the pressure rise, across a pump can be expressed as where D is the impeller diameter, p, is the fluid density, w is the rotational speed, adn q is the flowrate. determine a suitable set of dimensionless parameters
Answers
Answer:
hello your question is incomplete below is the complete question
The pressure rise Δp across a pump can be expressed as Δp = f(D, p, w, Q) where D is the impeller diameter, p is the fluid density, w is the rotational speed, and Q is the flowrate. determine a suitable set of dimensionless parameters
answer : Δp / D^2pw^2 = Ф (Q / D^3w )
Explanation:
k ( number of variables ) = 5
r ( number of reference dimensions ) = 3
applying the pi theorem
hence the number of pi terms = k - r = 5 - 3 = 2
Related Questions
Match the use of the magnetic field to its respective description.
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oooExplanation:
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A 550 kJ of heat quantity needed to increase water temperature from 32°C to 80°C. Calculate the mass
of the water when the specific heat capacity of water is 4200 J/kg °C.
Answers
Answer:
2.728 kg
Explanation:
The units help you keep the calculation straight.
[tex]\dfrac{550\text{ kJ}}{(80^\circ\text{C}-32^\circ\text{C})(4.200\text{ kJ/kg\,$^\circ$C})}=\dfrac{550}{48\cdot4.2}\text{ kg}\approx\boxed{2.728\text{ kg}}[/tex]
A differential amplifier is to have a voltage gain of 100. What will be the feedback resistance required if the input resistances are both 1 kΩ?
Answers
Answer:
required feedback resistance ( R2 ) = 100 k Ω
Explanation:
Given data :
Voltage gain = 100
input resistance ( R1 ) = 1 k ohms
calculate feedback resistance required
voltage gain of differential amplifier
[tex]\frac{Vout}{V2 - V1 } = \frac{R2}{R1}[/tex]
= Voltage gain = R2/R1
= 100 = R2/1
hence required feedback resistance ( R2 ) = 100 k Ω
1. Consider a solid cube of dimensions 1ft x 1ft x 1ft (=0.305m x 0.305m x 0.305m). Its top surface is 10
ft (=3.05 m) below the surface of the water. The density of water is pf=1000 kg/m3.
Consider two cases:
a) The cube is made of cork (pB=160.2 kg/m3)
b) The cube is made of steel (pB=7849 kg/m3)
In what direction does the body tend to move?
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Answer:
a) up
b) down
Explanation:
When the cube is less dense than water, it will tend to float (move upward). When it is more dense, it will sink (move downward).
a) 160.2 kg/m^3 < 1000 kg/m^3. The cube will move up.
__
b) 7849 kg/m^3 > 1000 kg/m^3. The cube will move down.
Liquid water at 300 kPa and 20°C is heated in a chamber by mixing it with superheated steam at 300 kPa and 300°C. Cold water enters the chamber at a rate of 2.6 kg/s. If the mixture leaves the mixing chamber at 60°C.
Required:
Determine the mass flow rate of the superheated steam required.
Answers
Answer:
0.154kg/s
Explanation:
From this question we have the following information:
P1 = 300kpa
T1 = 20⁰c
M1 = 2.6kg/s
For superheated system
P2 = 300kpa
T2 = 300⁰c
M2 = ??
T2 = 60⁰c
From saturated water table
h1 = 83.91kj/kg
h3 = 251.18kj/kg
From superheated water,
h2 = 3069.6kj/kg
The equation of energy balance
m1h1 + m2h2 = m3h3
When we input all the corresponding values:
We get
m2 = -434.902/-2818.42
m2 = 0.15430
m2 = 0.154kg/s
This is the mass flow rate of the superheated steam
Please check attachment for more detailed explanation.
thank you!
This question involves the concepts of energy balance and mass flow rate.
The mass flow rate of the superheated steam required is "0.15 kg/s".
Applying the energy balance in this situation, we get:
[tex]m_1h_1+m_2h_2=m_3h_3[/tex]
where,
m₁ = mass flow rate of liquid water at 300 KPa and 200°C = 2.6 kg/s
m₂ = mass flow rate of superheated at 300 KPa and 300°C = ?
h₁ = enthalpy of liquid water at 300 KPa and 200°C = 83.91 KJ/kg (from saturated steam table)
h₂ = enthalpy of superheated at 300 KPa and 300°C = 3069.6 KJ/kg (from superheated steam table)
h₃ = enthalpy of exiting fluid at 60°C = 251.18 KJ/kg (from saturated steam table)
m₃ = mass flow rate of exiting fluid = 2.6 kg/s + m₂
Therefore,
[tex](2.6\ kg/s)(83.91\ KJ/kg)+(m_2)(3069.6\ KJ/kg)=(2.6\ kg/s+m_2)(251.18\ KJ/kg)\\m_2(3069.6\ KJ/kg-251.18\ KJ/kg)=(2.6\ kg/s)(251.18\ KJ/kg-83.91\ KJ/kg)\\\\m_2=\frac{434.902\ KW}{2818.42\ KJ/kg}[/tex]
m₂ = 0.15 kg/s
Learn more about energy balance here:
https://brainly.com/question/9839609?referrer=searchResults
Describe how the fracture behavior would be different for a fiber-reinforced tape such as duct tape.
Answers
Answer:
A Normal tape is very weak under tensile force and when a small fracture is caused, it will affect the whole tape and the tape will fail easily. while a fiber-reinforced tape will not fail easily under same stress and this is the major advantage of a fiber-reinforced tape has over a normal tape
Explanation:
The fracture behavior would be different for a fiber-reinforced tape in the following way :
* It's behavior during tensile stress and its fracture behavior.
A Normal tape is very weak under tensile force and when a small fracture is caused, it will affect the whole tape and the tape will fail easily. while a fiber-reinforced tape will not fail easily under same stress and this is the major advantage of a fiber-reinforced tape has over a normal tape
Quadrilateral ABCD is a rectangle.
If m ZADB = 7k + 60 and mZCDB = -5k + 40, find mZCBD.
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Hope this helps...........
A spherical Gaussian surface of radius R is situated in space along with both conducting and insulating charged objects. The net electric flux through the Gaussian surface is:______
Answers
Answer:
Ф = [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]
Explanation:
Radius of Gaussian surface = R
Charge in the Sphere ( Gaussian surface ) = Q
lets take the radius of the sphere to be equal to radius of the Gaussian surface i.e. R
To determine the net electric flux through the Gaussian surface
we have to apply Gauci law
Ф = 4[tex]\pi r^2 E[/tex]
Ф = [tex]\frac{Q_{enc} |}{e_{0} }[/tex]
= [tex]\frac{Q}{e_{0} } [ \frac{\frac{4\pi }{3 }(R)^3 }{\frac{4}{3}\pi (R)^3 } ][/tex]