Why is the Pumping Power Calculation Resulting in a Negative Value?

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In summary, the conversation discusses the concept of pumping power and the use of energy balance equations in solving problems related to fluid mechanics. It is noted that the mechanical energy balance equation should be used instead of the thermal energy balance. The enthalpy change between outlet and inlet is affected by pressure drop and potential energy differences, and can be calculated using the open system version of the first law of thermodynamics. However, in some cases, such as when there is a pressure drop but no change in temperature, the change in internal energy may be non-zero due to viscous frictional heating.
  • #1
ussername
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Homework Statement


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(question one is irrelevant for my problem).

What is the pumping power? From energy balance I got a negative value - but it should be positive since the mechanical work from pump is given into system.

Homework Equations


The input+output velocity:
$$v=\frac{\dot{V}}{A}$$
The friction coefficient:
$$f=\frac{\left |\Delta p \right |}{\frac{L}{D}\frac{1}{2}\rho v^{2}}$$
Moody diagram (not shown here).
The energy balance of open system:
$$\dot{W}=\dot{m}\cdot \left (h_{out}-h_{in}+\frac{v^{2}_{out}-v^{2}_{in}}{2}+g\cdot (z_{out}-z_{in}) \right )$$
The specific enthalpy:
$$h=u+\frac{p}{\rho }$$

The Attempt at a Solution


The velocity of input and output is ##v=5\,m\cdot s^{-1}##.
From calculated Reynolds number and relative roughness I found friction coefficient ##f=0.02## in Moody diagram (that is a correct value).
The pressure loss due to friction was calculated from the given equation: ##p_{F,out}-p_{F,in}=-5MPa##.
The hydrostatic pressure loss is ##p_{H,out}-p_{H,in}=\rho \cdot g\cdot (z_{out}-z_{in})=-177kPa##.
The total pressure loss is ##p_{out}-p_{in}=p_{F,out}-p_{F,in}+p_{H,out}-p_{H,in}=-5000000-177000=-5177kPa##.
The enthalpy change between outlet and inlet is:
##h_{out}-h_{in}=\frac{p_{out}-p_{in}}{\rho }=\frac{-5177000}{1000}=-5177\,J\cdot kg^{-1}##.
The change of potential energy between outlet and inlet is:
##e_{p,out}-e_{p,in}=g\cdot (z_{out}-z_{in})=9.81\cdot 18=174.58\, J\cdot kg^{-1}##
The resulting power of the pump is:
##\dot{W}=1000\cdot 1\cdot (-5177+174.58)=-5MW##
 
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  • #2
You should be using the mechanical energy balance equation, not the thermal energy balance. The mechanical energy balance is a modified version of the Bernoulli equation that includes frictional energy dissipation.
 
  • #3
But I don't know why I cannot use the energy balance. There are no restrictions to use energy balance. I thought a shaft work in thermodynamics is defined ##dW_{shaft}=\vec{dF}_{ext}\cdot \vec{dl}##, i.e. ##W_{shaft}>0## for pressure forces acting on the system.
 
  • #4
ussername said:
But I don't know why I cannot use the energy balance. There are no restrictions to use energy balance. I thought a shaft work in thermodynamics is defined ##dW_{shaft}=\vec{dF}_{ext}\cdot \vec{dl}##, i.e. ##W_{shaft}>0## for pressure forces acting on the system.
Your problem is with the enthalpy change. Also, isn't zout higher than zin?
 
  • #5
The enthalpy change ##h_{out}-h_{in}## should be negative since ##p_{out}-p_{in}## is negative:
##h_{out}-h_{in}=\frac{p_{out}-p_{in}}{\rho }=\frac{-5177000}{1000}=-5177\,J\cdot kg^{-1}##.

Chestermiller said:
Also, isn't zout higher than zin?
It is, which means that hydrostatic pressure difference ##p_{H,out}-p_{H,in}## is negative and potential energy difference ##e_{P,out}-e_{P,in}## is positive.
 
  • #6
If you have a fluid flowing through an insulated pipe, and there is a pressure drop, according to the open system version of the first law, what is the enthalpy change?
 
  • #7
Chestermiller said:
If you have a fluid flowing through an insulated pipe, and there is a pressure drop, according to the open system version of the first law, what is the enthalpy change?
The enthalpy change is:
$$h_{out}-h_{in}=u_{out}-u_{in}+\frac{p_{out}-p_{in}}{\rho }$$

Since all power given to the system increases the mechanical energy of a fluid, the change of internal energy should be zero (there is no change of temperature nor compression of a fluid).
 
