Thermodynamics
Thermodynamics
(OP)
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.





RE: Thermodynamics
RE: Thermodynamics
RE: Thermodynamics
RE: Thermodynamics
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
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RE: Thermodynamics
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
RE: Thermodynamics
RE: Thermodynamics
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.
RE: Thermodynamics
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.
RE: Thermodynamics
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
RE: Thermodynamics
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...
RE: Thermodynamics
RE: Thermodynamics
jbwick, what values are you plotting? dT vs time?