Answer:
PERT Chart and GANTT Chart
As the project manager of a residential construction project, I will prefer the PERT chart to the GANTT chart because a PERT chart displays task dependencies unlike a Gantt chart. With the PERT chart, the sequence of tasks is clearly mapped out. Dependent tasks are carried out when other tasks that they depend on have been executed.
Explanation:
By definition, a Gantt chart is like a bar chart that lays out project tasks and timelines using bars. On the other hand, a PERT chart follows a structure in the form of flow charts or network diagrams. It displays all the project tasks in separate boxes. The boxes are then connected with arrows which clearly show the task dependencies.
As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial temperature of 500 C by suspending them in an oil bath at 30 C. If a convection coefficient of 500 W/m2 K is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of 50 C, at which point it is withdrawn from the bath
Answer:
Explanation:
Given that:
diameter = 100 mm
initial temperature = 500 ° C
Conventional coefficient = 500 W/m^2 K
length = 1 m
We obtain the following data from the tables A-1;
For the stainless steel of the rod [tex]\overline T = 548 \ K[/tex]
[tex]\rho = 7900 \ kg/m^3[/tex]
[tex]K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K[/tex]
[tex]\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657[/tex]
Here, we can't apply the lumped capacitance method, since Bi > 0.1
[tex]\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\[/tex]
[tex]0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81[/tex]
[tex]t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins[/tex]
However, on a single rod, the energy extracted is:
[tex]\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J[/tex]
Hence, for centerline temperature at 50 °C;
The surface temperature is:
[tex]T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}[/tex]
Compute the volume percent of graphite, VGr, in a 3.9 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Answer:
Vgr = 0.122 = 12.2 vol %
Explanation:
Density of ferrite = 7.9 g/cm^3
Density of graphite = 2.3 g/cm^3
compute the volume percent of graphite
for a 3.9 wt% cast Iron
W∝ = (100 - 3.9) / ( 100 -0 ) = 0.961
Wgr = ( 3.9 - 0 ) / ( 100 - 0 ) = 0.039
Next convert the weight fraction to volume fraction using the equation attached below
Vgr = 0.122 = 12.2 vol %
An industrial boiler consists of tubes inside of which flow hot combustion gases. Water boils on the exterior of the tubes. When installed, the clean boiler has an over all heat transfer coefficient of 300 W/m^2 . K. Based on experience, i is anticipated that the fouling factors on the inner and outer surfaces will increase linearly with time as Ra,t and Ryo-at where a, 2.5 x 10^-11 m2 K/W s and a,-1.0 x 10^-11 m^2 - K/W s for the inner and outer tube surfaces, respectively. If the boiler is to be cleaned when the overall heat transfer coeffi- cient is reduced from its initial value by 25%, how long after installation should the first cleaning be scheduled?
Answer:
the first cleaning be scheduled 1.006 years after installation
Explanation:
Given the data in the question;
U[tex]_{clean[/tex] = 300 W/m².K
first we determine the heat coefficient of the dirt surface;
overall heat transfer coefficient is reduced from its initial value by 25%
U[tex]_{dirt[/tex] = ( 1 - 25%) × U[tex]_{clean[/tex]
U[tex]_{dirt[/tex] = ( 1 - 0.25) × 300
U[tex]_{dirt[/tex] = 0.75 × 300
U[tex]_{dirt[/tex] = 225 W/m².K
next we find the inner fouling factor
[tex]R"_{f ,i[/tex] = [tex]a_it[/tex]
[tex]R"_{f ,o[/tex] = (2.5 × 10⁻¹¹)t
for the outer fouling water;
[tex]R"_{f ,o[/tex] = [tex]a_ot[/tex]
[tex]R"_{f ,o[/tex] = ( 1.0 × 10⁻¹¹ )t
now, we determine the total heat transfer coefficient
[tex]\frac{1}{U}[/tex] = [tex]R"_{f ,i[/tex] + [tex]R"_{f ,o[/tex]
we substitute
[tex]\frac{1}{U}[/tex] = (3.5 × 10⁻¹¹)t
so the first cleaning duration after insulation will be;
[tex]\frac{1}{U} = \frac{1}{U_{dirt}} - \frac{1}{U{clean}}[/tex]
we substitute
(3.5 × 10⁻¹¹)t = [tex]\frac{1}{225} - \frac{1}{300}[/tex]
(3.5 × 10⁻¹¹)t = 0.001111
t = 0.001111 / (3.5 × 10⁻¹¹)
t = 31742857.142857 seconds
t = 31742857.142857 / 3.154 × 10⁷
t = 1.006 years
Therefore, the first cleaning be scheduled 1.006 years after installation