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3. ^ car cnvironmcntal controllcr maintains thc tcmpcraturc at a constant levcl Too. Thc passcngcr compartmcnt dissipates (loscs) hcat according to Newton's la...

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3. ^ car cnvironmcntal controllcr maintains thc tcmpcraturc at a constant levcl Too. Thc passcngcr compartmcnt dissipates (loscs) hcat according to Newton's law. Thc car' s onboard systcm utilizes PI-controllcr to hold thc tcmpcraturc at the desired setting Too: According to Our class notes this meansd[' +kT = kR dtwith initial condition T(0) To and R satisfiesR = K 'Tx-T+d f( -Tlu)Jdu) with the controller parameters as the gain K (dimensionless) and integral time ti (units o

3. ^ car cnvironmcntal controllcr maintains thc tcmpcraturc at a constant levcl Too. Thc passcngcr compartmcnt dissipates (loscs) hcat according to Newton's law. Thc car' s onboard systcm utilizes PI-controllcr to hold thc tcmpcraturc at the desired setting Too: According to Our class notes this means d[' +kT = kR dt with initial condition T(0) To and R satisfies R = K 'Tx-T+d f( -Tlu)Jdu) with the controller parameters as the gain K (dimensionless) and integral time ti (units of time): Diffcrcntiatc our diffcrcntial cquation for T and rcwritc this pair of cqua- tions as singlc, sccond-ordcr diffcrcntial cquation for thc tcmpcraturc Note that wC have kR on the RHS while wC had R in the notes_ This choice means that k has no effect on R and R is an external temperature _ Write down the rules that distinguish when the systern is overdamped underdamped; and critically damped. What arc thc units of R? Let' $ assume that = 1, & = 1, and Tx = 1_ Choose the gain K that makes Our controller critically damped. Solve the problem for the temperature using the parameters in (3d). Use thc four-stcp approach: Stcp 1: Gct Yh, Stcp 2: Gct Yp; Stcp 3: Satisfy conditions using y = Yh + Yp, Stcp 4: Sketch:



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An object of mass $m_{1}$, specific heat $c_{1},$ and temperature $T_{1}$ is placed in contact with a second object of mass $m_{2},$ specific heat $c_{2},$ and temperature $T_{2}>T_{1}$. As a result, the temperature of the first object increases to $T$ and the temperature of the second object decreases to $T^{\prime}$. (a) Show that the entropy increase of the system is $$ \Delta S=m_{\mid} c_{1} \ln \frac{T}{T_{1}}+m_{2} c_{2} \ln \frac{T}{T_{2}} $$ and show that cncrgy conscrvation requires that $$ m_{1} c_{1}\left(T-T_{1}\right)=m_{2} c_{2}\left(T_{2}-T^{r}\right) $$ (b) Show that the cntropy change $\Delta S,$ considered as a function of $T$, is a maximum if $T=T^{\prime},$ which is just the condition of thermodynamic equilibrium. (c) Discuss the result of part (b) in terms of the idea of cntropy as a mcasurc of randomncss. 20.52 - CALC To heat 1 cup of water $\left(250 \mathrm{~cm}^{3}\right)$ to make coffee. you place an electric heating element in the cup. As the water temperature increases from $20^{\circ} \mathrm{C}$ to $78^{\circ} \mathrm{C},$ the temperature of the heating element remains at a constant $120^{\circ} \mathrm{C}$. Calculate the change in entropy of (a) the water; (b) the heating element; (c) the system of water and heating element. (Make the same assumption about the specific heat of water as in Example 20.10 in Section $20.7,$ and ignore the heat that flows into the ceramic coffee cup itself.) (d) Is this process reversible or irreversible? Explain. 20.53 e- DATA In your summer job with a venture capital firm, you are given funding requests from four inventors of heat cngincs. The inventors claim the following data for their operating prototypes: $$ \Delta S=m_{1} c_{1} \ln \frac{T}{T_{1}}+m_{2} c_{2} \ln \frac{T^{\prime}}{T_{2}} $$ and show that cnergy conscrvation requires that $$ m_{1} c_{1}\left(T-T_{1}\right)=m_{2} c_{2}\left(T_{2}-T^{\prime}\right) $$ (b) Show that the cntropy change $\Delta S$, considered as a function of $T$, is a maximum if $T=T^{\prime \prime},$ which is just the condition of thermodynamic equilibrium. (c) Discuss the result of part (b) in terms of the idea of cntropy as a measure of randomncss.

In this question we're going to be applying must transfer and transport phenomena for units and units conversion. So we've been given an expression of G. Dp. Why defended by D. is equality two plus 0 26. Um You over broad T. This to the power 1/3. Multiplied by DP. Again. You girl over meal multiplied are raised to the power of 1/2. So if we are to look at the right hand side we can see that two Plus 026 multiplied by one times 10 to the power negative five new turn. Second pair square meter divided by density which is one K. G. P. Big meter multiplied by D. Which is one times 10 to the negative five meta. Go ahead pair. two We raised this to the power of 1/3 and do the same for the other part of Which is 0.005 times this trip in matters times 10 meters per second. Want to play it by density of one kg p cubic meter. This should be divided by One terms to into the power negative five newton second pair square meter. And we raised this to the power of half so we simplify this. This becomes equal to 44.4. Now looking at the right hand side, we've got K. G. Agent multiplied by DP 7.0 05 m. Multiplied by 0.1. All of these divided by One times 10 to the power -5. This should be meters. Quite bad. Second and this should be equated To the right hand side, which is 44 sure. Now, if we make kg support formula, we've got our K G is equality 0.89 meters per second. Moving on. But we okay. The motto. The motto is more like an ideal state of law. So it could happen that the conditions under which the model was developed, for example temperature pressure or humidity. Just any parameter. It could happen that the conditions under which this mortal was developed. It's different from the real condition or it's out of range of the empirical data. So it can happen that the data, for example this motor could have been developed say for D. P. Ranging from 0.1 head up to maybe around one. So it could happen that in real life condition you can have a d. 0.001 Or. g. So if the motor was developed for this range and you have this you've got these conditions this or this that is out of range in which the model was developed. You could have different different solutions. Again When we're looking at two p. The Mortal assumes that particles, the damage of the particles are uniform, but this may not necessarily patrol in real life situations. Not all particles would be uniform. So again, this is what may because some defecation.


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