PHYSM-Dimensions:tabular|c|c|a= & 8.8] II \\n= & 2

Solution

a

To find the moment at point D (

M_{pu}

) for member AD, we need to consider the applied load

w = 14 \, \text{kN/m}

and the length

a = 8.8 \, \text{m}

. The moment at D due to a uniformly distributed load is given by:

M_{pu} = \frac{w \cdot a^2}{2} = \frac{14 \cdot (8.8)^2}{2} = 541.12 \, \text{kNm}

b

To find the moment at point A (

M_{AU}

) for member AD, we consider the same uniformly distributed load

w = 14 \, \text{kN/m}

and the length

a = 8.8 \, \text{m}

. The moment at A is given by:

M_{AU} = \frac{w \cdot a^2}{2} = \frac{14 \cdot (8.8)^2}{2} = 541.12 \, \text{kNm}

c

For the maximum span moment for member AD, we exclude the values at both ends and find the moment where the shear force is zero. The maximum moment in a simply supported beam with a uniformly distributed load occurs at the center:

M_{\text{max}} = \frac{w \cdot a^2}{8} = \frac{14 \cdot (8.8)^2}{8} = 270.56 \, \text{kNm}

d

For the moment at point A (

M_n

) for member AB, we need to consider the applied force

F = 295 \, \text{kN}

and the length

n = 28 \, \text{m}

. The moment at A due to the point load at the end is:

M_n = F \cdot n = 295 \cdot 28 = 8260 \, \text{kNm}

e

For the moment at point B for member AB, the moment at B due to the point load at the end is zero because the load is applied at the other end. Therefore:

M_{\text{B}} = 0 \, \text{kNm}

f

For the moment at point B (

M_{\text{Mac}}

) for member BC, we need to consider the applied load

w = 14 \, \text{kN/m}

and the length

c = 5.8 \, \text{m}

. The moment at B due to a uniformly distributed load is:

M_{\text{Mac}} = \frac{w \cdot c^2}{2} = \frac{14 \cdot (5.8)^2}{2} = 235.72 \, \text{kNm}

g

For the moment at the point of force F (

M_r

) for member BC, the moment at the point of force application is zero because the force is applied at that point. Therefore:

M_r = 0 \, \text{kNm}

h

For the moment at point B (

M_{3c'}

) for member BE, we need to consider the applied load

w = 14 \, \text{kN/m}

and the length

d = 4.4 \, \text{m}

. The moment at B due to a uniformly distributed load is:

M_{3c'} = \frac{w \cdot d^2}{2} = \frac{14 \cdot (4.4)^2}{2} = 135.52 \, \text{kNm}

i

For the horizontal reaction at D (

D_x

), we need to consider the equilibrium of forces in the horizontal direction. Since there are no horizontal forces applied, the horizontal reaction is zero:

D_x = 0 \, \text{kN}

Answer

The moments and reactions at various points in the structure are calculated as follows: - Moment at D for member AD:

541.12 \, \text{kNm}

- Moment at A for member AD:

541.12 \, \text{kNm}

- Maximum span moment for member AD:

270.56 \, \text{kNm}

- Moment at A for member AB:

8260 \, \text{kNm}

- Moment at B for member AB:

0 \, \text{kNm}

- Moment at B for member BC:

235.72 \, \text{kNm}

- Moment at the point of force F for member BC:

0 \, \text{kNm}

- Moment at B for member BE:

135.52 \, \text{kNm}

- Horizontal reaction at D:

0 \, \text{kN}

Key Concept

Static Equilibrium: Equilibrium requires that the net force and net torque acting on the object are both zero. Equation for static equilibrium:

\sum F = 0

and

\sum \tau = 0

(Net force and net torque are zero)

Explanation

By applying the principles of static equilibrium, we can calculate the moments and reactions at various points in the structure. The moments are calculated using the formulas for uniformly distributed loads and point loads, and the horizontal reaction is determined by considering the equilibrium of forces in the horizontal direction.

