6.D1 A large tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 13,300 N (3000 lbf). Determine the minimum required wire diameter, assuming a factor of safety of 2.0 and a yield strength of 860 MPa (125,000 psi) for the steel.
Solution
For this problem the working stress is computed using Equation 6.24 with N = 2, as
w=y
2 = 860 MPa
2 = 430 MPa (62,500 psi )
Since the force is given, the area may be determined from Equation 6.1, and subsequently the original diameter d0 may be calculated as
A0 = F
w= d0 2
2
And
d0= 4F
w=
(4)(13,300 N)
(
430 106N/m2)
6.3 103 m6.3 mm (0.25 in.)
6.D2 (a) Consider a thin-walled cylindrical tube having a radius of 65 mm is to be used to transport pressurized gas. If inside and outside tube pressures are 100 and 2.0 atm (10.13 and 0.2026 MPa), respectively, compute the minimum required thickness for each of the following metal alloys. Assume a factor of safety of 3.5.
(b) A tube constructed of which of the alloys will cost the least amount?
Alloy Yield Strength, y Density, Unit Mass Cost,
c
(MPa) (g/cm3) ($US/kg)
(a) To solve this problem we use a procedure similar to that used for Design Example 6.2. Alloy yield strength, tube radius and thickness, and the inside-outside pressure difference are related according to Equation 6.26;
furthermore, labels for the tube dimensions are represented in the sketch of Figure 6.21. For this problem all parameters in this equation are provided except for the tube thickness, t. Rearranging Equation 6.26 such that t is the dependent variable leads to
t Nrp
y (6.26b)
Tube wall thickness will vary from alloy to alloy since each alloy has a different yield strength, whereas N, r, and
p are constant. Their values are provided in the problem statement as follows:
N = 3.5
r = 65 mm = 65 103 m
p = 10.13 MPa 0.2026 MPa = 9.927 MPa
Using Equation 6.26b, we compute the tube wall thickness for these five alloys as follows:
t (plain steel)(3.5)(65 103 m)(9.927 MPa)
375 MPa 0.00602 m = 6.02 mm
t (alloy steel)(3.5)(65 103 m)(9.927 MPa)
(b) This portion of the problem asks that we determine at tube constructed of which of the alloys will cost the least amount. We begin by considering Equation 6.28, which gives tube volume as a function of t. Furthermore, when the expression for t, Equation 6.26b, is incorporated into Equation 6.28, the following equation for tube volume V results:
Now tube cost per unit length is dependent on volume, alloy density, and cost per unit mass values according to Equation 6.29. Substitution of the above expression for V into Equation 6.29 yields
Cost V
Using this equation, let us now determine the cost per unit length for the plain steel.
Cost (plain steel)
(2)(65103 m) (3.5)(65103 m)(9.927 MPa) 375 MPa
(3.5)(65103 m)(9.927 MPa) 375 MPa
2 (106 cm3/m3)(1 m)(7.8 g/cm3)
(1.65 $US/kg)
1000 g/kg
= $33.10
We will not present cost computations for the other alloys. However, the following table lists costs for all five alloys (along with wall thickness values calculated above.
Alloy t Cost
(mm) ($US)
Steel (plain) 6.02 33.10
Steel (alloy) 2.26 29.30
Cast iron 10.0 78.40
Aluminum 8.21 72.20
Magnesium 12.9 156.50
Hence, the alloy steel is the least expensive, the plain steel is next; the most expensive is the magnesium alloy.
6.D3 (a) Gaseous hydrogen at a constant pressure of 0.658 MPa (5 atm) is to flow within the inside of a thin-walled cylindrical tube of nickel that has a radius of 0.125 m. The temperature of the tube is to be 350°C and the pressure of hydrogen outside of the tube will be maintained at 0.0127 MPa (0.125 atm). Calculate the minimum wall thickness if the diffusion flux is to be no greater than 1.25 × 10–7 mol/m2.s. The concentration of hydrogen in the nickel, CH (in moles hydrogen per cubic meter of Ni), is a function of hydrogen pressure, PH2 (in MPa), and absolute temperature T according to
Furthermore, the diffusion coefficient for the diffusion of H in Ni depends on temperature as
DH(m2/s)(4.76 107)exp 39,560 J/mol RT
(6.35)
(b) For thin-walled cylindrical tubes that are pressurized, the circumferential stress is a function of the pressure difference across the wall (Δp), cylinder radius (r), and tube thickness (x) according to Equation 6.25—
that is
rp
x (6.25a)
Compute the circumferential stress to which the walls of this pressurized cylinder are exposed.
(Note: the symbol t is used for cylinder wall thickness in Equation 6.25 found in Design Example 6.2; in this version of Equation 6.25 (i.e., 6.25a) we denote wall thickness by x.)
