EB, Fluid Transport. Water is pumped from a reservoir to top of the mountain through a 1800m long 6-inch schedule 120 pipe at an average velocity of 5 m/s. The pipe discharges into a pressurized storage tank at a level 1500m above the level in the reservoir. The storage tank is maintained at a gauge pressure of 1520 mm Hg and has a total capacity of 2000 m^3. If the overall efficiency of pump is 70% and the cost of electricity to the motor is 5- cent/kW-h, (a) what is the total cost of pumping water to fill the storage tank? (b) Repeat part (a) if the storage tank pressure is lowered by 760 mm Hg (c) Repeat part (a) if the water is replaced with glycerol (SG = 1.26; µ = 1412 cP) and the pump efficiency is now 50%. (For turbulent flow regime, take f = 0.184Re^-0.2) expecting unit : (a) 10^2 $ (b) 10^2 $ (c) 10^3 $ SI constant Patm = 10^5 Pa; pwater - 1000 kg/m^3; pair - 1.2kg/m^3; μwater - 10^-3 N•s/m^2; uair - 2 x 10^-5 N•s/m^2 ; g = 9.8 m/s^2

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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**Fluid Transport in Engineering**

In this scenario, water is pumped from a reservoir to the top of a mountain through a 1800-meter long, 6-inch schedule 120 pipe at an average velocity of 5 m/s. The pipe discharges into a pressurized storage tank located 1500 meters above the reservoir level. This storage tank maintains a gauge pressure of 1520 mm Hg and has a total capacity of 2000 cubic meters.

Given that the overall efficiency of the pump is 70% and the electricity cost to the motor is 5 cents per kW-h:

**Questions:**
1. What is the total cost to pump water and fill the storage tank?
2. Repeat the calculation if the storage tank pressure is lowered by 760 mm Hg.
3. Redo part (a) if the water is replaced with glycerol (Specific Gravity = 1.26; Viscosity = 1412 cP) and the pump efficiency changes to 50%.

*Note: For turbulent flow, use the friction factor \(f = 0.184 \times \text{Re}^{-0.2}\).*

**Unit Expectations:**
- (a) \(10^2\) 
- (b) \(10^2\) 
- (c) \(10^3\) 

**SI Constants:**
- \( \text{Atmospheric pressure} = 10^5 \) Pa
- \( \text{Density of water} \approx 1000 \text{ kg/m}^3 \)
- \( \text{Density of air} \approx 1.2 \text{ kg/m}^3 \)
- \( \text{Viscosity of water} \approx 10^{-3} \text{ Ns/m}^2 \)
- \( \text{Viscosity of air} \approx 2 \times 10^{-5} \text{ Ns/m}^2 \)
- \( g = 9.8 \text{ m/s}^2 \)

**Diagram Explanation:**
- The diagram shows a reservoir at the base, connected to a vertical pipe reaching the storage tank 1500 meters above.
Transcribed Image Text:**Fluid Transport in Engineering** In this scenario, water is pumped from a reservoir to the top of a mountain through a 1800-meter long, 6-inch schedule 120 pipe at an average velocity of 5 m/s. The pipe discharges into a pressurized storage tank located 1500 meters above the reservoir level. This storage tank maintains a gauge pressure of 1520 mm Hg and has a total capacity of 2000 cubic meters. Given that the overall efficiency of the pump is 70% and the electricity cost to the motor is 5 cents per kW-h: **Questions:** 1. What is the total cost to pump water and fill the storage tank? 2. Repeat the calculation if the storage tank pressure is lowered by 760 mm Hg. 3. Redo part (a) if the water is replaced with glycerol (Specific Gravity = 1.26; Viscosity = 1412 cP) and the pump efficiency changes to 50%. *Note: For turbulent flow, use the friction factor \(f = 0.184 \times \text{Re}^{-0.2}\).* **Unit Expectations:** - (a) \(10^2\) - (b) \(10^2\) - (c) \(10^3\) **SI Constants:** - \( \text{Atmospheric pressure} = 10^5 \) Pa - \( \text{Density of water} \approx 1000 \text{ kg/m}^3 \) - \( \text{Density of air} \approx 1.2 \text{ kg/m}^3 \) - \( \text{Viscosity of water} \approx 10^{-3} \text{ Ns/m}^2 \) - \( \text{Viscosity of air} \approx 2 \times 10^{-5} \text{ Ns/m}^2 \) - \( g = 9.8 \text{ m/s}^2 \) **Diagram Explanation:** - The diagram shows a reservoir at the base, connected to a vertical pipe reaching the storage tank 1500 meters above.
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