Using the steady flow energy equation, find the power required

In summary: No, it would just be ##-\dot{W}##In summary, the pump requires 4 watts of power to move the water up a distance of 2 meters. Friction is accounted for by the steady flow energy equation, and the effects of friction are introduced into the problem by calculating the power required for the pump.
  • #1
JohnSmith236
6
0

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 = w
where
P = power (W)
= 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)


= 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
upload_2018-4-15_15-55-55.png
 

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  • #2
I would assume that the water is drawn from a reservoir, so essentially static initially.
Is it static where it exits the pipe?
 
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Likes JohnSmith236
  • #3
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?
 
  • #4
JohnSmith236 said:
Yes, it is static
Static initially, but you did not answer my question. What standard equation should you have quoted?
JohnSmith236 said:
how would I introduce the effects of friction into the problem?
We can come to that later.
 
  • #5
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
 
Last edited:
  • #6
JohnSmith236 said:

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 = w
where
P = power (W)
= 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)


= 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?
 
  • #7
Chestermiller said:
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.
 
  • #8
JohnSmith236 said:
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.
 
  • #9
Chestermiller said:
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))
 
  • #10
JohnSmith236 said:
So therefore, the power would be equal to -W+˙m(hin−hout+gzin−gzout))
No. It would just be ##-\dot{W}##
 
  • #11
Chestermiller said:
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.
 
  • #12
@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.
 

1. What is the steady flow energy equation?

The steady flow energy equation is a fundamental principle in thermodynamics that relates the flow of energy in a system to changes in its properties such as pressure, temperature, and velocity. It is expressed as: ΔH = Q - W + ΔKE + ΔPE, where ΔH is the change in enthalpy, Q is the heat added to the system, W is the work done on the system, and ΔKE and ΔPE are the changes in kinetic and potential energy, respectively.

2. How is the steady flow energy equation used to find power?

The steady flow energy equation can be rearranged to solve for the power required in a system. By isolating the work term (W), we can substitute it into the power equation (P = W/t) to get: P = (ΔH - Q + ΔKE + ΔPE)/t. This equation allows us to calculate the power required to maintain a steady flow of energy in a system.

3. What are the units of measurement for the steady flow energy equation?

The units of measurement for the steady flow energy equation depend on the specific variables being used. In general, enthalpy (H) is measured in joules (J), heat (Q) is measured in joules (J), work (W) is measured in joules (J), time (t) is measured in seconds (s), and kinetic and potential energy are measured in joules (J). It is important to use consistent units throughout the equation to ensure accurate results.

4. Can the steady flow energy equation be used for all types of systems?

Yes, the steady flow energy equation can be used for all types of systems as long as they are in a state of steady flow. This means that the properties of the system do not change with time, and energy is neither created nor destroyed. Examples of systems that can be analyzed using the steady flow energy equation include turbines, compressors, and heat exchangers.

5. How is the steady flow energy equation different from the first law of thermodynamics?

The steady flow energy equation is a specific application of the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred or converted. The steady flow energy equation applies this principle to systems in a state of steady flow, while the first law of thermodynamics can be applied to any type of system. Additionally, the steady flow energy equation takes into account changes in kinetic and potential energy, while the first law of thermodynamics only considers heat and work.

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