Numerical Methods for Engineers
7th Edition
ISBN: 9780073397924
Author: Steven C. Chapra Dr., Raymond P. Canale
Publisher: McGraw-Hill Education
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Chapter 32, Problem 8P
The flow through porous media can be described by the Laplace equation
whereh is head. Use numerical methods to determine the distribution of head for the system shown in Fig. P32.8.
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Chapter 32 Solutions
Numerical Methods for Engineers
Ch. 32 - Perform the same computation as in Sec. 32.1, but...Ch. 32 - 32.2 Develop a finite-element solution for the...Ch. 32 - Compute mass fluxes for the steady-state solution...Ch. 32 - Compute the steady-state distribution of...Ch. 32 - Two plates are 10cmapart, as shownin Fig.P32.5....Ch. 32 - 32.6 The displacement of a uniform membrane...Ch. 32 - 32.7 Perform the same computation as in Sec....Ch. 32 - The flow through porous media can be described by...Ch. 32 - 32.9 The velocity of water flow through the...Ch. 32 - 32.10 Perform the same computation as in Sec....
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- Four-bar linkage mechanism, AB=40mm, BC=60mm, CD=70mm, AD=80mm, =60°, w1=10rad/s. Determine the direction and magnitude of w3 using relative motion graphical method. A B 2 3 77777 477777arrow_forwardThe evaporator of a vapor compression refrigeration cycle utilizing R-123 as the refrigerant isbeing used to chill water. The evaporator is a shell and tube heat exchanger with the water flowingthrough the tubes. The water enters the heat exchanger at a temperature of 54°F. The approachtemperature difference of the evaporator is 3°R. The evaporating pressure of the refrigeration cycleis 4.8 psia and the condensing pressure is 75 psia. The refrigerant is flowing through the cycle witha flow rate of 18,000 lbm/hr. The R-123 leaves the evaporator as a saturated vapor and leaves thecondenser as a saturated liquid. Determine the following:a. The outlet temperature of the chilled waterb. The volumetric flow rate of the chilled water (gpm)c. The UA product of the evaporator (Btu/h-°F)d. The heat transfer rate between the refrigerant and the water (tons)arrow_forward(Read image) (Answer given)arrow_forward
- Problem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and (y2), respectively. Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s]. Givens: y1 = 4.112 m y2 = 0.387 m b = 0.942 m Answers: ( 1 ) 1880.186 lit/s ( 2 ) 4042.945 lit/s ( 3 ) 2553.11 lit/s ( 4 ) 3130.448 lit/sarrow_forwardProblem (14): A pump is being used to lift water from an underground tank through a pipe of diameter (d) at discharge (Q). The total head loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h where (V) is the flow velocity in the pipe. The elevation difference between the pump and tank surface is (h). Given the values of h [cm], d [cm], and K [-], calculate the maximum discharge Q [Lit/s] beyond which cavitation would take place at the pump entrance. Assume Turbulent flow conditions. Givens: h = 120.31 cm d = 14.455 cm K = 8.976 Q Answers: (1) 94.917 lit/s (2) 49.048 lit/s ( 3 ) 80.722 lit/s 68.588 lit/s 4arrow_forwardProblem (13): A pump is being used to lift water from the bottom tank to the top tank in a galvanized iron pipe at a discharge (Q). The length and diameter of the pipe section from the bottom tank to the pump are (L₁) and (d₁), respectively. The length and diameter of the pipe section from the pump to the top tank are (L2) and (d2), respectively. Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m], calculate total head loss due to friction (i.e., major loss) in the pipe (hmajor-loss) in [cm]. Givens: L₁,d₁ Pump L₂,d2 오 0.533 lit/s L1 = 6920.729 m d1 = 1.065 m L2 = 70.946 m d2 0.072 m Answers: (1) 3.069 cm (2) 3.914 cm ( 3 ) 2.519 cm ( 4 ) 1.855 cm TABLE 8.1 Equivalent Roughness for New Pipes Pipe Riveted steel Concrete Wood stave Cast iron Galvanized iron Equivalent Roughness, & Feet Millimeters 0.003-0.03 0.9-9.0 0.001-0.01 0.3-3.0 0.0006-0.003 0.18-0.9 0.00085 0.26 0.0005 0.15 0.045 0.000005 0.0015 0.0 (smooth) 0.0 (smooth) Commercial steel or wrought iron 0.00015 Drawn…arrow_forward
- The flow rate is 12.275 Liters/s and the diameter is 6.266 cm.arrow_forwardAn experimental setup is being built to study the flow in a large water main (i.e., a large pipe). The water main is expected to convey a discharge (Qp). The experimental tube will be built at a length scale of 1/20 of the actual water main. After building the experimental setup, the pressure drop per unit length in the model tube (APm/Lm) is measured. Problem (20): Given the value of APm/Lm [kPa/m], and assuming pressure coefficient similitude, calculate the drop in the pressure per unit length of the water main (APP/Lp) in [Pa/m]. Givens: AP M/L m = 590.637 kPa/m meen Answers: ( 1 ) 59.369 Pa/m ( 2 ) 73.83 Pa/m (3) 95.443 Pa/m ( 4 ) 44.444 Pa/m *******arrow_forwardFind the reaction force in y if Ain = 0.169 m^2, Aout = 0.143 m^2, p_in = 0.552 atm, Q = 0.367 m^3/s, α = 31.72 degrees. The pipe is flat on the ground so do not factor in weight of the pipe and fluid.arrow_forward
- Find the reaction force in x if Ain = 0.301 m^2, Aout = 0.177 m^2, p_in = 1.338 atm, Q = 0.669 m^3/s, and α = 37.183 degreesarrow_forwardProblem 5: Three-Force Equilibrium A structural connection at point O is in equilibrium under the action of three forces. • • . Member A applies a force of 9 kN vertically upward along the y-axis. Member B applies an unknown force F at the angle shown. Member C applies an unknown force T along its length at an angle shown. Determine the magnitudes of forces F and T required for equilibrium, assuming 0 = 90° y 9 kN Aarrow_forwardProblem 19: Determine the force in members HG, HE, and DE of the truss, and state if the members are in tension or compression. 4 ft K J I H G B C D E F -3 ft -3 ft 3 ft 3 ft 3 ft- 1500 lb 1500 lb 1500 lb 1500 lb 1500 lbarrow_forward
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