A 50-lbm iron casting, initially at 700o F, is quenched in a tank filled with 2121 lbm of oil, initially at 80o F. The iron cast
ing and oil can be modeled as incompressible with specific heats 0.10 Btu/lbm o R, and 0.45 Btu/lbm o R, respectively. For the iron casting and oil as the system, determine: a) The final equilibrium temperature (o F) b) The total entropy change for this process (Btu/ o R) (Hint: Total entropy change is the sum of entropy change of iron casting and oil.)
1 answer:
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Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
The pic shown can help to understand the question.
Answer:
a. 0.4544 N
b.
Explanation:
For computing the normality and molarity of the acid solution first we need to do the following calculations
The balanced reaction
= 0.27264 g
= 0.006816 mol
Now
Moles of needed is
= 0.003408 mol
= 0.333984 g
Now based on the above calculation
a. Normality of acid is
= 0.4544 N
b. And, the acid solution molarity is
= 0.00005112
=
We simply applied the above formulas