q8

Chris Nkinthorn $\texttt{20190723}$

Questions

1. Describe the reason(s) why the overall energy balance equations are not sufficient to solve the heat exchanger problem.

   Overall is what exits and enters, it is necessary to determine the total conductance of the heat exchanger to solve the two energy balances

   $$\begin{align} \dot{q} &=\dot{m}_{H} c_{H}\left(T_{H, i n}-T_{H, o u t}\right) \\ \dot{q} &=\dot{m}_{C} c_{C}\left(T_{C, o u t}-T_{C, j n}\right) \end{align}$$

   However, there are three unknowns: $T{\text {Cout}}, T{H, o u t}, \dot{q}$. To solve, we are required to solve an additional rate equation to determine the performance of a heat exchanger (HX). This requires analysis of the HX conductance’s dependence of the flow configuration.

2. Describe what is meant by the LMTD method of solution.

   Log Mean Temperature Difference (LMTD) is a method which solves the additional rate equation required in HX problems by a differential energy balance along the length of each welded pipe in the HX. It assumes time invariant and that temperature be the driving force in the energy exchange.

   $$\left.\begin{align} &\text{ }\frac{d T_{H}}{d x}=-\frac{U A}{L \dot{m}_{H} c_{H}}\left(T_{H}-T_{C}\right) \\ -&\left[\frac{d T_{C}}{d x}=-\frac{U A}{L \dot{m}_{C} c_{C}}\left(T_{H}-T_{C}\right)\right] \end{align}\right\}\texttt{combined}\\$$

   The individual energy balances are collected in to a single ordinary differential equation (ODE) dependent on a transformed variable, $\theta$.

$$\frac{d\overbrace{\left(T_{H}-T_{C}\right)}^{\theta}}{d x}=-\frac{U A}{L}\left(T_{H}-T_{C}\right)\left(\frac{1}{\dot{m}_{H} c_{H}}-\frac{1}{\dot{m}_{C} c_{C}}\right)$$

$$\frac{d \theta}{d x}=-\frac{U A}{L} \theta\left(\frac{1}{\dot{m}_{H} c_{H}}-\frac{1}{\dot{m}_{C} c_{C}}\right)$$

\label{eqn:ode} \ln \left(\frac{T_{H, o u t}-T_{C, i n}}{T_{H, i n}-T_{C, o u t}}\right)=-U A\left(\frac{1}{\dot{m}_{H} c_{H}}-\frac{1}{\dot{m}_{C} c_{C}}\right)

This solves energy balance as an expression of the boundary temperatures.

$$\dot{q} = UA \overbrace{\left[\frac{\left(T_{H, o u t}-T_{C, j n}\right)-\left(T_{H, j n}-T_{C, o u t}\right)}{\ln \left[\frac{\left(T_{H, o u t}-T_{C, i n}\right)}{\left(T_{H, i n}-T_{C, o u t}\right)}\right]}\right]}^{\Delta T_{LMTD}}$$

This a form reminiscent of convective heat transfer, where a heat transfer coefficient $h$ is substituted with $U$.

1. Discuss the conditions under which the LMTD solution is the same for parallel and counter-flow configurations.

   LMTD solution is always the same for parallel and counter-flow configurations due to how the expression of the temperature difference is the related by a factor of $-1$. This factor in the numerator is canceled by a recurrence in the denominator.

   $$\overbrace{ \frac{\left(\theta_{x=L}-\theta_{x=0}\right)}{\ln \left[\frac{\theta_{x=0}}{\theta_{x=L}}\right]} }^{\Delta T_{LMTD,pf}} = \frac{\cancel{(-1)}\left(\theta_{x=L}-\theta_{x=0}\right)}{\cancel{(-1)}\ln \left[\frac{\theta_{x=L}}{\theta_{x=0}}\right]} = \overbrace{ \frac{\left(\theta_{x=0}-\theta_{x=L}\right)}{\ln \left[\frac{\theta_{x=L}}{\theta_{x=0}}\right]} }^{\Delta T_{LMTD,cf}}$$

   The difference disappears when the L and 0 flip, so the expression is the same. Differences in performances due to the same boundary conditions are captured by a correlation or correction factor.

