Answer: 0.0033 m = 3.3 mm (0.13 in.)
Explanation:
since this problem requires us to calculate the maximum internal crack length allowable for the 7075-T651
aluminum alloy is given that it is loaded to a stress level equal to one-half of its yield strength.
For this alloy, KIc 24 MPa (22 ksi
); also, σ= σ_y/2 = (495 MPa)/2 = 248 MPa (36,000 psi).
Now solving for 2ac gives us;
2a_c=2/π (K_IC/γσ)^2=2/π [(24MPa √m)/(1.35)(248MPa) ]^2=0.0033m=3.3mm (0.13 in.)
Answer:
E=52000Hp.h
E=38724920Wh
E=1.028x10^11 ftlb
Explanation:
To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.
Finally, when you have the result Hp h / year you convert it to Ftlb and Wh
E=52000Hp.h
E=38724920Wh
E=1.028x10^11 ftlb
Answer:
If the heat engine operates for one hour:
a) the fuel cost at Carnot efficiency for fuel 1 is $409.09 while fuel 2 is $421.88.
b) the fuel cost at 40% of Carnot efficiency for fuel 1 is $1022.73 while fuel 2 is $1054.68.
In both cases the total cost of using fuel 1 is minor, therefore it is recommended to use this fuel over fuel 2. The final observation is that fuel 1 is cheaper.
Explanation:
The Carnot efficiency is obtained as:
Where is the atmospheric temperature and
is the maximum burn temperature.
For the case (B), the efficiency we will use is:
The work done by the engine can be calculated as:
where Hv is the heat value.
If the average net power of the engine is work over time, considering a net power of 2.5MW for 1 hour (3600s), we can calculate the mass of fuel used in each case.
If we want to calculate the total fuel cost, we only have to multiply the fuel mass with the cost per kilogram.