  • #8
ussername said:
The enthalpy change is:
$$h_{out}-h_{in}=u_{out}-u_{in}+\frac{p_{out}-p_{in}}{\rho }$$

Since all power given to the system increases the mechanical energy of a fluid, the change of internal energy should be zero (there is no change of temperature nor compression of a fluid).
I don't understand. What is your answer?
 
  • #9
The first law for open system is:
$$\dot{U}_{out}+\dot{E}_{k,out}+\dot{E}_{p,out}-\dot{U}_{in}-\dot{E}_{k,in}-\dot{E}_{p,in}=\dot{Q}+\dot{W}_{shaft}+p_{in}\cdot \dot{V}_{in}-p_{out}\cdot \dot{V}_{out}$$

That is:
$$\dot{H}_{out}+\dot{E}_{k,out}+\dot{E}_{p,out}-\dot{H}_{in}-\dot{E}_{k,in}-\dot{E}_{p,in}=\dot{Q}+\dot{W}_{shaft}$$
 
  • #10
ussername said:
The enthalpy change is:
$$h_{out}-h_{in}=u_{out}-u_{in}+\frac{p_{out}-p_{in}}{\rho }$$

Since all power given to the system increases the mechanical energy of a fluid, the change of internal energy should be zero (there is no change of temperature nor compression of a fluid).
So you are saying that ##\Delta h=0## because ##\Delta u=0## and ##\Delta p/\rho=0## in my pipe example?
 
  • #11
Chestermiller said:
So you are saying that ##\Delta h=0## because ##\Delta u=0## and ##\Delta p/\rho=0## in my pipe example?
No ##p_{out}-p_{in}## is nonzero which means that ##\Delta h\neq 0##.
 
  • #12
ussername said:
No ##p_{out}-p_{in}## is nonzero which means that ##\Delta h\neq 0##.
That is not correct.
 
  • #13
I don't understand now. You stated that "there is a pressure drop". That should have impact on enthalpy. Can you explain?
 
  • #14
ussername said:
I don't understand now. You stated that "there is a pressure drop". That should have impact on enthalpy. Can you explain?
Delta h is equal to zero. And, as you said, delta p/ rho is not equal to zero. So what does that tell you?
 
  • #15
It tells us that change of internal energy is nonzero. But how is it possible?
 
  • #16
ussername said:
It tells us that change of internal energy is nonzero. But how is it possible?
It is due to viscous frictional heating. For a pressure drop of 50 psi, how much of a temperature rise does that amount to, assuming the fluid is water?
 
  • #17
It will be much dissipated heat :) Also the situation with nonzero pressure drop but same temperature on inlet and outlet and adiabatic conditions - it does not make sense (the shaft work cannot be determined from energy balance)?
 
  • #18
ussername said:
It will be much dissipated heat :) Also the situation with nonzero pressure drop but same temperature on inlet and outlet and adiabatic conditions - it does not make sense (the shaft work cannot be determined from energy balance)?
I calculate a temperature rise of only 0.1 C. So it is nearly isothermal, but not quite. So on the thermal energy balance, the mechanical energy dissipation is practically negligible, but in the mechanical energy balance, the thermal term would play a huge role. This is why you can't use the version with h to solve a flow problem.
 
  • #19
If I neglect the friction with pipe wall, I get the shaft work:
##\dot{W}=(p_{in}-p_{out})\cdot A\cdot v=5177000\cdot 0.2\cdot 5=5177000\,W##
 
  • #20
See the most recent thread started by MexChemE. It is very relevant to all this.
 
  • #21
ussername said:
If I neglect the friction with pipe wall, I get the shaft work:
##\dot{W}=(p_{in}-p_{out})\cdot A\cdot v=5177000\cdot 0.2\cdot 5=5177000\,W##
This doesn't neglect the friction loss in the pipe. If you like, I will illustrate how I would go about solving this problem, including a few different (but equivalent) ways to determine the pump power. Would you like me to proceed?
 
  • #22
In my view the friction forces within the fluid don't play a role. If I count up work ##dW## done on elements within the pipe, the friction forces of every two adjacent elements are deducted (III. Newton's law). For the whole fluid there remains just a friction force with wall (neglected) and pressure forces on inlet and outlet.