2. Discuss the factors that may affect the evaluation of LMTD for a cross-flow heat exchanger.

   Performance of LMTD evaluation in the counterflow configuration is dependent on assumptions of physical parameters. The cross flow HX configuration is the most efficient given a specific set of operating conditions. Factors which affect the evaluation are those that would increase the required NTU to operate at a specific effectiveness, while holding the $C_R$ for cross flow constant. These factors include the fouling factor, which reflects the decrease in operational efficiency over time. An additional factor is friction factor which affects the pressure gradient, whose contribution was assumed negligible in the enthalpy conservation expression:

   $$0 = \dot{m}_{H} \left[ \left(\frac{\partial i_{H}}{\partial T}\right)_{P}\frac{d T_{H}}{d x} + \overbrace{ \cancel{\left(\frac{\partial i_{H}}{\partial P}\right)_{T}\frac{d p_{H}}{d x}} }^\texttt{neglect} \right]dx+d\dot{q}.$$
3. Discuss for which type of problems is the NTU method better than the LMTD method.

   The Number of Transfer Units (NTU) method is superior to the LMTD method for most problems. A specific example are simulation type problems where the outlet temps are no known. The process of solving temperatures would need to be done numerically, as the arguments are nestled in logarithmic expressions.

4. Describe the NTU method.

   The name of the method gives insight into the methodology: transfer units are a measure of the rate of heat transfer, normalized by the physical parameters. Similar to the LMTD method, the NTU method begins with a differential energy balance on each line, combined into a single conservation equation. This form is written in Eqn. $(\ref{eqn:ode})$, as the derivation is identical up to this point.

   $$\ln \left(\frac{T_{H, o u t}-T_{C, i n}}{T_{H, i n}-T_{C, o u t}}\right)=-U A\left(\frac{1}{\dot{C}_{H}}-\frac{1}{\dot{C}_{C}}\right)$$

   However, an expression for the effectiveness, an additional parameter correlating how the conductance ratio affects the outlet temperature on either line: 

   $$T_{C, a u t}=T_{C ; i n}+\frac{\dot{q}}{\dot{C}_{C}} \quad\&\quad T_{H, o u t}=T_{H, i n}-\frac{\dot{q}}{\dot{C}_{H}}\\ \rightarrow \ln \left(\frac{1-\varepsilon_{c f} C_{R}}{1-\varepsilon_{c f}}\right)=-N T U\left(C_{R}-1\right)$$

   ​ We cast the problem now as as a relationship between the new parameters NTU and $CR$ to find the $\varepsilon{eff}$: how far the temperature gets to the desired.

5. Physically, what is the meaning of having the capacity ratio, $C_R$, approaching zero

   The capacity ratio $C_R$ is bounded on the interval between $0$ and $1$. In the case that $C_R $ approaches zero, either the hot or cold line has a much higher capacitance than the other. When one is much larger, the additional energy required likely is the latent heat required for phase change. The physics remain the same: one line is has greater capacity for heat energy to be moved.

6. Physically, what is the meaning of having the NTU parameter approaching zero.

   Due to how NTU is defined as the heat transfer between the hot and cold normalized by the minimum capacitance, while the NTU parameter approaches zero, the heat transfer also approaches zero. This lack of heat transfer, means there is no driving force for temperature change: both inlet and outlet temperatures for both the the hot and cool will be equal.

7. Physically, what is the meaning of having the NTU parameter approaching infinity.

   An infinite amount of heat transfer is required to force one line’s temperature to be the same as the other line. So, as the NTU parameter approaches infinity, the temperatures approach each other. In counter flow, the temperature is the same along the length of the pipe whereas, parallel flow, there is a single pinch point at the outlet temperature.

8. Discuss why, in general, does the counter flow heat exchanger has a higher effectiveness when compared with a parallel HX.

   In the physical sense: the temperature gradient is more evenly distributed along the HX’s length in counter flow configuration. The parallel flow configuration has the hot and cold inlets adjacent to each other so the outlet temperatures approach an intermediate value which from the infinite NTU case, trend toward the same value. Along the length of each HX, the marginal benefit to the NTU parameter diminishes to zero in the parallel configuration whereas in the counterflow configuration the marginal benefit is constant.