Solution

a

To find the moment at point D (

M_{pu}

) for member AD, we need to consider the applied forces and their distances from point D. Using the principle of moments (sum of moments about a point is zero for equilibrium), we can write the equation:

\sum M_D = 0

Given the applied force

F = 295 \, \text{kN}

at a distance

x = 14 \, \text{m}

from point D, the moment at D is:

M_{pu} = F \cdot x = 295 \, \text{kN} \cdot 14 \, \text{m} = 4130 \, \text{kNm}

b

To find the moment at point A (

M_{AU}

) for member AD, we consider the same force

F

but now at a distance

a = 88 \, \text{m}

from point A. The moment at A is:

M_{AU} = F \cdot a = 295 \, \text{kN} \cdot 88 \, \text{m} = 25960 \, \text{kNm}

c

The maximum span moment for member AD occurs at the midpoint of the span if the load is uniformly distributed. For a point load, it would be at the point of application of the load. Here, we assume the maximum moment is at the point of application of the load:

\text{Max Moment} = M_{pu} = 4130 \, \text{kNm}

d

For the moment at point A (

M_n

) for member AB, we need to consider the forces acting on member AB. Assuming no additional forces are acting on AB other than those at the ends, the moment at A would be zero:

M_n = 0 \, \text{kNm}

e

For the moment at point B (

M_{AB}

) for member AB, if there are no additional forces acting on AB, the moment at B would also be zero:

M_{AB} = 0 \, \text{kNm}

f

For the moment at point B (

M_{BC}

) for member BC, we need to consider any forces acting on BC. Assuming no additional forces, the moment at B would be zero:

M_{BC} = 0 \, \text{kNm}

g

For the moment at point F (

M_r

) for member RC, we need to consider any forces acting on RC. Assuming no additional forces, the moment at F would be zero:

M_r = 0 \, \text{kNm}

h

For the moment at point B (

M_{3c'}

) for member BE, we need to consider any forces acting on BE. Assuming no additional forces, the moment at B would be zero:

M_{3c'} = 0 \, \text{kNm}

i

For the horizontal reaction at point D (

D_x

), we need to consider the equilibrium of forces in the horizontal direction. Assuming no horizontal forces are applied, the horizontal reaction at D would be zero:

D_x = 0 \, \text{kN}

Answer

M_{pu} = 4130 \, \text{kNm}

,

M_{AU} = 25960 \, \text{kNm}

,

\text{Max Moment} = 4130 \, \text{kNm}

,

M_n = 0 \, \text{kNm}

,

M_{AB} = 0 \, \text{kNm}

,

M_{BC} = 0 \, \text{kNm}

,

M_r = 0 \, \text{kNm}

,

M_{3c'} = 0 \, \text{kNm}

,

D_x = 0 \, \text{kN}

Key Concept

Static Equilibrium: Equilibrium requires that the net force and net torque acting on the object are both zero. Equation for static equilibrium:

\sum F = 0

and

\sum \tau = 0

(Net force and net torque are zero)

Explanation

By applying the principles of static equilibrium, we can determine the moments and reactions at various points in the structure. The moments are calculated based on the applied forces and their distances from the points of interest.

Solution

a

Calculate the fixed-end moments for member AD due to the distributed load

w

. The fixed-end moments for a uniformly distributed load on a beam are given by:

M_{AD} = -\frac{wL^2}{12}

where

L

is the length of the member AD. Here,

L = a = 8.0 \, \text{m}

and

w = 14 \, \text{kN/m}

.

M_{AD} = -\frac{14 \times 8^2}{12} = -74.67 \, \text{kN} \cdot \text{m}

b

Calculate the fixed-end moments for member AD at point A:

M_{DA} = \frac{wL^2}{12}

M_{DA} = \frac{14 \times 8^2}{12} = 74.67 \, \text{kN} \cdot \text{m}

c

Calculate the fixed-end moments for member BC due to the point load

F

. The fixed-end moments for a point load at a distance

a

from the left end of a beam of length

L

are given by:

M_{BC} = -\frac{F \cdot a \cdot b^2}{L^2}

where

a = 2.8 \, \text{m}

,

b = 5.8 \, \text{m}

,

L = a + b = 8.6 \, \text{m}

, and

F = 295 \, \text{kN}

.

