Answer:
410.7 K
Explanation:
Given data:
P1 ( initial air pressure ) = 100 kPa
Pi ( intercooler pressure ) = 400 kPa
P2( compressed air pressure ) = 1200 kPa
T1 = 300 k
Ti = 300 k
Calculate for the temperature at the exit of the second stage ( T2 )
we can calculate this using the relation below
T2 / Ti = ( P2/Pi )^(r-1/r)
∴ T2 = 300 ( 1200 / 400 )^( 1.4 - 1 / 1.4 )
= 300 ( 3 )^0.286
= 300 * 1.369 = 410.7 K
A continuous and aligned fiber-reinforced composite is manufactured using 80 vol% aramid fiber (a kevlar-like compound) embedded nylon 6-6. A part for a high-performance aircraft utilizes this composite. If the part experiences 953 lb-f (pounds force) along the fiber alignment axis, what is the force conveyed by the fibers ?
Answer:
the force conveyed by the fibers is 947.93 lb-f
Explanation:
Given the data in the question;
V_f = 80% = 0.8
V_m = 1 - V_f = 1 - 0.8 = 0.2
Now,
length of fibre L_f = length of Nylon L_n
V_f = A_f × L_f = 0.8
V_m = A_n × L_n = 0.2
so
V_f/V_m = A_f/A_n = 0.8/0.2
A_f/A_n = 4
now, the strains in fibre is equal to strains in nylon
(P/AE)f = (P/AE)n
P_f/A_fE_f = P_n/A_nE_n
P_f = (A_f/A_n)(E_f/E_n)(P_n)
P_f = ( 4 )( 131 / 2.8 )(Pn)
P_f = 187.14Pn
and P_n = Pf / 187.14
Hence
given that P_total = 953 lb-f
P_f + P_n = 953
P_f + ( P_f / 187.14 ) = 953
P_f( 1 + ( 1 / 187.14 ) ) = 953
P_f( 1.00534359 = 953
P_f = 953 / 1.00534359
P_f = 947.93 lb-f
Therefore, the force conveyed by the fibers is 947.93 lb-f
A cylindrical rod of copper (E = 110 GPa) having a yield strength of 240 MPa is to be subjected
to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an
elongation of 0.50 mm?
Answer:
"7.654 mm" is the correct solution.
Explanation:
According to the question,
[tex]E=110\times 10^3 \ N/mm^2[/tex][tex]\sigma_y = 240 \ mPa[/tex][tex]P = 6660 \ N[/tex][tex]L = 380 \ mm[/tex][tex]\delta = 0.5 \ mm[/tex]Now,
As we know,
The Elongation,
⇒ [tex]E=\frac{\sigma}{e}[/tex]
[tex]=\frac{\frac{P}{A} }{\frac{\delta}{L} }[/tex]
or,
⇒ [tex]\delta=\frac{PL}{AE}[/tex]
By substituting the values, we get
[tex]0.5=\frac{6660\times 380}{(\frac{\pi}{4}D^2)(110\times 10^3)}[/tex]
then,
⇒ [tex]D^2=58.587[/tex]
[tex]D=\sqrt{58.587}[/tex]
[tex]=7.654 \ mm[/tex]
Could anyone answer this, please? It's about solid mechanics. I will give you 100 points!!! It's due at midnight.
Answer:
sorry i don't know
Explanation:
Q2 [45 marks] Consider Ibra region where the installed solar panels cost on average 2 OMR /W.
[10 marks] What is the cost to install a 5kW PV system for a residence?
[10 marks] If the solar irradiance in Ibra is on average 800W/m2 and the installed panels have efficiency of 18%. How many panels are required if the panel’s area is 2m2?
[15 marks] Assume Ibra has an average of 10 day-hours, dusty environment which causes the efficiency of the solar system to drop by 10% on average, and 30 cloudy days/year which cause the efficiency of the solar panels drops by 50%. If electrical power cost per kWh is 0.05 OMR determine the break-even time for the 5kW PV system.
[10 marks] If the system to be off-grid, what would be the backup time if three 12-V batteries were selected each with a capacity of 200Ah. Assume that you can discharge the batteries up to 80% of their capacities.
Answer:
so hard it is
Explanation:
I don't know about this
please mark as brainleast
byýyy