1.0 SOLVED PROBLEMS

1.1 Does it take longer to cook a 4 min egg in Mexico city than on the beach ? Are you kidding 4 minutes is 4 minutes? I mean to get an egg to the same consistency as a typical 4 min egg, how long would it take in Mexico city vs. somewhere at sea level?

Different liquids boil at different temperatures for a given air pressure. For example, water boils at a temperature of 212 °F at an air pressure of 14.7 psia (the pressure at sea level). However, a temperature of 189 °F is required to boil water at a pressure of 11 psia which is the air pressure at 8,500 feet above sea level, the altitude of Mexico city. Just because water boils at a lower temperature in Mexico city doesn't mean that it takes a shorter time to boil an egg. The same amount of heat transfer is required to get the egg to the right consistency regardless of water temperature. It will take longer to transfer enough heat to cook the egg if the water is boiling at a lower temperature than a higher one. We are so used to water boiling at the same temperature that it is very surprising to find that it takes longer than 4 minutes to boil a 4 minutes egg in Mexico city. How much longer I don't know, I'd have to go to Mexico city. If there are any Mexico city residents out there on the web, please try the experiment and let me know.

1.2 In a multiple and identical pump system, if one pump is in poor running order what is the effect on the discharge header head and the flow to the system?

1.3 What happens if the damaged pump's performance curve has all points at a lower head than the good pump's performance curve?

The best that the damaged pump can do is to produce the head corresponding to its shut-off head HC (point 2) at 0 flow. Since the head produced by the good pump is higher, there will be flow through the damaged pump in the reverse direction. The flow however will be impeded since the pump can produce some head. The system behaves as a branch system. The branch flow sees a head drop which is the sum of the shut-off head of the damaged pump, plus any friction loss, plus the static head of the suction tank on the inlet of the damaged pump.

1.4 Ever wonder how it occurs that you can get partially full pipes in what appears to be a pressurized system? What would the difference in Total Head be for the same flow rate for a system with partially full pipes and a system with full pipes?

What conditions are required for a fluid to completely fill the piping throughout a system? Consider Figure B-1 a system with a long discharge pipe, sloping gradually down from point 3 to point 2. A fluid under positive pressure will completely fill the available volume in the pipes. However, certain areas may remain unfilled if the pressure drops to zero or less. This can occur when the fluid rises above its discharge point. Under these conditions, if air enters the system, an air pocket can establish itself at the high point of the line. From there, it can expand with the help of negative pressure towards the end of the pipe, collapsing the fluid and causing the line to be partially full. A combination of pipe size, flow rate and system geometry can produce conditions causing certain locations of the piping to be partially full.

Q.: What is the difference in the Total Head (at the same flow rate) when all the discharge piping is full compared to when it is partially empty? A.: The answer depends which of these two terms is greater: the elevation difference between the collapse point (point 3) and the end point (point 2, the discharge end of the pipe) and the friction head for a full pipe between these two same points. If the friction head is greater, then it requires more energy to pump the fluid if the pipe is full. If the friction head is smaller then it requires less energy to pump the fluid with a collapsed line for the same flow rate.

1. Determine the Total Head of the system with a full pipe.

The general system equation for a single inlet single outlet system is:

 (B1)

HF1-2: head loss due to friction between points 1 and 2

HEQ1-2: head loss due to equipment between points 1 and 2

v2: average pipe velocity at point 2

v1: average pipe velocity at point 1

z2: elevation at point 2

z1: elevation at point 1

Since the suction and discharge tanks are un-pressurized and there is no equipment in the line then H1 = 0, H2 = 0, HEQ1-2 = 0. In addition the velocity at the surface of the suction tank is quite low, v1=0. The Total Head for the system assuming that all the piping is completely full is:

 (B2)

If the fluid is going to collapse, the collapse will occur at a position where the pressure is zero or less. The zero pressure position is at point O (see Figure B-1). Point O is however in the vertical portion of the pipe run and the fluid will fill the pipe at this position. The collapse will have to occur further down at or near the change of direction to horizontal of the piping .

2. Determine the head at point 3.

The general equation for the head at any point in a single inlet single outlet system is:

 (B3)

where point X is the point where the head is required. By applying this equation and substituting 3 for X, the head at point 3 can be determined. The terms v1 = 0, H1 = 0,

HEQ1-X = 0, the above equation becomes:

 (B4)

Substitute HP-FULL from equation B2 in the above equation, with v2 = v3, we obtain:

 (B5)

3. Determine the Total Head of the system assuming the fluid has collapsed at point 3

If the fluid has collapsed at point 3, then the exit point of the system will be point 3. Point 3 is the exit point since the fluid runs downhill from 3 to 2 without the system providing energy. Or in other words, if the system were cut at point 3, there would be no effect on the remaining system. By applying equation B1, the subscript 2 which identifies the exit point 2 is replaced with 3, the new exit point. and H1 = 0, H3 = 0, HEQ1-3 = 0, v1 = 0. The Total Head for a system with a single inlet and outlet with the outlet at point 3:

