Hot engine oil with heat capacity rate of 4440 w/k (product of mass flow rate and specific heat) and an inlet temperature of 150°c flows through a double pipe heat exchanger. the double pipe heat exchanger is constructed of a 1.5-m-long copper pipe (k = 250 w/m·k) with an inner tube of inside diameter 2 cm and outside tube diameter of 2.25 cm. the inner diameter of the outer tube of the double pipe heat exchanger is 6 cm. oil flowing at a rate of 2 kg/s through inner tube exits the heat exchanger at a temperature of 50°c. the cold fluid, i. e., water enters the heat exchanger at 20°c and exits at 70°c. assuming the fouling factor on the oil side and water side to be 0.00015 m2 ·k/w and 0.0001 m2 ·k/w, respectively, determine the overall heat transfer coefficient on inner and outer surface of the copper tube.

Answers

Answer 1

Explanation:

fluid, i. e., water enters the heat exchanger at 20°c and exits at 70°c. assuming the fouling factor on the oil side and water side to be 0.00015 m2 ·k/w and 0.0001 m2 ·k/w, respectively, determine the overall heat transfer coefficient on inner and outer surface of the copper tube.xgjicbb .follow me


Related Questions

A jet aircraft is in level flight at an altitude of 30,000 ft with an airspeed of 500 ft/s. The aircraft has a gross weight of 19,815 lb, a wingspan of 53.3 ft, and an average chord length of 6 ft. The Oswald efficiency factor is 0.81 and the zero-lift drag coefficient is equal to 0.02. The jet has two turbofan engines, each producing a maximum thrust of 3,650 lb at sea level.

Required:
a. Create a plot of the drag polar for this aircraft for CL from 0 to 5. Plot CL on the vertical axis, CD on the horizontal axis, and do not include negative CL values.
b. What is the total drag coefficient at the flight condition described above?
c. What is the required thrust for level flight at this altitude in lb?
d. If the pilot runs the engines at maximum thrust, what is the instantaneous rate of climb at this altitude and velocity?

Answers

Answer:

a) attached below

b) 0.0337

c) 2730.206 Ib

d) 2320.338 ft/min

Explanation:

a) Plot of the drag polar for this aircraft

first we will calculate :

Wing area (s) = Wing span (b) * Average chord length(c)

                       = 53.3 * 6 = 319.8 ft^2

Aspect ratio =  b^2 / s = 8.883

K = 1 / [tex]\pi[/tex]eAR = 1 /

Drag polar ( Cd ) = 0.02 + 0.044 C^2L

attached below is a plot of the drag polar

Attached below is the detailed solution of the remaining part of the question

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