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marin [14]
2 years ago
5

A rigid tank contains 5 kg of saturated vapor steam at 100°C. The steam is cooled to the ambient temperature of 25°C. (a) Sketch

the process with respect to the saturation lines on a T-vdiagram. Indicatepressure valuesat each state. (5pt) (b) Determinethe entropy change of the steam, in kJ/K. (10 pt) (c) Determinethe totalentropy change associated with this process, in kJ/K. (15pt)

Engineering
1 answer:
luda_lava [24]2 years ago
4 0

Answer:

Explanation:

Part c:

(Q)out = m(u1 - u2) = 5(2506-193.67) = 11562KJ

Total Entropy change = Δs + (Qout/Tsurrounding) = -33.36 + (11562/393) = 5.44KJ/K

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A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s)
4vir4ik [10]

A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s) with a flow rate of 8.7 m^3/s, 6 mg/L BOD5 and 8.3 mg/L DO. Both streams are at 20°C. After mixing, the river is 3 meters deep and flowing at a velocity of 0.50 m/s. DOsat for this river is 9.0 mg/L. The deoxygenation constant is kd= 0.20 d^-1 and The reaction rate constant k at 20 °C is 0.27 d^-1.

The answer therefore would be the number 0.27 divided by two and then square while getting the square you would make it a binomial.

I wont give the answer but the steps

Your Welcome

8 0
3 years ago
Pedro holds a heavy science book over his head for 10 minutes. Petro is doing work during that time. True or False
algol [13]

Answer:

True because he is working his arms to lift and hold the weight

Explanation:

4 0
2 years ago
Can anyone help me ?
MrRa [10]
That’s too hard for me lol oof
3 0
3 years ago
A closed system consists of 0.3 kmol of octane occupying a volume of 5 m³. Determine (a) the weight of the system, in N, and (b)
Leni [432]

Answer:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

Explanation:

To calculate the mass of the octane(m):

Number of mole of octane (n) =0.3kmol(given)

Molarmass of octane (M) =114.23kg/kmol

m=n*M

m=(0.3kmol)*(114.23kg/kmol)

m=34.269kg

To calculate for the weight of octane(W):

W=g*m

W=(9.81m/s^2)*(34.269kg)

W=336.18N

b) For specific volumes of Vn and Vm:

Given volume of octane (V) =5m^3

Vm=V/m

Vm=5m^3/34.269kg

Vm=0.1459m^3/kg

And Vn will be :

Vn=V/m=5m^3/0.3kmol

Vn=16.67m/Kmol

Therefore, the answers are:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

7 0
3 years ago
Consider the following list. list = {24, 20, 10, 75, 70, 18, 60, 35} Suppose that list is sorted using the insertion sort algori
Greeley [361]

Answer:

Option B

10,20,24,75,70,18,60,35

Explanation:

The first, second and third iteration of the loop will be as follows

insertion sort iteration 1: 20,24,10,75,70,18,60,35

insertion sort iteration 2:10,20,24,75,70,18,60,35

insertion sort iteration 3: 10,20,24,75,70,18,60,35

8 0
3 years ago
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