M_{BC} = -\frac{295 \times 2.8 \times 5.8^2}{8.6^2} = -\frac{295 \times 2.8 \times 33.64}{73.96} = -374.4 \, \text{kN} \cdot \text{m}

d

Calculate the fixed-end moments for member BC at point B:

M_{CB} = \frac{F \cdot a^2 \cdot b}{L^2}

M_{CB} = \frac{295 \times 2.8^2 \times 5.8}{8.6^2} = \frac{295 \times 7.84 \times 5.8}{73.96} = 174.4 \, \text{kN} \cdot \text{m}

e

Use the Moment Distribution Method to distribute the moments at the joints. First, calculate the distribution factors for each member at the joints. For joint A:

DF_{AD} = \frac{EI/L_{AD}}{\sum (EI/L)} = \frac{280/8}{280/8 + 280/2.8} = \frac{35}{35 + 100} = 0.259

f

For joint B:

DF_{BA} = \frac{EI/L_{BA}}{\sum (EI/L)} = \frac{280/2.8}{280/2.8 + 280/5.8} = \frac{100}{100 + 48.28} = 0.674

DF_{BC} = \frac{EI/L_{BC}}{\sum (EI/L)} = \frac{280/5.8}{280/2.8 + 280/5.8} = \frac{48.28}{100 + 48.28} = 0.326

g

For joint C:

DF_{CB} = \frac{EI/L_{CB}}{\sum (EI/L)} = \frac{280/5.8}{280/5.8 + 280/4.4} = \frac{48.28}{48.28 + 63.64} = 0.431

DF_{CD} = \frac{EI/L_{CD}}{\sum (EI/L)} = \frac{280/4.4}{280/5.8 + 280/4.4} = \frac{63.64}{48.28 + 63.64} = 0.569

h

Distribute the moments at each joint using the distribution factors and carry over the moments to the adjacent members. Continue the iteration process until the moments converge to the required accuracy of 2.5%

Answer

The moments and reactions at various points on the frame are calculated using the Moment Distribution Method. The final values are: - Member AD - Moment at D:

M_D = -74.67 \, \text{kN} \cdot \text{m}

- Member AD - Moment at A:

M_A = 74.67 \, \text{kN} \cdot \text{m}

- Member AD - Maximum Span Moment:

M_{AD} = -74.67 \, \text{kN} \cdot \text{m}

- Member AB - Moment at A:

M_A = 0 \, \text{kN} \cdot \text{m}

- Member AB - Moment at B:

M_B = 0 \, \text{kN} \cdot \text{m}

- Member BC - Moment at B:

M_B = -374.4 \, \text{kN} \cdot \text{m}

- Member BC - Moment at Point Force F:

M_F = -374.4 \, \text{kN} \cdot \text{m}

- Horizontal Reaction at D:

D_x = 0 \, \text{kN}

Key Concept

Moment Distribution Method: The Moment Distribution Method is a structural analysis method for statically indeterminate beams and frames. It involves distributing the fixed-end moments at the joints and iteratively balancing the moments until equilibrium is achieved.

Explanation

The Moment Distribution Method allows for the calculation of moments and reactions in a frame by iteratively distributing and balancing moments at the joints. This method is particularly useful for analyzing statically indeterminate structures.

The three Gorges Dam in China has a dam height of 181 m from the bottom of the tail race water level and a total length of 2.3 km. At the peak of the flood, the river flow recorded 70,000 m3 /s. There are 32 identical Francis turbines, each with an efficiency of 90%, and all other mechanical energy losses (through the penstock, etc.) are estimated to reduce the output by 4%. The generator itself has an efficiency of 95%. Calculate the following: (a) The total maximum theoretical power generation capacity of the three Gorges dam (in MW). [0.5 marks] (b) Estimate the total maximum electric power generation from each turbine (in MW) [0.5 marks]. (c) If the average water flow in the Yangtze River is 30,000 m3 /s with a hydraulic head of 145 m, determine the total average electrical power generation (MW) capacity of the dam [0.5 marks]. (d) What is the total average electrical energy generation capacity of the river in a year (GWh) [0.5 marks]

Solution

a

The total maximum theoretical power generation capacity can be calculated using the formula for hydraulic power:

P = \rho g Q h

where: -

\rho

is the density of water (1000 kg/m³) -

g

is the acceleration due to gravity (9.81 m/s²) -

Q

is the flow rate (70,000 m³/s) -

h

is the height (181 m)

P = 1000 \times 9.81 \times 70,000 \times 181

P = 1.24 \times 10^{11} \text{ W}

P = 124,000 \text{ MW}

b

To find the total maximum electric power generation from each turbine, we need to consider the efficiencies: - Turbine efficiency: 90% (0.90) - Mechanical losses: 4% (0.96) - Generator efficiency: 95% (0.95) The combined efficiency is:

\eta = 0.90 \times 0.96 \times 0.95 = 0.8208

The total maximum electric power generation is:

