vertical depth to the centroid:
   $h_C = y_C \sin 60\degree = 1.25 \times 0.866 = 1.0825\,\text{m}$
3. Gate area:
   $A = 2 \times 2.5 = 5\,\text{m}^2$
4. Resultant force:
   $F_R = \gamma h_C A = 9{,}810 \times 1.0825 \times 5 = 53{,}095\,\text{N} \approx 53.1\,\text{kN}$

FE Reference Handbook 10.6, p. 184

Continuity at a junction: inflow equals total outflow. $Q_{\text{in}} = A_{\text{in}} v = \frac{\pi}{4}(0.3)^2(2) = 0.07069 \times 2 = 0.1414$ m³/s. The combined outlet area is $A_{\text{out}} = \frac{\pi}{4}(0.2^2 + 0.15^2) = \frac{\pi}{4}(0.0625) = 0.04909$ m². With equal velocities, $v_{\text{out}} = Q_{\text{in}}/A_{\text{out}} = 0.1414/0.04909 = 2.88$ m/s. The 2.0 m/s choice assumes the velocity is unchanged. The 4.5 m/s choice uses only the 200 mm pipe area. The 8.0 m/s choice uses only the 150 mm pipe area.

1. Inlet flow rate:
   $Q_{\text{in}} = \frac{\pi}{4}(0.3)^2(2) = 0.1414\,\text{m}^3/\text{s}$
2. Combined outlet area:
   $A_{\text{out}} = \frac{\pi}{4}(0.2^2 + 0.15^2) = 0.04909\,\text{m}^2$
3. Outlet velocity (equal in both pipes):
   $v_{\text{out}} = \frac{Q_{\text{in}}}{A_{\text{out}}} = \frac{0.1414}{0.04909} = 2.88\,\text{m/s}$

FE Reference Handbook 10.6, p. 185

First find the velocity in the siphon from the reservoir surface to the outlet (Torricelli over the full drop of 4 m): $v = \sqrt{2g(4)} = \sqrt{78.48} = 8.86$ m/s, so the velocity head is $v^2/2g = 4$ m. Because the pipe diameter is constant, this velocity exists everywhere, including the crest. Apply Bernoulli from the surface (datum, $P = 0$, $v \approx 0$, $z = 0$) to the crest ($z = +2$ m): $0 = P_{\text{crest}}/\gamma + v^2/2g + 2$, so $P_{\text{crest}}/\gamma = -2 - 4 = -6$ m, and $P_{\text{crest}} = -6 \times 9{,}810 = -58{,}860$ Pa $= -58.9$ kPa. The -19.6 kPa option includes only the elevation head (-2 m). The -39.2 kPa option includes only the velocity head (-4 m). The -78.5 kPa option uses the wrong velocity head from an 8 m drop.

1. Velocity head from the full 4 m drop to the outlet:
   $\frac{v^2}{2g} = 4\,\text{m}$
2. Bernoulli from surface (datum, $P=0$, $v\approx 0$) to the crest at $z = +2$ m:
   $0 = \frac{P_{\text{crest}}}{\gamma} + \frac{v^2}{2g} + 2$
3. Solve for the pressure head at the crest:
   $\frac{P_{\text{crest}}}{\gamma} = -2 - 4 = -6\,\text{m}$
4. Convert to pressure:
   $P_{\text{crest}} = -6 \times 9{,}810 = -58{,}860\,\text{Pa} \approx -58.9\,\text{kPa}$

FE Reference Handbook 10.6, p. 185

Hazen-Williams: $v = k_1 C R_H^{0.63} S^{0.54}$. For a full pipe, $R_H = D/4 = 0.5/4 = 0.125$ m. Then $R_H^{0.63} = 0.125^{0.63} = 0.270$ and $S^{0.54} = 0.002^{0.54} = 0.0349$. $v = 0.849 \times 120 \times 0.270 \times 0.0349 = 0.96$ m/s. The $1.49\,\text{m/s}$ option uses the USCS constant $k_1 = 1.318$ instead of the SI value 0.849. The $2.30\,\text{m/s}$ option uses $R_H = D$ instead of D/4. The $0.55\,\text{m/s}$ option raises the slope to the wrong power ($S^{0.63}$ instead of $S^{0.54}$).

