Problems 3.6,9,38,46,102,114,127,135,153; DP 3.1, Note: in 3.6, take the width of the slot to be "b"

3.6 The exit velocity from the slot at depth h is approximately u = (2g(h-z))^1/2 from Bernoulli's equation, where z = 0 at the slot centerline. The slot vertical height is 2L, with horizontal width b. The volume flow rate Q is an integral of v dot dA over the slot, giving int udA. Substitute for u, and integrate from z = -L to z = +L. This gives [2b/3](2g)^1/2 [(h+L)^3/2 - (h-L)^3/2 = Q (ans). From the definition of the derivative this gives Q = 2bL (2gh)^1/2 in the limit as L goes to zero.

3.9 Apply the Reynolds Transport Theorem to the tank containing seawater. Salinity S is rho_salt / rho_seawater . Salinity is "beta" for this problem; that is, the salt per unit mass of seawater.

3.38 Take a fixed cylindrical control volume enclosing the fluid between the two disks at time 0, with upper surface moving down with velocity V_o. Find V(r) by applying RTT for mass for the incompressible fluid. This is dM/dt_syst = dM/dt_CV + int rho v dot dA. dM/dt_syst = 0. dM/dt_CV = d( pi r^2 h) / dt , where h = h(t). int rho v dot dA = rho 2 pi r h V. Thus r^2 dh/dt = 2rhV = 0. But dh/dt = -V_o . Thus V = V_o r / 2h.

3.46 Take a control volume enclosing the input jet and the two inclined jets. The tangential force is taken to be zero. By the RTT for tangential momentum we have [0 = 0 + int rho v v dot dA ]_tangent. This is 0 = dot^m_2 - dot^m_3 - dot^m_1 V cos theta = alpha dot^m V -

(1 - alpha )dot^m V - dot^m V cos theta . Thus alpha = (1 + cos theta )/2.

3.102 Take a control volume of depth b enclosing a portion of the hydraulic jump. Apply the RTT for mass and momentum. The flow is steady, so from mass conservation we have rho V_1 h_1 b = rho V_2 h_2 b , where 1 is the input and 2 is the output. Thus V_2 = V_1 h_1 / h_2 . Inserting this in the momentum RTT result gives a quadratic equation for h_2 / h_1 , with solution -1/2 + (1/2)(1 + 8V_1^2 / gh_1 )^1/2 . (ans).

3.114 There is no torque on the sprinkler. The angular momentum equation is obtained by taking the cross product of the momentum equation with the radius vector. For the steady flow of this problem, 0 = 0 + int rho r x ( v - omega x r ) dot v dot dA , where omega is the angular velocity of the sprinkler and v is the velocity of the fluid relative to the nozzle. The speed of the fluid exiting each nozzle is Q/3A = 6.5 m/s. See example 3.15. The final omega is V_o cos theta / R, giving 414 r/min and 317 r/min.

3.127 Apply the RTT for energy for a control volume around the river. The heat in is 55,000,000 W, which is the difference between the internal energy convected out dot^m c_p T_out and the internal energy convected in dot^m c_p T_in . Solve for T_out = 23.14 deg C.

3.135 Apply Bernoulli's equation to the flows through the pump and the turbine. B_1 = B_2 + gh_t + gh_f for the turbine flow. In terms of head, 0 + 0 + 150 ft = 0 + 0 + 25 + 17 + h_t . Solve for h_t = 108 ft. Turbine power P = gamma Q h_t , where gamma is rho g . Q is 33.4 ft^3 /s. Thus P is 62.4x33.4x108 ft-lbf / s = 410 hp (check units). Similarly P_pump = 540 hp.

3.153 From Bernoulli's equation with no losses the velocity out of the hole is V_o = (2h(H-h))^1/2 , and the fall time is t_fall = (2h/g)^1/2 . Thus the horizontal distance is V_o t_fall =

2(h(H-h))^1/2 . Maximum x occurs at h = H/2 . etc.

