ME410 Heat Transfer, Spring 2025
Homework #6, Due 04/11/2025
Homework Format
1. Use this page as a cover sheet. For problems requiring extensive analysis, please show all
your work and highlight/underline/enclose the final answer. Do not forget to append your
detailed work.
2. Solution scheme: Prepare a schematic of the problem if not provided, list some
assumptions, if a control volume or energy balance is necessary, depict it, solve and
highlight/underline/enclose the final answer.
3. All parts must be legible. Points will be deducted for sloppy solutions.
4. You are encouraged to study with others, but your submission should be your own work.
5. Submit your assignment as a SINGLE PDF to GradeScope. You can either scan your
handwritten work solutions or work on a device that allows for handwriting input (iPad,
Surface, or touch screen computer).
6. Homework is due by 11:59 pm Eastern Time on the day that it is due. NO LATE
HOMEWORK is accepted.
Problem 1 (20-Points: Let’s test your conceptual understanding of internal convection)
Microchannels (ducts with hydraulic diameters in the order μm) have been used for electronics
cooling by embedding them into liquid-cooled heat sinks. Due to their small size, the flow regime
in microchannels is usually laminar (ReD < 2000). Consider a 2.5 cm long microchannel with a
hydraulic diameter of 50 μm, where a flow of Re = 500 is applied, and answer the following
questions.
1. A microchannel model that assumes negligible hydrodynamic entrance effects for this
channel is:
a. Controversial because you always must consider the entrance region.
b. Accurate, as the hydrodynamic entrance length is negligible in such small channels.
c. Not possible to formulate.
d. Not a real model.
2. The thermal entrance length will be Pr times the hydrodynamic entrance length:
a. True
b. False
3. If the flow rate in this microchannel is doubled and we assume that the flow is fully
developed in temperature and velocity:
a. The heat transfer coefficient will double.
b. The heat transfer coefficient will be half.
c. The heat transfer coefficient remains the same.
d. It is impossible to know without more information.
4. Liquid-cooled heat sinks using microchannels offer the following pros/cons under laminar
flow operation compared to larger channels:
a. Pro: microchannels are easy to manufacture/Con: microchannels are brittle.
b. Pro: h goes up/Con: pressure losses go up.
c. Pro: h goes up/Con: pressure losses go down:
ME410 Heat Transfer, Spring 2025
Homework #6, Due 04/11/2025
d. None of the above
5. For the same Re and hydraulic diameter, a water channel will be longer than an air duct for
us to see thermally developed conditions:
a. True.
b. False.
Problem 2 (40-Points: External and internal convection in a heat pump pipe) Heat is
rejected from a heat pump by circulating an ethylene glycol mixture (f = 1022 kg/m3, f = 0.00145
Pa-s, cf = 3905 J/kg K, kf = 0.5132, and Prf = 11.03) through a copper tube (kCu = 396 W/m K) that
is L = 10 m long with an outer diameter Dout = 1.7 cm and wall thickness  = 2 mm. The ethylene
glycol enters the tube at Tin = 17°C with a mass flow rate 𝑚̇𝑓 = 0.25 kg/s. Water (w = 996 kg/m3,
w = 0.854×10-3 Pa-s, kw = 0.6111, and Prw = 5.841) at Tw = 5°C flows at an average velocity uw =
3.2 m/s over the copper tube, in a cross-flow configuration, to cool down the ethylene glycol
stream.
(a) Determine if the flow inside the pipe is laminar or turbulent. If the flow is turbulent,
calculate the heat transfer coefficient for internal flow using the Gnielinski correlation, and
assuming that the pipe is smooth (smooth pipe friction coefficient calculation).
(b) Determine if the external water flow around the cylinder is laminar or turbulent. Use the
Churchill and Bernstein correlation to calculate the external heat transfer coefficient.
(c) Calculate the conduction resistance of the copper pipe and draw an equivalent thermal
resistance for this problem.
(d) Estimate the ethylene glycol’s outlet temperature.
(e) Calculate the total heat transfer rate from the ethylene glycol to the water.
Problem 3 (40-Points: Let’s use EES to design a heat exchanger… pipe!) We are
designing an air heater for cheap! The idea is to wrap a film heater around a copper pipe to increase
the temperature of air flowing at 0.01 kg/s from 20°C to 80°C assuming the inlet air pressure is
100 psi. To get the EES model started, calculate all pertinent air properties at the inlet-to-outlet
average temperature using an ideal gas model in the material properties database of EES. Also,
let’s assume that the pipe is 1.2 m long and has an outer diameter of 6.35 mm and 0.89 mm thick
(what do you think the conduction thermal resistance is?)
(a) Calculate the heat transfer rate and heat flux needed from the strip heater.
(b) Calculate the pressure drop in this pipe using a useful EES function. The syntax is CALL
PipeFlow_ND(Re,Pr,L/D_i,e/D_i: Nusselt_bar_T, Nusselt_bar_H, f_bar). Where the
function inputs are: Re is the Reynolds number, Pr is the Prandtl number, D_i is the pipe’s
inner diameter, and e = 0.0015 mm is copper’s roughness. The outputs are Nusselt_bar_T
is Nu assuming a constant surface temperature, Nusselt_bar_H is Nu assuming a constant
heat flux, and f_bar is the friction factor you need for this calculation.
(c) The previous function outputs Nu numbers regardless of being fully developed,
developing,
laminar,
or
turbulent.
Additionally,
CALL
PipeFlow_ND_local(Re,Pr,L/D_i,e/D_i: Nusselt_bar_T, Nusselt_bar_H, f_bar) has the
same inputs and outputs as the function in Part (b), but the outputs are local values. Propose
ME410 Heat Transfer, Spring 2025
Homework #6, Due 04/11/2025
an equation to find the maximum temperature in the pipe (using the heat flux and local heat
transfer coefficient), and then calculate it using EES.
(d) Now that your code is running, our design objectives are that we want to have a pressure
drop of only 10 psi and a maximum pipe temperature of 100°C. Comment out the definition
of length and outer pipe diameter in your EES code while you add the value of the desired
pressure drop and max temperature difference.