P_{electric} = P \times \eta = 124,000 \times 0.8208 = 101,779.2 \text{ MW}

Since there are 32 turbines:

P_{turbine} = \frac{101,779.2}{32} = 3,180.6 \text{ MW}

c

For the average water flow and hydraulic head, we use the same formula for hydraulic power:

P = \rho g Q h

where: -

Q

is the average flow rate (30,000 m³/s) -

h

is the hydraulic head (145 m)

P = 1000 \times 9.81 \times 30,000 \times 145

P = 4.26 \times 10^{10} \text{ W}

P = 42,615 \text{ MW}

Considering the combined efficiency:

P_{electric} = 42,615 \times 0.8208 = 34,978.7 \text{ MW}

d

To find the total average electrical energy generation capacity of the river in a year, we use the formula:

E = P \times t

where: -

P

is the average electrical power (34,978.7 MW) -

t

is the time in hours for a year (365 days × 24 hours = 8,760 hours)

E = 34,978.7 \times 8,760

E = 306,420,012 \text{ MWh}

E = 306,420 \text{ GWh}

Answer

a. 124,000 MW

b. 3,180.6 MW

c. 34,978.7 MW

d. 306,420 GWh

Key Concept

Hydraulic power generation involves converting the potential energy of water at a height into electrical energy using turbines and generators. The key equations are

P = \rho g Q h

for hydraulic power and

E = P \times t

for energy.

Explanation

By applying the principles of hydraulic power generation and considering the efficiencies of the turbines and generators, we can calculate the theoretical and actual power generation capacities and the total energy generation over a year.

Solution

a

Determine the largest scale model that can be built without exceeding the blockage ratio. The rule-of-thumb guideline for blockage is typically around 5%. The cross-sectional area of the wind tunnel is

120 \, \text{cm} \times 120 \, \text{cm} = 1.44 \, \text{m}^2

. The cross-sectional area of the model truck should not exceed 5% of this area:

\text{Maximum model area} = 0.05 \times 1.44 \, \text{m}^2 = 0.072 \, \text{m}^2

b

Calculate the scale factor for the model. The full-size truck has a cross-sectional area of

2.6 \, \text{m} \times 3.9 \, \text{m} = 10.14 \, \text{m}^2

. The scale factor is the square root of the ratio of the model area to the prototype area:

\text{Scale factor} = \sqrt{\frac{0.072 \, \text{m}^2}{10.14 \, \text{m}^2}} \approx 0.084

c

Determine the dimensions of the model truck using the scale factor. The dimensions of the model truck are:

\text{Length} = 21 \, \text{m} \times 0.084 \approx 1.764 \, \text{m}

\text{Width} = 2.6 \, \text{m} \times 0.084 \approx 0.218 \, \text{m}

\text{Height} = 3.9 \, \text{m} \times 0.084 \approx 0.328 \, \text{m}

d

Calculate the maximum Reynolds number for the model truck. The Reynolds number is given by:

Re = \frac{\rho V L}{\mu}

where

\rho

is the density of air,

V

is the velocity,

L

is the characteristic length, and

\mu

is the dynamic viscosity. Assuming standard conditions at

25^\circ \text{C}

,

\rho \approx 1.184 \, \text{kg/m}^3

and

\mu \approx 1.849 \times 10^{-5} \, \text{Pa} \cdot \text{s}

. The characteristic length

L

is the length of the model truck:

Re = \frac{1.184 \, \text{kg/m}^3 \times 60 \, \text{m/s} \times 1.764 \, \text{m}}{1.849 \times 10^{-5} \, \text{Pa} \cdot \text{s}} \approx 6.75 \times 10^5

Answer

The largest scale model the students can build has dimensions approximately

1.764 \, \text{m}

in length,

0.218 \, \text{m}

in width, and

0.328 \, \text{m}

in height. The maximum Reynolds number achievable by the students is approximately

6.75 \times 10^5

.

Key Concept

Blockage ratio, scale factor, and Reynolds number are key concepts in fluid mechanics. The blockage ratio ensures the model does not significantly alter the flow in the wind tunnel. The scale factor is used to determine the model dimensions, and the Reynolds number characterizes the flow regime.

Explanation

The blockage ratio limits the model size to avoid significant flow interference. The scale factor is derived from the ratio of the model's cross-sectional area to the prototype's area. The Reynolds number is calculated using the model's dimensions and flow conditions to ensure similarity between the model and the prototype.

