Using the steady flow energy equation, find the power required

[JohnSmith236]

Homework Statement

A pump is used to move water through a pipe of diameter 150mm, Figure below. The water has a temperature of 20 Celsius and an absolute pressure of 100KPa. The pump moves the water up a vertical distance of 2m and the water exists at atmospheric pressure.
Q1)
Assuming the process is adiabatic and friction less, and the required mass flow rate is ṁ= 2kg/s , determine, using the steady flow energy equation, the power required for the pump.

Q2)
Discuss how the effects of friction are accounted for by the steady flow energy equation. How would you introduce the effects of friction into the problem described above?
2. Homework Equations
0 = ∆H + ∆K.E.+ ∆P.E.

P = ṁ w
where
P = power (W)
ṁ = mass flow rate (kg/s)
w = specific work (Nm/kg, J/kg)

Specific work - w - can be expressed:
w = g h
where
h = head (m)
g = acceleration due to gravity (9.81 m/s2)

ṁ= ρ Q
where
ρ = density (kg/m3)
Q = volume flow rate (m3/s)

Power= ρ Q g h

The Attempt at a Solution

Q = ṁ/ρ
= 2/1000
=0.002

h= 2m ?? Not sure this answer is correct do not know the formula for head

P = ρ Q g h

= (1,000 kg/m3) (2*10-3 m3/s) (9.81 m/s2) (2)

= 4W

 

[haruspex]

I would assume that the water is drawn from a reservoir, so essentially static initially.
Is it static where it exits the pipe?

 

[JohnSmith236]

Thank you for your response. Yes, it is static however I am not sure what the formula is for the pressure head (h) and whether the formula I have used is correct since I haven't used the diameter given. Furthermore, how would I introduce the effects of friction into the problem?

 

[haruspex]

Yes, it is static

Static initially, but you did not answer my question. What standard equation should you have quoted?

how would I introduce the effects of friction into the problem?

We can come to that later.

 

[JohnSmith236]

Yes, at the bottom of the pipe where the water exits it is static, meaning the change in height is 2m-0m is that right?
p1 - p2 = γ (h2 - h1) where γ is the specific weight (N/m³) which is γ = ρ g, γ = 1000*9.81=9810
Not sure if this is the equation i should've used need help on this part of the question

 

[Chestermiller]

Homework Statement

A pump is used to move water through a pipe of diameter 150mm, Figure below. The water has a temperature of 20 Celsius and an absolute pressure of 100KPa. The pump moves the water up a vertical distance of 2m and the water exists at atmospheric pressure.
Q1)
Assuming the process is adiabatic and friction less, and the required mass flow rate is ṁ= 2kg/s , determine, using the steady flow energy equation, the power required for the pump.

Q2)
Discuss how the effects of friction are accounted for by the steady flow energy equation. How would you introduce the effects of friction into the problem described above?
2. Homework Equations
0 = ∆H + ∆K.E.+ ∆P.E.

P = ṁ w
where
P = power (W)
ṁ = mass flow rate (kg/s)
w = specific work (Nm/kg, J/kg)

Specific work - w - can be expressed:
w = g h
where
h = head (m)
g = acceleration due to gravity (9.81 m/s2)

ṁ= ρ Q
where
ρ = density (kg/m3)
Q = volume flow rate (m3/s)

Power= ρ Q g h

The Attempt at a Solution

Q = ṁ/ρ
= 2/1000
=0.002

h= 2m ?? Not sure this answer is correct do not know the formula for head

P = ρ Q g h

= (1,000 kg/m3) (2*10-3 m3/s) (9.81 m/s2) (2)

= 4W
View attachment 224077

You omitted the shaft work from your equation for the steady flow energy equation. It should read:
$$0=\dot{Q}-\dot{W}+\dot{m}(h_{in}-h_{out}+gz_{in}-gz_{out)})$$
How is the rate of shaft work $\dot{W}$ related to the power supplied by the pump?

 

[JohnSmith236]

You omitted the shaft work from your equation for the steady flow energy equation. It should read:
$$0=\dot{Q}-\dot{W}+\dot{m}(h_{in}-h_{out}+gz_{in}-gz_{out)})$$
How is the rate of shaft work $\dot{W}$ related to the power supplied by the pump?

That is what I am trying to figure out.

 

[Chestermiller]

That is what I am trying to figure out.

The pump power is the rate at which the pump does work on the fluid. This is the same as the negative rate of doing shaft rate.

 

[JohnSmith236]

The pump power is the rate at which the pump does work on the fluid. This is the same as the negative rate of doing shaft rate.

So therefore, the power would be equal to -W+˙m(hin−hout+gzin−gzout))

 

[Chestermiller]

So therefore, the power would be equal to -W+˙m(hin−hout+gzin−gzout))

No. It would just be $-\dot{W}$

 

[JohnSmith236]

No. It would just be $-\dot{W}$

Oh yes of course, thank you very much Chester I really appreciate the fact that you pointing that out to me I completely left out that equation by mistake.

 

[Chestermiller]

@haruspex has pointed out that there is a kinetic energy term that should be included in the steady state energy balance too for this problem. For the equation in post #6, that would mean that there should be $\left(\frac{v^2}{2}\right)_{in}-\left(\frac{v^2}{2}\right)_{out}$ within the parenthesis in the equation.