Question

please I need your help for this question.

A tank containing 45kg of liquid water initially at 45°C has one inlet and one exit with equal mass flow rates. Liquid water enters at 45°C and a mass flow rate of 270kg/hr. A cooling coil immersed in the water removes energy at a rate of 7.6kW. The water is well mixed by a paddle wheel so that the water temperature is uniform throughout. The power input to the water from the paddle wheel is 0.6kW. The pressures at the inlet and exit are equal and all kinetic and potential energy effects can be ignored. Determine the variation of water temperature

Answer 1

I'm not fluent in Thermo but I found a Thermo book on Google Books that can help you:

http://books.google.com/books?id=7VJlgo7AescC&pg=PA80&lpg=PA80&dq=a+tank+containing+45+kg+of+liquid+water+initially&source=bl&ots=clUf09Rvhf&sig=k-w7q-VvF433z6YCQUaTvsWmlVI&hl=en&sa=X&ei=QIu-T6SzJ8LW6gH_pvlh&ved=0CEsQ6AEwBQ#v=onepage&q=a%20tank%20containing%2045%20kg%20of%20liquid%20water%20initially&f=false

You may have to scroll up to find the examples.

Answer 2

There may be a way to approach this question directly with temperature, but i decided to consider the Joules in the fluid as a concentration and attacked it much like a brine solution question:

Answer 3

I considered that since there is a cooling effect whereby Watts (J/s) are leaving the fluid then you can model the fluid coming in as adding watts, and fluid leaving as subtracting watts
\[Joules_{entering}=mass_{water} *specificHeat_{Water} *\Delta T\]
\[Joules_{entering}=270 (\frac{kg}{hr}) (\frac{hr}{3600s}) (\frac{4190J}{kg ^oC})(45^oC)\]
Since the specificheat of water changes at zero degree i choose this temperature as the abatrary temperature reference

Answer 4

The Joules leaving as a result of the cooling exchange can be determined in two parts:
7kW = 7kJ/s, as given in the question and then the following:
let 'y' be the total number of Joules in 45kg of mixture at any instant. The rate at which joules are leaving the mixture can be given by finding the concentration and multipling by the fluid rate:
\[(\frac{Joules_{inMixture}}{Amount_{ofMixture}})(Rate_{ofExit})\]
\[(\frac{y}{45kg})(\frac{270kg}{hr})(\frac{hr}{3600s})\]

Answer 5

If you look at the units on all of these rates you'll see they're all in Joules/s which represents the first derivative of Joules. Now we can form our differential equation:
\[\frac{dy}{dt} = (Joules_{Entering}) - (Joules_{Exiting})\]
\[\frac{dy}{dt} = (14141.25\frac{J}{s}) - (7000\frac{J}{s})-(y\frac{1}{600}\frac{J}{s})\]
\[\frac{dy}{dt} = 7141.25- y \frac{1}{600}\]
\[\frac{dy}{dt} + y \frac{1}{600}= 7141.25\]

From here you can proceed with integrating factor to solve y(the quantity of Joules in the 45kg mixture at any time t). By dividing the resulting solution by Specific heat capacity of water and then the total mass you will arrive at a formula that descibes the temperature.

Answer 6

After reviewing this can be achieved readily with temperature as well, converting the 7000J/s to a change in temperature you must use(as in the previous example)
\[Q=mc\Delta T\]
temperature loss do to cooling element:
\[7000 \frac{J}{s} \rightarrow 0.037125431 \frac{^oC}{s}\]
which yields:
\[\frac{dT}{dt}=(\frac{270kg}{hr})(\frac{hr}{3600sec})(\frac{1}{45kg})(45^oC-T) - 0.037125431 \frac{^oC}{s}\]
\[\frac{dT}{dt}+ \frac{1}{600}T =0.037874569\]

Which will yield your equation in Temperature, but still requires Integrating Factor

Answer 7

Very simple, write an energy balance and find enthalpies at different inlet and exit states.

Mass flow is given and velocity and height are ignored.