Let's say I have a simple oil pump connected in a closed system with a heat exchanger. The pump adds 127,000 Btu/hr of heat through friction and the heat exchanger removes the heat at 46,140 Btu/hr. What else do I need to find the steady state operating conditions of this system, i.e., the max oil temperature? Is there a simple way to convert the net heat input of 80860 Btu/hr into a final system temperature? Yes, my thermo is that rusty.
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Thermodynamics 2
- Thread starter texag
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What ivy says is correct, all the heat. At some point, the suction and discharge lines will radiate heat, the heat exchanger will remove heat because the delta lg mean of the the oil being hotter will make it remove heat. The oil may vaporize which will remove heat ( if there is a vent). Lots of thing could happen before the system takes in enough heat to melt the steel...
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- #4
I think I need to redefine the situation slightly based on the responses so far. Assume that all piping is adiabatic and heat is only removed by the heat exchanger and only added by the pump. When I start the pump the temperature will rise until the heat exchanger is removing heat at the same rate the pump is creating it due to friction. What I do not know is the time that will happen and what the resulting temperature will be. For example, imaging the cooling system in your vehicle. Simply stated it is a heat source (engine) and heat exchanger (radiator). My situation is identical to a vehical's system minus the thermostat. The oil temperature will rise to some finite level, I need to know what that is. Am I oversimplifying this situation?
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- #5
No, you appear to be overcomplicating the problem. The amount of heat removed by a heat exchange is not fixed; it is a function of the temperature delta.
Therefore, since your given value of heat removal is based on some temperature delta, simply divide the heat removed by the temperature delta to get an effective heat transfer coefficient. Divide the desired heat removal by the effective heat transfer coefficient to get the temperature delta at which the heat exchanger will remove the desired amount of heat.
TTFN
FAQ731-376
Therefore, since your given value of heat removal is based on some temperature delta, simply divide the heat removed by the temperature delta to get an effective heat transfer coefficient. Divide the desired heat removal by the effective heat transfer coefficient to get the temperature delta at which the heat exchanger will remove the desired amount of heat.
TTFN
FAQ731-376
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- #6
sailoday28
Mechanical
If the heat generated by the pump is dependent on its fluid inlet temperature (T in p) and flow rates, w1, wp along with specific heats are known-
for steady state conditions
subscript 1 for fluid 1 and p for pumpthen
Q=Q(T in p) 1
Q=w1 cp1 (Tin 1 - T out 1) 2
Q=wp Cpp (Tout p - T in p) 3
Qhtx =Q= UAFt(Log mean delta T) 4
Log mean delta T
= function of (Tin1, Tout 1, Tout p, T in p 5
UFt=UFt(Tin1, Tout 1, Tout p, T in p) 6
There are 6 unknowns and 6 equations
Unknowns, Tout 1, Tin p, Tout p, Q , UFt, Log mean delta T
The heat exchanger and pump vendors should be able to provide additional data to simplify and perhaps eliminate some of the variables listed above.
Regards
for steady state conditions
subscript 1 for fluid 1 and p for pumpthen
Q=Q(T in p) 1
Q=w1 cp1 (Tin 1 - T out 1) 2
Q=wp Cpp (Tout p - T in p) 3
Qhtx =Q= UAFt(Log mean delta T) 4
Log mean delta T
= function of (Tin1, Tout 1, Tout p, T in p 5
UFt=UFt(Tin1, Tout 1, Tout p, T in p) 6
There are 6 unknowns and 6 equations
Unknowns, Tout 1, Tin p, Tout p, Q , UFt, Log mean delta T
The heat exchanger and pump vendors should be able to provide additional data to simplify and perhaps eliminate some of the variables listed above.
Regards
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- #7
Either i am missing something really basic or there is no steady state here... using rounded numbers to simplify and considering average properties:
The pump adds Qp = Wp Cp (T2 - T1) = 127,000 Btu/h
The exchanger removes Qe = We Ce (t2 - t1) = 46,000 Btu/h
The neat heat input per hour = 81,000 Btu/hr
Without going into differential equations, just following the first cycle:
If only 46,000 Btu are removed... the Oil will not go back to the temperature T1 to start over again...
The exit temperature of the oil from the exchanger will be:
T2 - (46000/WpCp) = T1'
T1' > T1
The new T2 temperature (T2') will be:
T2'= T1'+ (127,000/WpCp) = T2 +(127,000 - 46,000)/WpCp
Unless the heat added to the system is removed from the system the oil temperature will increase until something breaks... usually the pump seals first.