I would be happy to see if you get different result from mine.
 
  • #23
ussername said:
In my view the friction forces within the fluid don't play a role. If I count up work ##dW## done on elements within the pipe, the friction forces of every two adjacent elements are deducted (III. Newton's law). For the whole fluid there remains just a friction force with wall (neglected) and pressure forces on inlet and outlet.
Oh really? Well here is the macroscopic mechanical energy balance equation presented in Transport Phenomena by Bird, Stewart, and Lightfoot (a classic that has stood the test of time, and was updated to a second edition in ~ 2000), Chapter 7, Eqn. 7.5-10:

For steady turbulent incompressible flow,
$$\frac{1}{2}(v_b^2-v_a^2)+g(z_b-z_a)+\frac{(p_b-p_a)}{\rho}=\dot{W}_m-\sum{\frac{1}{2}v^2f\frac{L}{D}}-\sum{\frac{1}{2}v^2}e_v$$
where a and b are downstream and upstream locations in the conduit, respectively, ##\dot{W}_m## is the work done per unit mass by the pump, the first summation represents the sum of the frictional losses over all straight sections of conduit, and the second summation represents the sum of the frictional losses over all fittings, valves, meters, etc.
I would be happy to see if you get different result from mine.
For your problem,

let location 0 be the location immediately before the pump
let location 1 be the location immediately after the pump
let location 2 be the location of the final downstream location

We are going to work here with the locations 0 and 2 in your system. The pressures at locations 0 and 2 are equal, since they are open to the atmosphere. The velocities in the piping system at locations 0 and 2 are equal. The resistance in the valves and fittings are assumed negligible. So, the BSL equation for your problem reduces to: $$\dot{W}_m=\frac{1}{2}v^2f\frac{L}{D}+g(z_2-z_0)$$
The first term on the right hand side is the frictional energy loss per unit mass, and the second term is the potential energy increase per unit mass. I have calculated the frictional and potential energy terms in this equation using your methodology, and have obtained the following:
$$\frac{1}{2}v^2f\frac{L}{D}=5084\ (m/s)^2$$
$$g(z_2-z_0)=176.4\ (m/s)^2$$
So, $$\dot{W}_m=5260\ (m/s)^2$$
If I multiply this by the mass flow rate, I obtain the power:
$$\dot{W}=\dot{m}\dot{W}_m=(1000)(5260)=5.26\times 10^6\ watts=5.26/ Megawatts$$
Since the pressure change in your particular system between points 1 and 2 is given by $$(p_1-p_2)=\frac{1}{2}\rho v^2f\frac{L}{D}+\rho g(z_2-z_0)$$
for your particular system, the power is also given by $$\dot{W}=\dot{m}\frac{(p_1-p_2)}{\rho}$$
And since, in your particular system, the pressure at point 2 is equal to the pressure at point 0 (the pump inlet), for your particular system, the power is also equal to $$\dot{W}=\dot{m}\frac{(p_1-p_0)}{\rho}$$where ##(p_1-p_0)## is the pressure increase across the pump.
 

Related to Why is the Pumping Power Calculation Resulting in a Negative Value?

What is pumping power?

Pumping power is the amount of energy required to move a fluid through a system, typically measured in watts or horsepower. It is an important consideration in the design and operation of pumps.

How do you calculate pumping power?

Pumping power can be calculated using the equation: Power = (Flow Rate x Pressure) / Efficiency. Flow rate is the volume of fluid moved per unit time, pressure is the force needed to move the fluid, and efficiency takes into account any losses in the system.

What factors affect pumping power?

The main factors that affect pumping power are flow rate, pressure, fluid properties (such as viscosity and density), and the efficiency of the pump. Other factors, such as system design and operating conditions, can also play a role.

Why is it important to accurately calculate pumping power?

Accurately calculating pumping power is important because it allows for the proper sizing and selection of pumps, which can impact the efficiency and cost of a system. It also helps to ensure that the pump is operating at its optimal performance and can prevent issues such as pump failure or excessive energy consumption.

What are some common mistakes when calculating pumping power?

Some common mistakes when calculating pumping power include not considering all factors that affect power (such as efficiency or fluid properties), using incorrect units of measurement, and not accounting for changes in operating conditions. It is important to carefully review all inputs and double check calculations to ensure accuracy.

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