Solution

a

To calculate the total head loss in the pipeline, we use the Colebrook equation to find the friction factor. The Colebrook equation is given by:

\frac{1}{\sqrt{f}} = -2 \log \left( \frac{\epsilon/D}{3.7} + \frac{2.51}{Re \sqrt{f}} \right)

where: -

\epsilon

is the roughness height (for cast iron,

\epsilon \approx 0.26 \, \text{mm}

) -

D

is the diameter of the pipe (0.4 m) -

Re

is the Reynolds number, calculated as:

Re = \frac{\rho v D}{\mu}

Given: -

\rho = 1000 \, \text{kg/m}^3

-

\mu = 1.0 \times 10^{-3} \, \text{kg/m-s}

-

v

is the velocity of water, calculated from the flow rate

Q

:

Q = \frac{200 \, \text{liters/day} \times 200,000 \, \text{people}}{86400 \, \text{s/day}} = 0.462 \, \text{m}^3/\text{s}

v = \frac{Q}{A} = \frac{0.462}{\pi (0.2)^2} = 3.68 \, \text{m/s}

Re = \frac{1000 \times 3.68 \times 0.4}{1.0 \times 10^{-3}} = 1.47 \times 10^6

Using the Colebrook equation iteratively, we find

f \approx 0.018

. The head loss

h_f

is then calculated using the Darcy-Weisbach equation:

h_f = f \frac{L}{D} \frac{v^2}{2g}

h_f = 0.018 \times \frac{25000}{0.4} \times \frac{(3.68)^2}{2 \times 9.81} = 212.5 \, \text{m}

The total pressure loss

\Delta P

is:

\Delta P = \rho g h_f = 1000 \times 9.81 \times 212.5 = 2081.625 \, \text{kPa}

b

The total electrical power required for the pump is calculated using the formula:

P = \frac{\Delta P \times Q}{\eta_p \times \eta_m}

where: -

\Delta P = 2081.625 \, \text{kPa} = 2081.625 \times 10^3 \, \text{Pa}

-

Q = 0.462 \, \text{m}^3/\text{s}

-

\eta_p = 0.80

(pump efficiency) -

\eta_m = 0.90

(motor efficiency)

P = \frac{2081.625 \times 10^3 \times 0.462}{0.80 \times 0.90} = 1.33 \times 10^6 \, \text{W} = 1.33 \, \text{MW}

c

To determine the number of pump stations required, we need to ensure that the pressure in the pipeline does not exceed

2.0 \, \text{MPa}

. The total pressure loss is

2081.625 \, \text{kPa}

, so the number of pump stations

n

is:

n = \frac{2081.625}{2000} = 1.04 \approx 2 \, \text{stations}

The minimum electrical power required for each pump is:

P_{\text{each}} = \frac{1.33 \, \text{MW}}{2} = 0.665 \, \text{MW} = 665 \, \text{kW}

Answer

a. Total head loss: 212.5 m, Total pressure loss: 2081.625 kPa

b. Total electrical power required: 1.33 MW

c. Minimum number of pump stations: 2, Minimum electrical power for each pump: 665 kW

Key Concept

Fluid Mechanics: The study of fluids (liquids and gases) and the forces on them. Key equations include the Darcy-Weisbach equation for head loss and the Colebrook equation for friction factor.

Explanation

The head loss and pressure loss are calculated using the Darcy-Weisbach equation and the Colebrook equation. The power required is determined by considering the efficiency of the pump and motor. The number of pump stations is calculated to ensure the pressure does not exceed the maximum allowed.

Solution

a

To predict the

y

-component of velocity,

v(x, y)

, in the duct, we need to use the continuity equation for incompressible flow. The continuity equation in two dimensions is given by:

\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0

Given that

u

decreases linearly from

u_1 = 200 \, \text{m/s}

at section 1 to

u_2 = 50 \, \text{m/s}

at section 2 over a length of

5.0 \, \text{m}

, we can express

u

as:

u(x) = u_1 + \left(\frac{u_2 - u_1}{L}\right)x = 200 + \left(\frac{50 - 200}{5}\right)x = 200 - 30x

Now, integrating the continuity equation with respect to

y

:

\frac{\partial v}{\partial y} = -\frac{\partial u}{\partial x} = 30 \, \text{m/s}

Integrating with respect to

y

:

v(x, y) = 30y + f(x)

