March 4 Homework Solutions
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.
Using the standard density of air = 0.00238 slugs/ft3 and the given unit width, w = 1 ft, gives the final answer as
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.
Solving this equation for D2 in terms of Q1 gives.
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?
This is a transient continuity problem. If we assume that the density and area are constant we can write the equation as follows
If we assume that the flow rates are constant so that the derivative becomes a finite difference we have the following result.
h = 0.0510 ft