Transient Behavior of an Air-Cooling System An air-cooled engine generates heat at a constant rate Q gen = 8530 Btu/min. The air in the engine housing is circulated rapidly enough for its temperature to be considered uniform and equal to the temperature of the outlet air. Air passes through the

Question

Transient Behavior of an Air-Cooling System
An air-cooled engine generates heat at a constant rate [latex]\dot{Q}_{gen}[/latex] = 8530 Btu/min.

The air in the engine housing is circulated rapidly enough for its temperature to be considered uniform and equal to the temperature of the outlet air. Air passes through the engine housing at a rate of 6.00 lb-mole/min, entering at a temperature of 65°F, and an average of 0.200 lb-mole of air is contained within the engine housing.
(We will neglect the variation of this quantity with the changing temperature of the gas.) Heat is lost from the housing to its surroundings at a rate

[latex]\dot{Q}_{lost}[/latex](Btu/min) = [33.0 Btu/(°F·min)](T - 65°F)

Suppose the engine is started with the inside air temperature equal to 65°F.
1. Calculate the steady-state air temperature if the engine runs continuously for a very long time, assuming

C[latex]_{v}[/latex] = 5.00 Btu/(lb-mole·°F)

2. Derive a differential equation for the variation of the outlet temperature with time from startup and solve it. Calculate how long it will take for the outlet temperature to reach one degree below its steady-state value.

Accepted Answer

System = the air within the engine housing.
1. The steady-state energy balance equation may be obtained by setting dT/=dt equal to zero in Equation 10.3-12.

Open System: [latex]\boxed{MC_{v}\frac{dT}{dt} =\dot{m}C_{p}(T_{in}- T) +\dot{Q} +\dot{W}_{s}}[/latex] (10.3-12)

[latex]0=\dot{m}C_{p}(T_{in} - T) +\dot{Q}+\dot{W}_{s}\ \left. \Large{\Downarrow} ight. \begin{matrix} T_{in} = 65°F (given)\ T=T_{s}(°F)(steady-state outlet temperature) \ \dot{W}_{s} =0 (no moving parts) \ \dot{Q}=\dot{Q}_{gen} - \dot{Q}_{lost}\end{matrix} \ \dot{m}C_{p}(T_{s} - 65.0°F) = 8530 Btu/min - 33.0 (T_{s} - 65.0°F)[/latex]

Gases:

[latex]\boxed{\Delta \hat{H}\left(\frac{\mathrm{kJ}}{\mathrm{mol}} ight) \approx \int_{T_1}^{T_2} C_{\mathrm{p}}(T) d T= ext { Enthalpy("Species", } T1, \mathrm{~T} 2, " \mathrm{~T} ext { unit","g"]}}[/latex] (8.3-12a)

[latex]\boxed{ \begin{aligned} \Delta \hat{U}\left(\frac{\mathrm{kJ}}{\mathrm{mol}} ight) & \approx \int_{T_1}^{T_2} C_{\mathrm{v}}(T) d T \approx \int_{T_1}^{T_2} C_{\mathrm{p}}(T) d T-R\left(T_2-T_1 ight) \quad\left(R=8.314 imes 10^{-3} \frac{\mathrm{kJ}}{\mathrm{mol} \cdot \mathrm{K}} ight) \ & \approx ext { Enthalpy("Species", T1,T2,"T unit","g") }-8.314 \mathrm{e}-3^(\mathrm{~T} 2-\mathrm{T} 1)\end{aligned}}[/latex] (8.3-12b)

Liquids and Solids:

[latex] \boxed{ \begin{aligned}\Delta \hat{H}\left(\frac{\mathrm{kJ}}{\mathrm{mol}} ight) &\approx \hat{V} \Delta P+\int_{T_1}^{T_2} C_{\mathrm{p}}(T) d T \ &\approx 1.013 \mathrm{e}-4^ \mathrm{MW}(" ext { species") }(\mathrm{P} 2-\mathrm{P} 1) / \mathrm{SG}(" ext { species") } \& + ext { Fnthalpy("Species", } \mathrm{T} 1, \mathrm{~T} 2, " \mathrm{~T} ext { unit","state") }\end{aligned}}[/latex] (8.3-12c)

[latex]\boxed{ \begin{aligned} \Delta \hat{U}\left(\frac{\mathrm{kJ}}{\mathrm{mol}} ight) & \approx \int_{T_1}^{T_2} C_{\mathrm{v}}(T) d T \approx \int_{T_1}^{T_2} C_{\mathrm{p}}(T) d T \ & \approx ext { Enthalpy("Species", T1,T2,"T unit", "state") }\end{aligned}}[/latex] (8.3-12d)

Assuming ideal-gas behavior

[latex]C_{p}=C_{v}+R= (5.00 + 1.99) Btu/(lb-mole\cdot °F) \quad\quad\quad(Equation 8.3-12) \ \dot{m}C_{p} = \begin{array}{c|c}6.00 lb-mole &6.99 Btu \ \hline min &lb-mole\cdot °F\end{array} = 41.9 \frac{Btu}{min\cdot °F} [/latex]

The energy balance then becomes

[latex]41.9(T_{s} - 65.0°F)= 8530 Btu/min - 33.0 (T_{s} - 65.0°F) \ \left. \Large{\Downarrow} ight. \ \boxed{T_{s}=179°F}[/latex]

2. The unsteady-state balance equation (10.3-12) is, for our system,

[latex]\begin{matrix}MC_{v} \frac{dT}{dt} =\dot{m}C_{p}(65°F - T) + \dot{Q}_{gen} - \dot{Q}_{lost} \\ \left. \Large{\Downarrow} ight.\begin{matrix}M = 0.200 lb-mole \ C_{v}= 5.00 Btu/(lb-mole\cdot °F) \ \dot{m}C_{p} = 41.9 Btu/(min\cdot °F) [from Part 1] \ \dot{Q}_{gen}= 8530 Btu/min \\dot{Q}_{lost}= 33.0(T - 65°F) Btu/(min\cdot °F) \end{matrix} \\ \boxed{\begin{matrix} \frac{dT}{dt} =- 74.9T + 13,400°F /min \ t=0, T= 65°F\end{matrix} }\end{matrix}[/latex]

You can check this result by setting dT/dt = 0 and solving the resulting equation for the steady-state value of T. The result is (13,400/74.9)°F = 179°F, in agreement with the result of Part 1.
The solution of the equation is obtained by separating variables and integrating:

[latex]\int_{65°F}^{T}{\frac{dT}{(13,400 - 74.9T)}} = \int_{0}^{t}{dt}\ \left. \Large{\Downarrow} ight.\- \frac{1}{74.9} \ln (13,400 - 74.9T) \left. \Large{]} ight. ^{T}_{65°F} = t \\left. \Large{\Downarrow} ight.\ \ln \frac{(13,400 - 74.9T)}{[13,400 - (74.9)(65)]} =- 74.9t \\left. \Large{\Downarrow} ight.\ 13,400 - 74.9T =8530 \exp (-74.9t) \\left. \Large{\Downarrow} ight. \ \boxed{T(°F) = 179 - 114\exp (- 74.9t)}[/latex]

A check on the solution may be obtained by substituting t = 0 and verifying that T equals the specified initial value, 65°F.
If you evaluate T from this equation for a number of values of t(s), a plot appears as follows:

The plot starts from the initial condition of 65°F and asymptotically approaches the steady-state value of 179°F. It takes 3.8 seconds for the temperature to reach 178°F, one degree short of its final value.

Question 10.3.2: Transient Behavior of an Air-Cooling System An air-cooled en......

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