assume SI units, & 1 atm = 105 Pa; ρwater = 1000 kg/m3; ρair = 1.2 kg/m3; g = 9.8 m/s2 **EB, Loop Problem Explanation:**Water is pumped in a closed loop with a total length of \( L = 40 \, \text{m} \) and a constant pipe area \( A = 0.01 \, \text{m}^2 \). The pump performance is described by the head versus flow rate, where:- \( h_p = 10 - 1000 \, Q^2 \)The total piping system friction head loss is given by:- \( h_f = 4 \, \frac{V^2}{g} \)Where:- \( Q \) is the flow rate and \( V \) is the water velocity in the pipe.**Tasks:**(a) **Find the flow rate \( Q \).**(b) **Calculate the required pump power.**(c) **Determine the pressure increase above the pump \(\Delta p = p_2 - p_1\).** - Assume the pump is 100% efficient, implying no friction losses inside the pump.**Answers:**- **(a)** The flow rate \( Q = 10^{-2} \, \text{m}^3/\text{s} \).- **(b)** The pump power \( W_p = 10^1 \, \text{kW} \).- **(c)** The pressure increase \( \Delta p = 10^1 \, \text{kPa} \). The diagram depicts a pipeline system with a pump. The flow direction is indicated by arrows, and the system includes a pipeline of length \( L = 40 \, \text{m} \) and a cross-sectional area \( A = 0.01 \, \text{m}^2 \). The system features a pump located between two points marked as 1 and 2, with the pump work output denoted by \( w_p \).Key elements:- **Flow Path**: The arrows indicate the direction of fluid flow through the pipeline.- **Pipeline**: The length of the pipeline is 40 meters, and it has a cross-sectional area of 0.01 square meters.- **Pump**: Positioned between points 1 and 2, the pump aids in moving the fluid through the system by adding work, denoted by \( w_p \). This diagram can be used to study fluid dynamics, particularly focusing on the interaction between the fluid flow and the pump within a closed system.

Answered: Image /qna-images/answer/c32f2c06-ff5e-4eae-8a78-ebad45ab0f32.jpg

EB, loop. Water is pumped in a closed loop of total length L = 40m and constant pipe area A = 0.01 m². T pump performance is given as head versus flow rate: hp = 10-1000 Q², and the total piping system friction loss is given as h = 4 V²/g, where Q is flow rate and V is the water velocity in the pipe. (a) Find the flow ; (b) Find the required pump power; (c) Find the Ap increase above the pump, Ap = P2-p₁. Assume the pur 100% efficient - no friction losses inside the numn Ans OM: (a)O: 10-² m³/s: (b)W- 10⁰ kW: (c)AP: 10

EB, loop. Water is pumped in a closed loop of total length L = 40m and constant pipe area A = 0.01 m². T pump performance is given as head versus flow rate: hp = 10-1000 Q², and the total piping system friction loss is given as h = 4 V²/g, where Q is flow rate and V is the water velocity in the pipe. (a) Find the flow ; (b) Find the required pump power; (c) Find the Ap increase above the pump, Ap = P2-p₁. Assume the pur 100% efficient - no friction losses inside the numn Ans OM: (a)O: 10-² m³/s: (b)W- 10⁰ kW: (c)AP: 10

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Applied Fluid Mechanics

A fluid is a substance that is continuously deformed by the application of shear stress. The rate of deformation of a fluid depends on its viscosity. The fluid tries to flow when it interacts with some other system.

Question

assume SI units, & 1 atm = 105 Pa; ρwater = 1000 kg/m3
; ρair = 1.2 kg/m3
; g = 9.8 m/s2

**EB, Loop Problem Explanation:**

Water is pumped in a closed loop with a total length of \( L = 40 \, \text{m} \) and a constant pipe area \( A = 0.01 \, \text{m}^2 \). The pump performance is described by the head versus flow rate, where:

- \( h_p = 10 - 1000 \, Q^2 \)

The total piping system friction head loss is given by:

- \( h_f = 4 \, \frac{V^2}{g} \)

Where:
- \( Q \) is the flow rate and \( V \) is the water velocity in the pipe.

**Tasks:**

(a) **Find the flow rate \( Q \).**

(b) **Calculate the required pump power.**

(c) **Determine the pressure increase above the pump \(\Delta p = p_2 - p_1\).**
- Assume the pump is 100% efficient, implying no friction losses inside the pump.

**Answers:**

- **(a)** The flow rate \( Q = 10^{-2} \, \text{m}^3/\text{s} \).
- **(b)** The pump power \( W_p = 10^1 \, \text{kW} \).
- **(c)** The pressure increase \( \Delta p = 10^1 \, \text{kPa} \).

The diagram depicts a pipeline system with a pump. The flow direction is indicated by arrows, and the system includes a pipeline of length \( L = 40 \, \text{m} \) and a cross-sectional area \( A = 0.01 \, \text{m}^2 \). The system features a pump located between two points marked as 1 and 2, with the pump work output denoted by \( w_p \).

Key elements:
- **Flow Path**: The arrows indicate the direction of fluid flow through the pipeline.
- **Pipeline**: The length of the pipeline is 40 meters, and it has a cross-sectional area of 0.01 square meters.
- **Pump**: Positioned between points 1 and 2, the pump aids in moving the fluid through the system by adding work, denoted by \( w_p \).

This diagram can be used to study fluid dynamics, particularly focusing on the interaction between the fluid flow and the pump within a closed system.

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