1. Hydraulic radius of a full circular pipe:
   $R_H = \frac{D}{4} = \frac{0.5}{4} = 0.125\,\text{m}$
2. Evaluate the power terms:
   $R_H^{0.63} = 0.125^{0.63} = 0.270, \quad S^{0.54} = 0.002^{0.54} = 0.0349$
3. Hazen-Williams velocity:
   $v = 0.849 \times 120 \times 0.270 \times 0.0349 = 0.96\,\text{m/s}$

FE Reference Handbook 10.6, p. 189

Check the regime first: $Re = vD/\nu = (0.5)(0.04)/(8 \times 10^{-5}) = 250$, which is laminar ($Re < 2{,}100$), so $f = 64/Re = 64/250 = 0.256$. Head loss: $h_f = f(L/D)(v^2/2g) = 0.256 \times (50/0.04) \times (0.5^2/19.62) = 0.256 \times 1{,}250 \times 0.01274 = 4.08$ m. Convert to pressure: $\Delta P = \rho g h_f = 900 \times 9.81 \times 4.08 = 36{,}020$ Pa $\approx 36.0$ kPa. The 4.08 kPa choice reports the head loss in meters (4.08 m) instead of converting to pressure. The 40.0 kPa choice uses $\rho = 1{,}000$ (water) instead of the oil density. The 2.81 kPa choice wrongly assumes turbulent flow with $f = 0.02$.

1. Reynolds number:
   $Re = \frac{vD}{\nu} = \frac{0.5 \times 0.04}{8 \times 10^{-5}} = 250 \;(\text{laminar})$
2. Laminar friction factor:
   $f = \frac{64}{Re} = \frac{64}{250} = 0.256$
3. Head loss:
   $h_f = f \frac{L}{D} \frac{v^2}{2g} = 0.256 \times 1{,}250 \times 0.01274 = 4.08\,\text{m}$
4. Pressure drop:
   $\Delta P = \rho g h_f = 900 \times 9.81 \times 4.08 = 36{,}020\,\text{Pa} \approx 36.0\,\text{kPa}$

FE Reference Handbook 10.6, p. 187

For a plate moving away from the jet, the relevant velocity is the RELATIVE velocity $(v - u)$. Both the mass flow striking the plate and the momentum change scale with this relative velocity, so $F = \rho A (v - u)^2 = 1{,}000 \times 0.002 \times (30 - 10)^2 = 1{,}000 \times 0.002 \times 400 = 800$ N. The 1,800 N choice uses the absolute jet velocity $v$ (ignores the plate motion): $\rho A v^2 = 1{,}800$ N. The 40 N choice forgets to square the relative velocity: $\rho A (v-u) = 40$ N. The 1,200 N choice mixes a relative velocity in one factor with the absolute velocity in the other: $\rho A (v-u) v = 1{,}200$ N.

1. Relative velocity of the jet with respect to the plate:
   $v_{\text{rel}} = v - u = 30 - 10 = 20\,\text{m/s}$
2. Force on the moving flat plate:
   $F = \rho A (v - u)^2 = 1{,}000 \times 0.002 \times 20^2$
3. Compute:
   $F = 1{,}000 \times 0.002 \times 400 = 800\,\text{N}$

FE Reference Handbook 10.6, p. 193

For a $180\degree$ reversal, the velocity changes from $+v$ to $-v$, a change of 2v, so $F = \rho Q (v - (-v)) = 2\rho Q v$. Area $A = \frac{\pi}{4}(0.04)^2 = 0.0012566$ m². $Q = A v = 0.0012566 \times 25 = 0.031416$ m³/s. $F = 2 \times 1{,}000 \times 0.031416 \times 25 = 1{,}571$ N $= 1.57$ kN. The 0.785 kN choice uses the flat-plate result $\rho Q v$ (factor 1 instead of 2). The 3.14 kN choice applies a factor of 4. The 6.28 kN choice uses the radius in place of the diameter when computing the area (an area four times too large).