(c) The room-temperature yield strength of Ni is 100 MPa (15,000 psi), and σy diminishes about 5 MPa for every 50°C rise in temperature. Would you expect the wall thickness computed in part (b) to be suitable for this Ni cylinder at 350°C? Why or why not?
(d) If this thickness is found to be suitable, compute the minimum thickness that could be used without any deformation of the tube walls. How much would the diffusion flux increase with this reduction in thickness?
However, if the thickness determined in part (c) is found to be unsuitable, then specify a minimum thickness that you would use. In this case, how much of a decrease in diffusion flux would result?
Solution
(a) This portion of the problem asks for us to compute the wall thickness of a thin-walled cylindrical Ni tube at 350C through which hydrogen gas diffuses. The inside and outside pressures are, respectively, 0.658 and 0.0127 MPa, and the diffusion flux is to be no greater than 1.25 107 mol/m2-s. This is a steady-state diffusion problem, which necessitates that we employ Equation 5.2. The concentrations at the inside and outside wall faces
may be determined using Equation 6.34, and, furthermore, the diffusion coefficient is computed using Equation (b) Now we are asked to determine the circumferential stress:
= r p
x
=(0.125 m)(0.658 MPa 0.0127 MPa) 0.00366 m
= 22.0 MPa
(c) Now we are to compare this value of stress to the yield strength of Ni at 350C, from which it is possible to determine whether or not the 3.66 mm wall thickness is suitable. From the information given in the problem, we may write an equation for the dependence of yield strength (y) on temperature (T) as follows:
y = 100 MPa 5 MPa
50C
T Tr
where Tr is room temperature and for temperature in degrees Celsius. Thus, at 350C
y = 100 MPa 5 MPa 50C
(350C 20C) = 67 MPa
Inasmuch as the circumferential stress (22.0 MPa) is much less than the yield strength (67 MPa), this thickness is entirely suitable.
(d) And, finally, this part of the problem asks that we specify how much this thickness may be reduced and still retain a safe design. Let us use a working stress by dividing the yield stress by a factor of safety, according to Equation 6.24. On the basis of our experience, let us use a value of 2.0 for N. Thus
w =y
N = 67 MPa
2 = 33.5 MPa Using this value for w and Equation 6.25a, we now compute the tube thickness as
x = rp
w
(0.125 m)(0.658 MPa0.0127 MPa) (33.5 MPa)
= 0.00241 m = 2.41 mm
Substitution of this value into Fick's first law we calculate the diffusion flux as follows:
J =DC
6.D4 Consider the steady-state diffusion of hydrogen through the walls of a cylindrical nickel tube as described in Problem 6.D3. One design calls for a diffusion flux of 2.5 × 10–8 mol/m2.s, a tube radius of 0.100 m, and inside and outside pressures of 1.015 MPa (10 atm) and 0.01015 MPa (0.1 atm), respectively; the maximum allowable temperature is 300°C. Specify a suitable temperature and wall thickness to give this diffusion flux and yet ensure that the tube walls will not experience any permanent deformation.
Solution
This problem calls for the specification of a temperature and cylindrical tube wall thickness that will give a diffusion flux of 2.5 108 mol/m2-s for the diffusion of hydrogen in nickel; the tube radius is 0.100 m and the inside and outside pressures are 1.015 and 0.01015 MPa, respectively. There are probably several different approaches that may be used; and there is not one unique solution. Let us employ the following procedure to solve this problem: (1) assume some wall thickness, and, then, using Fick's first law for diffusion (which also employs Equations 5.2 and 6.35), compute the temperature at which the diffusion flux is that required; (2) compute the yield strength of the nickel at this temperature using the dependence of yield strength on temperature as stated in Problem 6.D3; (3) calculate the circumferential stress on the tube walls using Equation 6.25a; and (4) compare the yield strength and circumferential stress values--the yield strength should probably be at least twice the stress in order to make certain that no permanent deformation occurs. If this condition is not met then another iteration of the procedure should be conducted with a more educated choice of wall thickness.
As a starting point, let us arbitrarily choose a wall thickness of 2 mm (2 103 m). The steady-state diffusion equation, Equation 5.2, takes the form
J =DC
Solving this expression for the temperature T gives T = 500 K = 227C; this value is satisfactory inasmuch as it is less than the maximum allowable value (300C).
The next step is to compute the stress on the wall using Equation 6.25a; thus
= rp
x
= (0.100 m)(1.015 MPa0.01015 MPa)
(
2 103m)
50.2 MPa
Now, the yield strength (y) of Ni at this temperature may be computed using the expression
y = 100 MPa 5 MPa
50C
T Tr
where Tr is room temperature. Thus,
y = 100 MPa – 0.1 MPa/C (227C – 20C) = 79.3 MPa
Inasmuch as this yield strength is about 50% greater than the circumferential stress, a wall thickness of 2 mm is a satisfactory design parameter; furthermore, 227C is a satisfactory operating temperature.