NAME COLLEGE OF ENGINEERING AND COMPUTER SCIENCE MECHANICAL ENGINEERING

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Solutions toMidterm Examination

1 . (37 points) The gate shown in the figure at the right pivots about the hinge and is held in place by the 2000 lb_f counterweight, W. Find the water depth, h. Assume the weight of the gate is negligible.

For the vertical part of the gate the resultant force of the water is given by the equation F_R = h_cA where h_c = h/2 and A = (4 ft)h. This gives F_R = .h/2)(4 ft)h = h^(2 ft).

The distance from the top of the water to the point where this resultant force acts, y_R is given by the following equation

The moment of inertia about the centroid is given by the equation I_xc = ab³/12 where a is the width of the area and b is the height. In our case, I_xc = (4 ft)h³/12. Substituting value of I_xc into the equation for y_R gives

A moment balance about the hinge is shown in the figure at the left. The distance, d, from the hinge to the location of the resultant force is simply d = h – y_R – h = 2h/3 = h/3.

T he moment balance about the hinge has only two forces, F_R and W. This balance requires that F_Rd = W(3 ft).

Substituting d = h/3 and F_R = h^(2 ft) into this moment balance with W = 2000 lb_f and  for water = 62.4 lb_f/ft³ gives

h = 5.24 ft

2

(3)

(4)

. (37 points) Air flows through a Venturi channel with a rectangular cross section. The width of the channel is 0.06 m; the heights at point (1) and the exit are 0.04 m. The height at the throat is 0.02 m. Compressibility and viscous effects are negligible.

(a) What is the flow rate when the water in the small tube attached to the static pressure tap at the throat is drawn up 0.10 m as shown?

Since the flow can be assumed to be inviscid and incompressible, we can use the Bernoulli equation to relate the various state points in the flow. In particular we can relate conditions at the throat (3) to conditions at the exit (4).

Since both locations are at the same elevation, z₄ – z₃ = 0. We also have p₄ = 0 because this point is a free jet (open to the atmosphere.) At the throat, the manometer equation for the water height of 0.10 m gives p₃ + (0.1 m)_w = 0 or p₃ = –(0.1 m)_w. Finally we can use the continuity equation, V₄A₄ = V₃A₃, to give V₄(0.06 m)(0.04 m) = V₃(0.06 m)(0.02 m) or V₃ = 2V₄. Making these substitutions in the Bernoulli equation gives.

We are not given any data on the atmospheric pressure of the temperature of the flowing air. We can assume that the air is at standard conditions for which we find  = 1.23 kg/m³ in the inside front cover of the text. Using this datum and a value of 9800 N/m³ for _w (also from the tables in the inside front cover) we can solve for the exit velocity, V₄.

The flow rate, Q = V₄A₄ = (23.1 m/s)(0.06 m)(0.04 m) or Q = 0.0554 m³/s.

(b) Determine the height, h₂, at section (2) for the flow rate you found in part (a).

We can again apply Bernoulli’s equation, this time between section (2) and the exit to find the desired height, h₂. In this case Bernoulli’s equation becomes

Since both locations are at the same elevation, z₄ – z₂ = 0. As before, we have p₄ = 0, and, the manometer equation for the water height h₂ at station (2) is p₂ + _w(0.05 m) = 0 or p₂ = ‑(0.05 m)_w. Making these substitutions and the value of V₄ = 23.1 m/s found above gives.

Solving this equation for V₂ and substituting data for  and _w as before gives

Taking the square root gives V₂ = 36.5 m/s. Finally, we can apply the continuity equation to find the height, h₂

h₂ = 0.02534 m

(c) Determine the pressure at section (1) required to produce this flow.

This time we apply the Bernoulli equation between points (1) and (4).

Since both locations are at the same elevation, z₄ – z₁ = 0. As before, we have p₄ = 0, Since the flow areas at both points (1) and (4) are the same, the two velocities are the same. These substitutions in the Bernoulli equation give

p₁ = 0

3 . (37 points) Water flows through the 20^o reducing bend shown in the figure at the right at a rate of 0.025 m³/s. The flow is frictionless, gravitational effects are negligible, and the pressure at section (1) is 150 kPa. Determine the x and y components of the force required to hold the bend in place.

We can use the general momentum balance in both the x and y directions to get two equations for the two required force components. The general equation is shown below.

Here we have one inlet and one outlet and we assume steady flow. This gives the following balance equations in the two directions. Note that the pressure force is in the +x direction at station 1 with no y component and we have to resolve the components of the pressure force at station 2; the x component is –p₂A₂cos 20^o, because it acts in the minus-x direction; the y component is p₂A₂sin20^o, because it acts in the +y direction. F_x and F_y denote the x and y components of force required to hold the bend in place.

For this system, V_x,1 = V₁, V_y,1 = 0, V_x,2 = V₂cos20^o, and V_y,2 = –V₂sin20^o. The negative sign for V_y,2 is because it is in the minus-y direction.

The continuity equation tells us that the mass flow rate is the same at both inlet and outlet. In addition, the water has constant density so the volume flow rate is constant. We can compute the mass flow rate and the velocities at inlet and outlet from continuity relations.

Using the density of water as 998 kg/m³ we find the mass flow rate as

We can substitute this mass flow rate and the following velocity components into our momentum equations: V_x,1 = V₁, V_y,1 = 0, V_x,2 = V₂cos20^o, and V_y,2 = –V₂sin20^o. This gives:

We are given P₁ = 150 kPa, but we do not know the value of P₂. Since the flow is frictionless and we have already assumed that it has constant density, we can use Bernoulli’s equation to relate points (1) and (2) and solve for P₂.

We are told that gravitational effects are negligible so we can ignore the gz term. This gives

We now have the required information to compute the resultant forces.

F_x = –882 N

F_y = –156 N

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