Show complete and correct solution, thank you. With the schematic diagram of a pump, tanks, and piping determine the following:1) The velocity of the liquid at point 2 (ft/s)2) Pump work used by the liquid (Wpump U) (hp)3) Electric motor power in out (kVW)4) In figure 3, the shaded (pink) in the U tube is mercury; determine y. specific gravity of mercury 13.75) Mass flow rateAssume the whole system is steady flow with no friction loss in the pipes and tank.Use only the following constant and conversiong= 32.2 ft/s1 hp= 33,000 ft-Ibf/min1 hp = 0.746 kWSpecific gracity of mercury 13.7In computing for the density of mercury use the density of water =62.4 lbm/ft³Do not convert English units to SI units except horsepower to kWGiven:ininftpsiapsiaininft13.12.10.70.5114.145604.12.125Seat NumberDiameter of pipe in point 1Diameter of pipe in point 2Specific gravity of the liquidh (ft) (see figure 2)Pressure at point 1Pressure at point 2Pump and motor efficiencyDiameter of pipe in point 4Diameter of pipe in point 5Z3 (see figure 1) Figure 1Pressurized tankZ3pumpDifferent diametersSee figure 3See figure 2motorAt point 2Figure 2V1StagnationpointAt points 4 and 5Figure 3D5D4y

Name: Show complete and correct solution, thank you. With the schematic diag
Uploaded: 2024-03-20T06:56:55.000Z
Channel: Bartleby
Description: Show complete and correct solution, thank you. With the schematic diagram of a pump, tanks, and piping determine the following:1) The velocity of the liquid at point 2 (ft/s)2) Pump work used by the liquid (Wpump U) (hp)3) Electric motor power in out

Answered: With the schematic diagram of a pump,…

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter4: Numerical Analysis Of Heat Conduction

Section: Chapter Questions

Problem 4.29P

See similar textbooks

Related questions

Topic Video

Question

Show complete and correct solution, thank you.

With the schematic diagram of a pump, tanks, and piping determine the following:
1) The velocity of the liquid at point 2 (ft/s)
2) Pump work used by the liquid (Wpump U) (hp)
3) Electric motor power in out (kVW)
4) In figure 3, the shaded (pink) in the U tube is mercury; determine y. specific gravity of mercury 13.7
5) Mass flow rate
Assume the whole system is steady flow with no friction loss in the pipes and tank.
Use only the following constant and conversion
g= 32.2 ft/s
1 hp= 33,000 ft-Ibf/min
1 hp = 0.746 kW
Specific gracity of mercury 13.7
In computing for the density of mercury use the density of water =62.4 lbm/ft³
Do not convert English units to SI units except horsepower to kW
Given:
in
in
ft
psia
psia
in
in
ft
1
3.1
2.1
0.7
0.51
14.1
45
60
4.1
2.1
25
Seat Number
Diameter of pipe in point 1
Diameter of pipe in point 2
Specific gravity of the liquid
h (ft) (see figure 2)
Pressure at point 1
Pressure at point 2
Pump and motor efficiency
Diameter of pipe in point 4
Diameter of pipe in point 5
Z3 (see figure 1)

Figure 1
Pressurized tank
Z3
pump
Different diameters
See figure 3
See figure 2
motor
At point 2
Figure 2
V1
Stagnation
point
At points 4 and 5
Figure 3
D5
D4
y

Expert Solution

Step 1

“Since you have posted a question with multiple sub-parts, we will solve first three subparts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.”

Given Data:

Diameter of pipe at point 1 $d_{1} = 3.1 in = \frac{3.1}{12} = 0.258 ft$

Diameter of pipe at point 2 $d_{2} = 3.1 in = \frac{2.1}{12} = 0.175 ft$

The specific gravity of the liquid $S_{L} = 0.7$

Difference between the height of liquid in both pipe $h = 0.51 ft$

Pressure at point 1 $P_{1} = 14.1 psi = 14.1 \times 144 = 2030.4 \frac{lbm}{{ft}^{2}}$

Pressure at point 2 $P_{2} = 45 psi = 45 \times 144 = 6480 \frac{lbm}{{ft}^{2}}$

Head at point 3 $z_{3} = 25 ft$

Efficiency of pump and motor $η = 0.60$

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.