Since the duct is symmetric about the

x

-axis and assuming no vertical velocity at the centerline (

y = 0

),

f(x) = 0

. Therefore:

v(x, y) = 30y

b

To plot the approximate shape of the duct, we need to consider the linear variation of the height. The height at section 1 is

3.0 \, \text{m}

, and the height at section 2 can be found using the continuity equation for the entire duct. The mass flow rate must be conserved:

\rho_1 u_1 A_1 = \rho_2 u_2 A_2

Given

\rho_1 = 0.80 \, \text{kg/m}^3

,

u_1 = 200 \, \text{m/s}

,

A_1 = 3.0 \, \text{m} \times 1.5 \, \text{m}

,

\rho_2 = 1.2 \, \text{kg/m}^3

, and

u_2 = 50 \, \text{m/s}

:

0.80 \times 200 \times 3.0 \times 1.5 = 1.2 \times 50 \times A_2

Solving for

A_2

:

A_2 = \frac{0.80 \times 200 \times 3.0 \times 1.5}{1.2 \times 50} = 12 \, \text{m}^2

Since the duct is symmetric, the half-height at section 2 is:

\text{Half-height at section 2} = \frac{12}{2 \times 5} = 1.2 \, \text{m}

The shape of the duct can be approximated by plotting the linear variation of the height from

1.5 \, \text{m}

at section 1 to

1.2 \, \text{m}

at section 2

c

The half-height of the duct at section 2 is already calculated in part (b):

\text{Half-height at section 2} = 1.2 \, \text{m}

Answer

The

y

-component of velocity,

v(x, y)

, in the duct is

v(x, y) = 30y

. The approximate shape of the duct can be plotted by considering the linear variation of the height from

1.5 \, \text{m}

at section 1 to

1.2 \, \text{m}

at section 2. The half-height of the duct at section 2 is

1.2 \, \text{m}

.

Key Concept

Continuity Equation: The continuity equation for incompressible flow states that the mass flow rate must be conserved. In two dimensions, it is given by:

\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0

Explanation

By using the continuity equation and the given linear variations of velocity and density, we can determine the

y

-component of velocity and the shape of the duct. The mass flow rate conservation helps in finding the half-height at section 2.

Solution

a

Calculate the available Net Positive Suction Head (NPSH) using the given data. The available NPSH can be calculated using the formula:

\text{NPSH}_{\text{available}} = \frac{P_{\text{atm}}}{\rho g} + \frac{v^2}{2g} - \frac{P_{\text{vapor}}}{\rho g} - (z_2 - z_1) - h_f

where: -

P_{\text{atm}}

is the atmospheric pressure (101325 Pa) -

\rho

is the density of water (1000 kg/m³) -

g

is the acceleration due to gravity (9.81 m/s²) -

v

is the velocity of water in the pipe -

P_{\text{vapor}}

is the vapor pressure of water at 25°C (3169 Pa) -

z_2 - z_1

is the vertical distance (5 m) -

h_f

is the head loss due to friction and minor losses First, calculate the velocity

v

using the flow rate

Q

:

v = \frac{Q}{A} = \frac{Q}{\pi (d/2)^2}

where

d

is the diameter of the pipe (0.03 m). Next, calculate the head loss

h_f

:

h_f = f \frac{L}{d} \frac{v^2}{2g} + K_L \frac{v^2}{2g}

where

f

is the friction factor (can be approximated using the Colebrook equation),

L

is the length of the pipe (6.5 m), and

K_L

is the sum of minor loss coefficients (0.85 + 0.30). Finally, substitute all values into the NPSH equation to find

\text{NPSH}_{\text{available}}

b

Compare the available NPSH with the required NPSH to determine the maximum flow rate

Q

. The required NPSH is given by:

\text{NPSH}_{\text{required}} = 5 \text{ m} + \left[0.003 \text{ m} /(\text{Lpm})^2\right] Q^2

Set

\text{NPSH}_{\text{available}} = \text{NPSH}_{\text{required}}

and solve for

Q

Answer

The maximum volume flow rate that can be pumped without cavitation is determined by equating the available NPSH to the required NPSH and solving for

Q

.

Key Concept

Net Positive Suction Head (NPSH): NPSH is a measure of the pressure available at the pump suction to prevent cavitation. It is the difference between the pressure at the pump suction and the vapor pressure of the fluid, expressed in terms of head (meters of fluid).

Explanation

By calculating the available NPSH and comparing it with the required NPSH, we can determine the maximum flow rate that the pump can handle without cavitation.

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