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alex41 [277]
3 years ago
15

A surveyor knows an elevation at Catch Basin to be elev=2156.77 ft. The surveyor takes a BS=2.67 ft on a rod at BM Catch Basin a

nd an FS=6.72 ft on a rod at point X.
What is the elevation at point X? Express your answer to six significant figures.
Engineering
1 answer:
fenix001 [56]3 years ago
3 0

Answer:

the elevation at point X is 2152.72 ft

Explanation:

given data

elev = 2156.77 ft

BS = 2.67 ft

FS = 6.72 ft

solution

first we get here height of instrument that is

H.I = elev + BS   ..............1

put here value

H.I =  2156.77 ft + 2.67 ft  

H.I = 2159.44 ft

and

Elevation at point (x) will be

point (x)  = H.I - FS   .............2

point (x)  = 2159.44 ft  - 6.72 ft

point (x)  = 2152.72 ft

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Consider a mixing tank with a volume of 4 m3. Glycerinflows into a mixing tank through pipe A with an average velocity of 6 m/s,
Svetach [21]

This question is incomplete, the complete question as well as the missing diagram is uploaded below;

Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.

Take p_o = 880 kg/m³ and p_{glycerol = 1260 kg/m³    

 

Answer:

the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³  

Explanation:

Given that;

Inlet velocity of Glycerin, V_A = 6 m/s

Inlet velocity of oil, V_B = 3 m/s  

Density velocity of glycerin, p_{glycerol = 1260 kg/m³

Density velocity of glycerin, Take p_o = 880 kg/m³

Volume of tank V = 4 m

from the diagram;

Diameter of glycerin pipe, d_A = 100 mm = 0.1 m

Diameter of oil pipe, d_B = 80 mm = 0.08 m

Diameter of outlet pipe d_C = 120 mm = 0.12 m

Now, Appling the discharge flow equation;

Q_A + Q_B = Q_C

A_Av_A + A_Bv_B = A_Cv_C

π/4 × (d_A)²v_A + π/4 × (d_B )²v_B = π/4 × (d_C)²v_C

we substitute

π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)²v_C

0.04712 + 0.0150796 = 0.0113097v_C

0.0621996 = 0.0113097v_C

v_C = 0.0621996 / 0.0113097

v_C  = 5.5 m/s

Now we apply the mass flow rate condition

m_A + m_B = m_C

p_{glycerin}A_Av_A + p_0A_Bv_B = pA_Cv_C  

so we substitute

1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5

1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335

59.3712 + 13.27 = 0.06220335p  

72.6412 = 0.06220335p    

p = 72.6412 / 0.06220335

p =  1167.8 kg/m³  

Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³  

4 0
3 years ago
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