Question: A concentric tube ( parallel flow ) heat exchanger is built and operated as shown in the illustration be - low. e purpose of the

A concentric tube (parallel flow) heat exchanger is built and operated as shown in the illustration be-
low. e purpose of the heat exchanger is to use heat from a hot-water waste stream to heat a stream
of cold water from the municipal supply. e cold tube-side flow (0.35 kg/s) passes through the entire
heat exchanger, while the hot shell-side flow (0.92 kg/s) enters at the middle and splits evenly
(0.46 kg/s in each direction).e two halves of the heat exchanger are identical, except for the flow
direction on the shell side.
e cold inlet temperature is 8C, and the hot inlet temperature is 92C.e hot outlet temperature
from the side with countercurrent flow is 73C. Use the heat capacity at the average of the hot inlet
and cold inlet temperatures throughout your calculations, for both the hot- and cold-sides.
(a) Determine the cold-side temperature at the mid-point in the tube, by considering only the coun-
tercurrent side of the heat exchanger. Determine the log-mean temperature dierence on that side
and estimate the overall heat transfer coecient for that side of the heat exchanger.
(b) Explain why you expect that the overall heat transfer coecient for the other half of the heat ex-
changer will almost identical to the value above. en explain separately why the overall heat
transfer coecient for the other half of the heat exchanger will be very slightly dierent.
(c) Assuming that the overall heat transfer coecients are the same for both sides of the heat ex-
changer, estimate the outlet temperatures from the other half of the heat exchanger using the
NTU method (on the second half of the heat exchanger).
(d) An outlet cold temperature of 40C is desired, which can be achieved by altering the right-le
split of flow on the hot side. Consider the worst case for heat exchange with the same total flow
rates and inlet temperatures. Assuming that the overall heat transfer coecient for each half does
not change: what is the minimum cold-side outlet temperature that can be achieved with such an
approach? Hint: using the NTU method again is the easiest approach.
A concentric tube ( parallel flow ) heat

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