1. Jet area:
   $A = \frac{\pi}{4}(0.04)^2 = 1.2566 \times 10^{-3}\,\text{m}^2$
2. Flow rate:
   $Q = A v = 1.2566 \times 10^{-3} \times 25 = 0.031416\,\text{m}^3/\text{s}$
3. Force for a $180\degree$ reversal:
   $F = 2\rho Q v = 2 \times 1{,}000 \times 0.031416 \times 25 = 1{,}571\,\text{N} \approx 1.57\,\text{kN}$

FE Reference Handbook 10.6, p. 191

The rise of water in the stagnation tube is the velocity head: $h = v^2/2g$. Solving, $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.5} = \sqrt{9.81} = 3.13$ m/s. The 2.21 m/s option drops the factor of 2 and uses $\sqrt{gh} = 2.21$ m/s. The 6.27 m/s option doubles the correct velocity. The 9.81 m/s option forgets the square root entirely and reports $2gh = 9.81$.

1. The rise equals the velocity head:
   $h = \frac{v^2}{2g}$
2. Solve for velocity:
   $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.5}$
3. Compute:
   $v = \sqrt{9.81} = 3.13\,\text{m/s}$

FE Reference Handbook 10.6, p. 200

For an inclined venturi the head term includes elevation: $\frac{P_1 - P_2}{\gamma} + z_1 - z_2 = \frac{30{,}000}{9{,}810} - 0.3 = 3.058 - 0.3 = 2.758$ m. Throat area $A_2 = \frac{\pi}{4}(0.075)^2 = 0.004418$ m². Area ratio: $(D_2/D_1)^4 = (0.5)^4 = 0.0625$, so $1 - 0.0625 = 0.9375$. $Q = C_v A_2 \sqrt{\frac{2g(2.758)}{0.9375}} = 0.98 \times 0.004418 \times \sqrt{57.7} = 0.98 \times 0.004418 \times 7.60 = 0.0329$ m³/s. The 0.0346 choice ignores the elevation difference (uses 3.058 m). The 0.0363 choice adds the elevation instead of subtracting it. The 0.132 choice uses the upstream area $A_1$ instead of the throat area.

1. Effective head (throat is 0.3 m above the inlet):
   $\frac{P_1 - P_2}{\gamma} + z_1 - z_2 = \frac{30{,}000}{9{,}810} - 0.3 = 2.758\,\text{m}$
2. Throat area:
   $A_2 = \frac{\pi}{4}(0.075)^2 = 4.418 \times 10^{-3}\,\text{m}^2$
3. Area ratio correction:
   $1 - \left(\frac{D_2}{D_1}\right)^4 = 1 - 0.0625 = 0.9375$
4. Flow rate:
   $Q = C_v A_2 \sqrt{\frac{2g(2.758)}{0.9375}} = 0.98 \times 4.418 \times 10^{-3} \times 7.60 = 0.0329\,\text{m}^3/\text{s}$

FE Reference Handbook 10.6, p. 200

Under Froude similitude, the time scale is the square root of the length scale: $t_r = \sqrt{L_r} = \sqrt{1/36} = 1/6$. So $t_m = t_p \times \sqrt{L_r} = 60 \times (1/6) = 10$ s. The 1.67 s option scales by the length ratio directly ($60/36 = 1.67$). The 360 s option multiplies by 6 instead of dividing. The 60 s option assumes no time scaling. The model runs faster because both length and velocity are scaled down, and velocity scales as $\sqrt{L_r}$.

1. Froude time scale:
   $t_r = \sqrt{L_r} = \sqrt{\frac{1}{36}} = \frac{1}{6}$
2. Model time:
   $t_m = t_p \sqrt{L_r} = 60 \times \frac{1}{6} = 10\,\text{s}$

FE Reference Handbook 10.6, p. 202

Count the variables, including the dependent one: $\Delta P, v, \rho, \mu, D, L, \varepsilon$ gives $n = 7$. The base dimensions are mass, length, and time (M, L, T), so $r = 3$. Buckingham Pi: $k = n - r = 7 - 3 = 4$. The choice 3 subtracts one too many (treats $r = 4$). The choice 5 uses $r = 2$, forgetting that the variables span all three of M, L, and T. The choice 7 reports $n$ alone without subtracting the dimensions.

1. Variables: $\Delta P, v, \rho, \mu, D, L, \varepsilon$:
   $n = 7$
2. Base dimensions (M, L, T):
   $r = 3$
3. Buckingham Pi theorem:
   $k = n - r = 7 - 3 = 4$

FE Reference Handbook 10.6, p. 202