In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a closed tank containing 100 kg of water, ini
tially at 295 K. There is negligible heat transfer between the contents of the tank and their surroundings. Modeling the metal part and water as incompressible with constant specific heats 0.5 kJ/kg K and 4.4 kJ/kg K, respectively, determine the final equilibrium temperature after quenching, in K.
1 answer:
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Answer:
Explanation:
Hello,
a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:
- and
: extent of the reactions (2).
- ,
,
,
and
: Molar flows at the second stream (5).
On the other hand, we've got the following equations:
- : oxygen mole balance.
- : methane mole balance.
- : water mole balance.
- : formaldehyde mole balance.
- : carbon dioxide mole balance.
Thus, the degrees of freedom are:
It means that we need two additional equations or data to solve the problem.
b. Here, the two missing data are given. For the fractional conversion of methane, we define:
And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:
In such a way, one realizes that the output formaldehyde's molar flow is:
Which is equal to the first reaction extent , therefore, one computes the second one from the fractional conversion of methane as:
Now, one computes the rest of the output flows via:
-
-
-
-
-
The total output molar flow is:
Therefore the output stream composition turns out into:
Best regards.
Answer:
6.37 inch
Explanation:
Thinking process:
We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.
To determine the pressure drop in the pipe:
Using the Bernoulli equation for mass conservation:
thus
The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.
Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F
from the tables
Re= 2.01 × 10⁵
Hence, f = 0.018
Therefore, pressure drop, (P1-P2)/p = 2.70 ft
This occurs at ae presure change of 1.17 psi
Correlating with the chart, we find that the diameter will be D= 0.513
= <u>6.37 in Ans</u>