I've always used a "heat balance" to resolve a heat exchange problems.
If the only heat input and removal is at the pump and HE... am i missing something?
saludos.
a.
The pump adds Qp = Wp Cp (T2 - T1) = 127,000 Btu/h
The exchanger removes Qe = We Ce (t2 - t1) = 46,000 Btu/h
The neat heat input per hour = 81,000 Btu/hr
Without going into differential equations, just following the first cycle:
If only 46,000 Btu are removed... the Oil will not go back to the temperature T1 to start over again...
The exit temperature of the oil from the exchanger will be:
T2 - (46000/WpCp) = T1'
T1' > T1
The new T2 temperature (T2') will be:
T2'= T1'+ (127,000/WpCp) = T2 +(127,000 - 46,000)/WpCp
Unless the heat added to the system is removed from the system the oil temperature will increase until something breaks... usually the pump seals first.
I've always used a "heat balance" to resolve a heat exchange problems.
If the only heat input and removal is at the pump and HE... am i missing something?
saludos.
a.
Steady state will be achieved because at some temperature the fluid being circulated will boil and the the energy required to vaporize the oil will match the energy input.
because there is a heat exchanger, its Q= UA dTlm will increase and remove heat. Some trial end error calculations can be performed to find where the system reaches steady state.
because there is a heat exchanger, its Q= UA dTlm will increase and remove heat. Some trial end error calculations can be performed to find where the system reaches steady state.
A pump that adds 127,000 BTU/hr through friction? A heat exchanger associated with that pump that can only remove 46,000 BTU/hr? Piping that is truly adiabatic?
On the planet that I live on, the pump would die within the first hour either through bearing failure from excessive heat or it would cavitate to death. Is this a classroom problem, because no one that has ever run a pump would pose it?
David
On the planet that I live on, the pump would die within the first hour either through bearing failure from excessive heat or it would cavitate to death. Is this a classroom problem, because no one that has ever run a pump would pose it?
David
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- #11
I was trying to avoid getting really detailed with my problem, but I owe to those who have taken the time to try to help me to better describe the system. Hope this clears up some confusion.
In reality 90W oil is pumped (pump adds some heat) from a 20gal reservoir to the heat exchanger (46,140 Btu/hr rating), then it travels into a plunger pump to lubricate moving parts (this is where most of the heat is added through friction), and finally back into the reservoir. I based the heat added off of the energy conversion efficiency of the plunger pump, I am making the assumption that all the lost power is lost through friction and thus heat. In reality the plunger pump loses some heat through convection along with piping, filters, and reservoir exposure to atmosphere.
We have empirical data on the max temperature the oil reached for extended periods of operation, steady state temp is highly dependant on plunger pump rpm. I would like to make a mathematical model of the system to know T steady state for any combination of ambient temp, efficiency (heat added), rpm, heat removal rate, conduit length, etc...
In reality 90W oil is pumped (pump adds some heat) from a 20gal reservoir to the heat exchanger (46,140 Btu/hr rating), then it travels into a plunger pump to lubricate moving parts (this is where most of the heat is added through friction), and finally back into the reservoir. I based the heat added off of the energy conversion efficiency of the plunger pump, I am making the assumption that all the lost power is lost through friction and thus heat. In reality the plunger pump loses some heat through convection along with piping, filters, and reservoir exposure to atmosphere.
We have empirical data on the max temperature the oil reached for extended periods of operation, steady state temp is highly dependant on plunger pump rpm. I would like to make a mathematical model of the system to know T steady state for any combination of ambient temp, efficiency (heat added), rpm, heat removal rate, conduit length, etc...
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- #13
Too incomplete to fill in the unknowns sailoday28 listed. My company sells only the plunger pump in this system, we make recommendations for the other components but ultimately they are the customers responsibility. Originally I intended to do exactly what you had mentioned, trend empirical data then build an approximate model off of that. That works great for my system but does not address a specific situation the customer will have. The problem is that sometimes customers ask me if they will need a heat exchanger (size?) with their specific application, which by the way wildly fluctuates. That is why I was wanting to build a separate theoretical model before gathering test data and then compare the two and refine my math model. Get close enough to real world and then I can plug and chug for any system variable.
jbwick, what values are you plotting? dT vs time?
jbwick, what values are you plotting? dT vs time?
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