Heat Exchangers Questions and Answers

Answer: A

The key says that oil coolers are an application of straight tube, fixed tubesheet heat exchangers. In this type the tubes are welded to tubesheets fixed in the shell, so the outside of the tubes cannot be cleaned mechanically, and the type is used where the shell side fluid is clean, as with oil. A fluid that is flammable or toxic in the tubes, or a cooler with water inside the tubes, is not the keyed application, and jacket water coolers are not the keyed application either.

Answer: D

The advantage of a removable bundle U-tube type heat exchanger is that the shell circuit can be inspected and mechanically cleaned, since the whole bundle can be pulled out. A single tube pass is not possible in a U-tube exchanger, because the fluid has to go down one limb and back up the other. The tube wall thickness at the U-bend is thinner and not thicker, and the tubes in the nest cannot be easily replaced individually, so the other statements are not advantages.

Answer: C

The correction factor applies to a plate heat exchanger.

The LMTD formula is exact only for pure counter-flow or parallel-flow arrangements. For multi-pass and cross-flow arrangements, such as most plate heat exchangers with their several passes and channel patterns, the true mean temperature difference is the counter-flow LMTD multiplied by a correction factor \(F \le 1\).

In a steam-jacketed kettle and in evaporator tubes, one side is condensing steam at constant temperature, so the flow arrangement does not matter and \(F = 1\). A single spiral tube is effectively a simple counter- or co-current exchanger and is not where the correction factor is characteristically used.

Answer: B

No correction factor is needed for a 1 shell pass and 1 tube pass heat exchanger.

The LMTD formula is exact only for pure parallel flow or pure counterflow. A 1-1 shell-and-tube exchanger is one of these: the shell fluid and tube fluid each make a single pass, so the exchanger is analysed directly as counterflow (or parallel flow) with F = 1.

In 1-2 and 2-4 shell-and-tube exchangers the tube fluid reverses direction, so part of the exchanger runs in parallel flow and part in counterflow; the true mean temperature difference is less than the counterflow LMTD and a correction factor F below 1 is applied. Plate heat exchangers are also normally treated with a correction factor, because their pass and channel arrangements depart from pure counterflow; the key takes the 1-1 shell-and-tube exchanger as the case that needs none.

Answer: A

The dimensionless one is heat exchanger effectiveness.

Effectiveness is the actual heat transfer rate divided by the maximum possible rate for the same inlet conditions, \(\varepsilon = q/q_{max}\). Both are in watts, so the ratio has no unit and lies between 0 and 1.

The others all carry units. Humid heat is the heat needed to warm 1 kg of dry air and its vapour by 1 K, in kJ.kg⁻¹.K⁻¹. Humid volume is the volume of moist air per kg of dry air, in m³.kg⁻¹. The overall heat transfer coefficient is in W.m⁻².K⁻¹.

Answer: C

The LMTD is 10 °C.

The heat capacity rates are equal:

\[ \begin{aligned} C_{w} &= 0.5 \times 4.18 = 2.09\ \text{kW/K} \\ C_{a} &= 2.09 \times 1 = 2.09\ \text{kW/K} \end{aligned} \]

With effectiveness 0.8 and a maximum temperature difference of 80 - 30 = 50 °C, each stream changes by 0.8 × 50 = 40 °C: water leaves at 40 °C and air at 70 °C. In a counterflow exchanger with equal capacity rates, the temperature difference is the same at both ends, 80 - 70 = 40 - 30 = 10 °C, so the LMTD equals that constant difference.

40 °C is the temperature change of each stream, 20 °C and 5 °C have no basis here. A check: Q = 2.09 × 40 = 83.6 kW, and UA = Q/LMTD = 8.36 kW/K, so NTU = 4 and \(\varepsilon = 4/5 = 0.8\).

Answer: A

The heat transfer area is \(0.707\ \text{m}^{2}\).

Heat taken up by the water, with \(\dot{m} = 1500/3600\) kg/s:

\[ \begin{aligned} Q &= \tfrac{1500}{3600} \times 4.187 \times (80 - 30) \\ &= 87.23\ \text{kW} \end{aligned} \]

With condensing steam at 120°C, the temperature differences at the two ends are 90 and 40 degrees, so the log mean temperature difference is

\[ \begin{aligned} \Delta T_{m} &= \frac{90 - 40}{\ln(90/40)} = 61.7^\circ\text{C} \\ A &= \frac{87230}{2000 \times 61.7} = 0.707\ \text{m}^{2} \end{aligned} \]

The other options are the same digits shifted by powers of ten, the usual result of mixing kW with W or kg/hr with kg/s, and 141.4 doubles one of them.

Answer: B

An automobile radiator is a cross-flow heat exchanger.

In a radiator the hot coolant flows through tubes from the top tank to the bottom tank, while air is driven across the tubes and fins by the vehicle's motion and the fan. The two fluids move at right angles to each other, which is the definition of cross flow. The air side is unmixed, channelled between fins.

In parallel flow and counter flow the two streams run along the same axis, in the same or opposite directions. A regenerator stores heat in a matrix that the hot and cold fluids pass through alternately, as in some gas turbine plants; a radiator transfers heat directly through the tube walls.

Answer: A

The effectiveness is \(1 - \exp(-\text{NTU})\).

In a condenser the condensing vapour stays at its saturation temperature while it gives up latent heat, so its effective heat capacity rate is infinite. That makes \(C_{max} \to \infty\), and so the capacity ratio \(C_{r} = C_{min}/C_{max} = 0\).

Putting \(C_{r} = 0\) in any of the standard effectiveness relations, parallel or counter flow, gives the same result:

\[ \varepsilon = 1 - e^{-\text{NTU}} \]

Option B is the parallel-flow relation, which reduces to this only when \(C_{r} = 0\); option C is a garbled counter-flow form; and \(\frac{\text{NTU}}{1 + \text{NTU}}\) is the counter-flow result for \(C_{r} = 1\), the opposite limit.

Answer: C

The LMTD is 60°C.

In counterflow, the hot inlet faces the cold outlet and the hot outlet faces the cold inlet:

\[ \begin{aligned} \Delta T_{1} &= 140 - 80 = 60^{\circ}\text{C} \\ \Delta T_{2} &= 80 - 20 = 60^{\circ}\text{C} \end{aligned} \]

The formula \(\text{LMTD} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln(\Delta T_{1}/\Delta T_{2})}\) becomes \(\frac{0}{0}\) when the two ends are equal, but its limit is simply the common value: the temperature difference is the same all along the exchanger, 60°C.

That is the trap in the first option: 0/0 here is an indeterminate form with a definite limit, not a sign that the LMTD cannot be found. 80°C and 110°C are not differences between any facing pair of temperatures.