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The problem is to select a pump with minimum cost

with given performance characteristics. It must deliver no

less than 1.0 ft^3 / s of water, but now rotate slower than 10

rev/s. The cost is assumed to be proportional to the power

input to the pump. The relevant dimensional parameters of

the pump performance are the pump work rate per mass flow

rate gh_p, the volume flow rate Q, the water density rho, the

shaft rotation rate n, the impeller diameter D, the power to

the water rho Q g h_p , and the power input to the pump.

The relevant dimensionless parameters for the pump are

functions of the dimensionless volume flow rate zeta =

Q/nD^3 . The dimensionless pump head is phi = gh_p / n^2

D^2 = 6.04 - 161 zeta . The efficiency eta = 70 zeta - 91500

zeta^3. The dimensionless power to the water pi_W = P_W

/ (rho n^3 D^5) = zeta phi . The dimensionless power input

P_I is pi_I / eta which should be minimized as a function of

zeta . Since Q and n are fixed, the pump impeller diameter to

give minimum power can be determined.

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from the lower to the upper reservoir. Pipe friction losses

are h_f = 27 V^2 / 2g , where V is the pipe velocity Q/A . If

the pump is 75% efficient, what horsepower is needed to drive

it? The pipe diameter is 6 inches. Efficiency eta = P_water / P_$

Q = 1500 gpm / 448.8 gpm/ ft^3/s = 3.34 ft^3 / s .

V = Q/A = 3.34 / pi (3/12)^2 = 17 ft/s

h_f = 27 17^2 / 2x32.2 = 121 ft.

From Bernoulli's equation:

p1/ rho g + V1^2 / 2 + z1 = p2/ rho g + V2^2 / 2 + z2 + h_f - h_p

0 + 0 + 50 = 0 + 0 + 150 + 121 - h_p

Thus, h_p = 221 ft.

P_pump = P_$ = P_water / eta . P_water = rho g Q h_p

P_pump = 62.4 lbf / ft^3 x 3.34 ft^3 / s x 221 ft / 0.75 = 61,600 ft lbf / s

= 61,600 / 550 = 112 hp

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rotation rate, varies with the flow rate, resulting in a pump performance

curve. Suppose that this pump is 75% efficient and is used for the system

of 3.141. Estimate (a) the flow rate in gpm, and (b) the horsepower.

The pump head h_p is 300 - 50 Q from fig p3.142. Thus, Bernoulli's

equation becomes:

p1/ rho g + V1^2 / 2 + z1 = p2/ rho g + V2^2 / 2 + z2 + h_f - h_p

0 + 0 + 50 = 0 + 0 + 150 + 27 V^2 / 2g - (300 - 50 Q)

Substitute V = Q / A to get a quadratic equation for Q:

Q^2 + 4.6 Q - 18.4 = 0 . Solve to give Q = 2.57 ft^3 / s .

Thus, P_$ = rho g Q h_p / eta = 62.4x2.57x[300 - 50x2.57] / .75 = 36,700 ft lbf / s

= 36,700 / 550 = 67 hp

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your design in terms of the dimensionless pump head phi, the dimensionless flow

rate zeta , and the efficiency eta . The cost C for the pump is to be minimized

with the assumption that C is proportional to P_$ only (is this realistic?). You

can express P_water as rho n^3 D^5 from dimensional analysis, so pi_water

= P_water / rho n^3 D^5 is dimensionless pump power delivered, and pi_$ =

pi_water / eta is dimensionless power cost. You have the same system as in

3.142 except for the pump, so you can use the same expression for frictional

head loss. Pump costs can presumably be reduced by reducing the flow rate,

since this reduces the frictional head, but there may be a limit due to decreased

efficiency besides the given constrain on minimum flow rate of 1 ft^3 / s . Set

up a spreadsheet and do sample calculations of C for the given constraints, and

then do "cut and try" calculations within the constraints to see if you can reduce

C further. Report your results and conclusions as a technical memo; that is, a

brief summary of the problem and your design, followed by supporting theory,

calculations, plots, etc. (minimum cost occurs for about D = 1.7 ft impeller

diameter).