Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
Answer:
A. 8.4
Explanation:
[OH⁻] = 2.6 × 10⁻⁶ Take the negative log of each side
-log[OH⁻] = pOH = 5.59 Apply the pH/pOH relation
pH + pOH = 14.00 Insert the value of pOH
pH + 5.59 = 14.00 Subtract 5.59 from each side
pH = 14.00 -5.59 = 8.41
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For #1 the answer is dependent relationship . 2. answer is independent relationship . the last question the answer is dependent.
