4.56 Water flows in a branching pipe shown in figure P4.56 at the right, with uniform velocity at each inlet and outlet. The fixed control volume coincides with the system at time t – 20 s. Make a sketch to indicate (a) the boundary of the system at time t = 20.2s, (b) the fluid that left the control volume during that 0.2 s interval, and (c) the fluid that entered the control volume during that interval.
The fluid travels a distance Vt during the time periodt = 0.2 s. At point 1, this distance is V1t = (2 m/s)(0.2 s) = 0.4 m. At point 2, this distance is V2t = (1 m/s)(0.2 s) = 0.2 m. At point 3, this distance is V3t = (1.5 m/s)(0.2 s) = 0.3 m. Using these dimensions we can prepare the sketch of the system and the control volume shown below.
3.61 The wind blows across a field with an approximate velocity profile shown in Figure P4.61 shown at the right. Use equation 4.16 with the parameter b equal to the velocity to determine the momentum flow rate across the vertical surface A-B which is of unit depth into the paper.
Equation 4.16 gives the following expression for the flow of a quantity b
In this case, b is actually a vector quantity, V = Vi, where i is the unit vector in the x direction. From the graph we see that V = (1.5/s)y between 0 and 10 feet and V = 15 ft/s between 10 and 20 ft. We can thus write our integral as follows noting that n = i, and b is the vector quantity, V
Here we have used the fact that i▪I = 1 and the differential area, dA = wdy where w is the width of one ft into the page. Substituting the relationship between V and y given above gives the following result for the integral.
5.7 Water flows along the centerline of a 50-mm-diameter pipe with an average velocity of 10 m/s and out radially between two large circular disks as shown in Figure P 5.7 at the right. The disks are parallel and spaced 10 mm apart. Determine the average velocity of the water at a radius of 300 mm in the space between the disks.
This is a basic continuity equation problem with constant density so that Q = V1A1 = V2A2. The desired velocity is found as follows
5.11 At cruise conditions air flows into a jet engine at a steady rate of 65 lbm/s. Fuel enters the engine at a steady rate of 0.60 lbm/s. The average velocity of the exhaust gases is 1500 ft/s relative to the engine. If the engine exhaust effective cross section is 3.5 ft2, estimate the density of the exhaust gases in lbm/ft3.
This is a problem of mass conservation and use of the continuity equation. The exhaust mass is he sum of the air mass flow rate and the fuel mass flow rate: 65 lbm/s + 0.60 lbm/s = 56.60 lbm/s. This mass flow rate of exhaust must be equal to the density-velocity-area product, VA, for the exhaust. We can thus solve for the unknown density.
= 0.0125 lbm/ft3
5.12 Air at standard atmospheric conditions is drawn into a compressor at the steady rate of 30 m3/min. The compressor pressure ratio, pexit/pinlete, is 10 to 1. Through the compressor p/n remains constant with n = 1.4. If the average velocity in the compressor discharge pipe is not to exceed 30 m/s calculate the minimum discharge pipe diameter required.
We can apply the continuity equation to this problem with 1 as the inlet and 2 as the outlet.
Since p/n = constant, p1/1n = p2/2n. Making this substitution and substituting numerical data gives.
. D2 = 0.064 m.
5.23 The Hoover Dam backs up the Colorado River and creates Lake Mead, which is approximately 115 miles long and has a surface area of approximately 225 square miles. If during flood conditions the Colorado River flows into the lake at a rate of 45,000 cfs and the outflow from the dam is 8000 cfs, how many feet per 24-hour day will the lake level rise?