Thermofluids: Change in energy

[johnsonc007]

1. The problem at hand: A gas is contained in a closed rigid tank. An electric resistor in the tank transfers energy to the gas at a constant rate of 1000W. Heat transfer between the gas and the surroundings occurs at a rate of Q(dot) = -50t, where Q (dot) is in watts, and t is in time, minutes.
(a) Plot the time rate of change of energy of the gas for 0<t<20 min in watts. (actually t is greater than or equal to 0, and t is less than or equal to 20)
(b) Determine the net change in energy of the gas after 20 minutes, in kJ


2. I am not following on how to complete this task. I am confused by the Q(dot) equation. How can it equal watts when multiplied by minutes, as a watt is a J/s? Would get Q(dot)=Jmin/s...


3. I have tried to solve this numerous times, all with the wrong solution. I tried starting part b first as I need that data to do part a. I know that dE/dt = Q(dot) - W(dot). I drew two closed systems, one with the gas isolated with the resistor work entering and Q=0, and then another with the resister in the system and Q leaving the system, so that W=0. Am I on the right path? If so, what can I do with that? I found W=1200kJ but Q being some huge number and not working towards the correct solution.
Also took derivatives, but that got me nowhere fast too :(

 

[edgepflow]

1. The problem at hand: A gas is contained in a closed rigid tank. An electric resistor in the tank transfers energy to the gas at a constant rate of 1000W. Heat transfer between the gas and the surroundings occurs at a rate of Q(dot) = -50t, where Q (dot) is in watts, and t is in time, minutes.
(a) Plot the time rate of change of energy of the gas for 0<t<20 min in watts. (actually t is greater than or equal to 0, and t is less than or equal to 20)
(b) Determine the net change in energy of the gas after 20 minutes, in kJ


2. I am not following on how to complete this task. I am confused by the Q(dot) equation. How can it equal watts when multiplied by minutes, as a watt is a J/s? Would get Q(dot)=Jmin/s...


3. I have tried to solve this numerous times, all with the wrong solution. I tried starting part b first as I need that data to do part a. I know that dE/dt = Q(dot) - W(dot). I drew two closed systems, one with the gas isolated with the resistor work entering and Q=0, and then another with the resister in the system and Q leaving the system, so that W=0. Am I on the right path? If so, what can I do with that? I found W=1200kJ but Q being some huge number and not working towards the correct solution.
Also took derivatives, but that got me nowhere fast too :(

As time goes on, the tank surface temperature increases and so does the heat that escapes. Thus, the power input as a function of time is:

Q(t) = 1000 watt - 50 (watt / min ) t

For example, when t = 5 min, the energy that is absorbed inside the tank is:

Q(5 min) = 1000 watt - 50 (watt / min) 5 min = 750 watt

The total energy absorbed is: Energy = power X time or E = P t or dE = P dt

and E = Integral (Q (t) ) dt

 

[johnsonc007]

Ok. Totally makes sense for part 'a', actually figured it out last night with a study partner.

Now for part 'b': I was considering that the resistor input was work transfer, not heat transfer. Why is the resistor considered a heat transfer source?

Your assistance helped me finish the problem, thank you! However I would like a better understanding of the relationship of the resistor to the energy of the gas and the system. Can you clarify? (The textbook associated with this class showed a resistor as a work source, not heat. It must have been just for that example.)

 

[edgepflow]

Ok. Totally makes sense for part 'a', actually figured it out last night with a study partner.

Now for part 'b': I was considering that the resistor input was work transfer, not heat transfer. Why is the resistor considered a heat transfer source?

Your assistance helped me finish the problem, thank you! However I would like a better understanding of the relationship of the resistor to the energy of the gas and the system. Can you clarify? (The textbook associated with this class showed a resistor as a work source, not heat. It must have been just for that example.)

The resister creates heat by what is called "I squared R" heating. In other words, when you pass a current I (in amps) through a resistance R (in ohms) you release heat Q = I^2 R in watts. This is why most of the power consumption of an incandescent light bulb goes into heat.

In this problem, the 1000 watt source could come from passing 10 amps of current through a 10 ohm resister (or any other combination of I^2 R that gives 1000 watt). As far as the energy is concerned, 1000 w is important.

Now from the 1st law of thermo, we simply have (ignoring changes in kinetic and potential energy): Qheat = m (hf - hi) = m cp (Tf - Ti). We could then find the temperature of the gas, Tf, as a function of time by solving the first law equation for Tf and with Qheat = Integral Qheat(t) dt.

Usually a work source will have shaft rotation (e.g turbine, pump, fan, windmill, and so on).

 

[DeeNos]

Dear scientist,

Thank you for reaching out with your questions and concerns about the problem provided. I understand that you are struggling with understanding the Q(dot) equation and how to solve the task at hand. Let me try to explain it in a way that may be easier for you to understand.

First, let's define some terms. Thermofluids is the study of the behavior of fluids and their interactions with heat and energy. In this problem, we are dealing with a gas, which is a type of fluid, and the transfer of energy through heat.

Now, let's look at the Q(dot) equation. Q(dot) is the rate of heat transfer, which is measured in watts. Watts is a unit of power, which is the rate at which energy is transferred. So, Q(dot) is essentially telling us how much heat energy is being transferred per unit of time (in this case, per minute). So, Q(dot) = -50t means that the rate of heat transfer is decreasing by 50 watts for every minute that passes.

Moving on to part (a) of the problem, we are asked to plot the time rate of change of energy of the gas for 0<t<20 min. This means we need to plot the change in energy over time for the given time interval. To do this, we can use the formula dE/dt = Q(dot) - W(dot), where dE/dt is the time rate of change of energy, Q(dot) is the rate of heat transfer, and W(dot) is the rate of work done on the system. In this case, there is no work being done on the system, so W(dot) = 0. Therefore, the equation becomes dE/dt = Q(dot). We can then plug in the given values for Q(dot) and solve for dE/dt. This will give us the time rate of change of energy, which we can then plot over the given time interval.

For part (b), we need to determine the net change in energy of the gas after 20 minutes. To do this, we can use the formula E2 - E1 = ∫Q(dot)dt, where E2 is the final energy of the gas, E1 is the initial energy of the gas, and ∫Q(dot)dt is the integral of the rate of heat transfer over time. We can plug in the given values and solve for E2 -