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nikitadnepr [17]
3 years ago
11

Three return steam lines in a chemical processing plan enter a collection tank operating at a steady state at 1 bar. Steam enter

s inlet 1 with flow rate of 0.8 kg/s and a quality of 0.9. Steam enters inlet 2 with flow rate of 2kg/s at 200 degrees C. Steam enters inlet 3 with flow rate 1.2 kg/s at 95 degrees C. Steam exits the tank at 1 bar. The rate of heat transfer from the collection tank is 40 kW. Neglecting kinetic and potential energy effects, determine for the steam exiting the tank:(a)Mass flow rate in kg/s (b) the temperature in degrees C.
Engineering
1 answer:
gavmur [86]3 years ago
4 0

Answer:

a. 4kg/s

b. 99.97 °C

Explanation:

Hello,

a. The resulting mass balance turns out into:

F_1+F_2+F_3=F_{out}\\F_{out}=0.8kg/s+2kg/s+1.2kg/s=4kg/s

b. Now, the energy balance is:

F_1h_1+F_2h_2+F_3h_3-Q_{out}=F_{out}h_{out}

In such a way, the first enthalpy is taken as a liquid-vapor mixture at 1 bar and 0.9 quality, it means:

h_1=hf(1bar)+xhfg(1bar)\\h_1=419.06kJ/kg+0.9*2256.5kJ/kg=2449.91kJ/kg

Second enthalpy is taken by identifying that stream as an overheated vapor at 1 bar and 200 °C, thus, the resulting enthalpy is:

h_2=2875.5kJ/kg

Then, the third enthalpy is taken by considering that at 95°C and 1 bar the water is a saturated liquid, thus:

h_3=hf(95^0C)=398.09kJ/kg.

Now, by solving for h_{out}, we've got:

h_{out}=\frac{0.8kg/s*2449.91kJ/kg+2kg/s*2875.5kJ/kg+1.2kg/s*398.09kJ/kg-40kW}{4kg/s} \\h_{out}=\frac{8148.612kJ/s}{4kg/s} \\h_{out}=2037.153kJ/kg

Finally, by searching for that value of enthalpy, one sees that at 1 bar, the exiting stream is a liquid-vapor mixture that is at 99.97 °C and has a 72%- quality.

(NOTE: all the data was extracted from Cengel's book 7th edition).

Best regards.

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For a flow rate of 212 cfs find the critical depth in (a) a rectangular channel with ????=6.5 ft, (b) a triangular channel with
Fofino [41]

Answer:

A. 3.21ft

B. 3.51ft

C. 2.95ft

D. 1.5275ft

Explanation:

A) Q =212 cu.f/s

Formula for critical depth of rectangular section is: dc =[(Q^2) /(b^2(g))]^1/3

Where dc =critical depth, ft

Q= quantity of flow or discharge, ft3/s

B= width of channel, ft (m)

g = acceleration due to gravity which is 9.81m/s2 or 32.185ft/s2

Now, from the question,

Q = 212 cu.f/s and b=6.5ft

Therefore, the critical depth is: [(212^2)/(6.5^2 x32. 185)]^(1/3)

To give ; critical depth= (44,944/1359.82)^(1/3) = 3.21ft

B. Formula for critical depth of a triangular section; dc = (2Q^2/gm^2)^(1/5)

From the question, Q =212 cu.f/s and m=1.6ft while g= 32.185ft/s2

Therefore, critical depth = [(212^2) /(1.6^2 x32. 185)] ^(1/5) = (44,944/84.466)^(1/5) = 3.51ft

C. For trapezoidal channel, critical depth(y) is derived from (Q^2 /g) = (A^3/T)

Where A= (B + my)y and T=(B+2my)

Now from the question, B=6.5ft and m=5ft.

Therefore, A= (6.5 + 2y)y and T=(6. 5 + 2(5y))= 6.5 + 10y

Now, let's plug the value of A and T into the initial equation to derive the critical depth ;

(212^2 /32.185) = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Which gives;

1396.43 = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Multiply both sides by 6.5 + 10y to get;

1396.43(6.5 + 10y) = [((6.5 + 2y)^3)y^3]

Factorizing this, we get y = 2. 95ft

D) Formula for critical depth of a circular section; dc =D/2[1 - cos(Ѳ/2)]

Where D is diameter of pipe and Ѳ is angle at critical depth in radians.

Angle not given, so we assume it's perpendicular angle is 90.

Since angle is in radians, therefore Ѳ/2 = 90/2 = 45 radians ; converting to degree, = 2578. 31

Therefore, dc = (6.5/2) (1 - cos (2578.31))

dc = 3.25(1 - 0.53) = 3.25 x 0.47 = 1.5275ft

8 0
3 years ago
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