 (B6)

4. Establish the difference between the Total Head for a full line and a collapsed line.

The difference between the Total Head for a full line and a collapsed line is equation B2 - equation B6:

 (B7)

Our analysis compares the two cases at equal volumetric flow rates. This means that velocities are equal (v3 = v2),that is that v3 in the collapse fluid case equals v2 in the full pipe case. By rearranging the terms in equation B5:

and substituting into equation B7 then:

 (B8)

Therefore, the difference between pumping into a non-full vs. a full discharge line is affected by the value of the head at point 3. By analyzing the value of H3 , the conditions that make H3 negative, positive or zero will tell us if it is more difficult to pump into a full or non-full line.

 (B9)

Since HF1-2 - HF1-3 = HF3-2, equation B5 becomes:

H3 can be negative or positive depending on the value of the two terms in equation B9. If the friction component is greater than the static head between points 2 and 3:

If the friction component is less than the static head:

If it is necessary to avoid collapse, then a simple remedy is to add a restriction on the end of the pipe. This will add friction to the system and ensure that the complete pipe can be pressurized.

1.5 How long does it take to empty a tank?

To answer this question, we must first write the system equation assuming that the tank is kept full and that we have a constant flow (see Figure 1).

Figure 1

A simple system with a pump has the following system equation:

 (1)

Since there is no pump in this system then, HP = 0,and

 (2)

Let's assume that the entire friction loss for this system is an exit loss and the friction loss of a manual valve (butterfly). Therefore :

 (3)

There is no equipment so that HEQ = 0.

VP is the velocity through the pipe which is so short that we can neglect the friction loss. There is a relationship between v1, v2 and vP.

 (4)

where Q is the flow rate, A1, A2 and AP are respectively the cross-sectional areas of the tank, the outlet pipe and the pipe velocity. In this case, the outlet velocity equals the pipe velocity v2 = vP.

The system is open to atmosphere therefore H1= H2 = 0. We rewrite equation (2) taking into account equation (3):

 (5)

and since v2 = v1A1/A2 equation (5) becomes:

 (6)

and after simplification:

 (7)

This is the point at which we have to shift gears and adapt equation (7) to the system which we really want to study, that is one where the velocities v1 and v2 are constantly changing as well as the height z1. At each instant in time, we know that equation (7) is valid, therefore if we take the differential with respect time we will be able to take into account the variation of v1 and z1 with time.

If we differentiate both sides of equation (7) we get:

 (8)

since we know that v1 = dz1/dt and then

 (9)

which means that we can divide both sides of the equation by dz1/dt and equation (9) becomes:

 <(10)

We will select our vertical coordinate z to start at the level of point 2, and z1 becomes the variable z which will vary from 0 to h.

 (11)

Equation (11) is a simple 2nd order differential equation which after integrating twice will lead to the solution for z. However, integrating twice will force us to introduce two constants. The values of these constants depends on the initial conditions. The initial conditions are: at time t = 0, z = h and at time t = 0, v = 0 or dz/dt = 0.

After the first integration we get:

 (12)

since dz/dt = v1 = 0 at time t = 0, A = 0. After a second integration equation (12) (with A=0) becomes:

 (13)

since z = h at time t = 0 then B = h, and

 (14)

assuming we have a round tank which is typical then and, therefore and equation (14) becomes:

The graph below shows the time required to empty a 3 feet diameter tank with 3 feet of fluid with either a 2" discharge pipe or a 2.5" pipe.

1.6 How to calculate the pump flow, total head or efficiency based on power consumed by the motor.

This procedure will provide a method by which one can use the power generated by the motor to determine the total head, flow or efficiency of a pump. The power generated by the motor can be calculated by measuring the motor amperage taking into account the power factor of the specific motor.

Total Head (HP) and pump flow (Q) in an existing system are measurable quantities. However, sometimes one of these terms may be difficult to determine. Why? The devices required for the measurements may be difficult to install and or expensive, or the equipment may not be shut down long enough to install the devices. However, the power consumed by the motor can be easily measured based on the current flow (amperes).

The power consumed by the pump (at the pump shaft) is:

where Ppump: power consumed at the pump shaft

SG: specific gravity of the fluid

Q: flow through pump

pump : pump efficiency>

An induction motor has a reactive component which uses power to generate a magnetic field, this power cannot be used at the shaft. The power factor determines how much of the total power input to the motor can be transferred t the shaft. The power factor and motor efficiency is given by the motor manufacturers in the form of tables for various loads and different motor sizes (see table). The power delivered to the pump shaft based on the flow of current to the motor is:

where V: motor supply voltage, often 575volts for motors 250 hp and less

A: motor supply amperage

motor : motor efficiency

P.F.: power factor

By combining the 2 equations above, we can determine the Total Head, the flow or the pump efficiency.