A heat exchanger wall consists of a copper plate 2 cm thick. The heat transfer coefficients on the two sides of the plate are 2700 and 7000 W/( m2 K), corresponding to fluid temperatures of 92 and 32°C, respectively. Assuming that the thermal conductivity of the wall is 375 W/ ( m K), (a) compute the surface temperatures in °C, and (b) calculate the heat flux in W/ m2.
GIVEN
• Heat exchanger wall, thickness (L) = 2 cm = 0.02 m
• Heat transfer coefficients
? hc1 = 2700 W/(m2 K)
? hc2 = 7000 W/(m2 K)
• Fluid temperatures
? Tf1 = 92°C
? Tf2 = 32°C
• Thermal conductivity of the wall (k) = 375 W/( m K) FIND (a) Surface temperatures (Tw1, Tw1) in °F (b) The heat flux (q/A) in W/m2. ASSUMPTIONS
• One dimensional heat transfer prevails
• The system has reached steady state
• Radiative heat transfer is negligible
SKETCH
The thermal circuit for the wall is shown below
The surface temperatures can only be calculated after the heat flux has been established, therefore part
(b) will be solved before part (a).
(b) The resistances are in series, therefore the total resistance is
The total rate of heat transfer is given by
Therefore the heat flux (q/A) is
(a) Equation (1.10) can be applied to the convective heat transfer on the fluid 1 side
Similarly, on the fluid 2 side
You might also like to view...
Mercury experiences extreme high and low temperatures between night and day because:
A) it is so close to the Sun. B) its dense atmosphere creates a runaway greenhouse. C) its oceans are much hotter than ours. D) Mercury has no axial tilt, with its equator always exposed to direct sunlight. E) it has no true atmosphere to moderate temperatures over the globe.
A proton moving eastward with a velocity of 5.0 × 103 m/s enters a magnetic field of 0.20 T pointing northward. What is the magnitude and direction of the force that acts on the proton?
A) 0 N B) 1.6 × 10-16 N upwards C) 1.6 × 10-16 N downwards D) 1.1 × 10-16 N eastwards E) 4.4 × 10-16 N westwards
Write the equation of a wave, traveling along the +x axis with an amplitude of 0.02 m, a frequency of 440 Hz, and a speed of 330 m/sec
a. y = 0.02 sin [880?(x/330 ? t)] b. y = 0.02 cos [880? x/330 ? 440t] c. y = 0.02 sin [880?(x/330 + t)] d. y = 0.02 sin [2?(x/330 + 440t)] e. y = 0.02 cos [2?(x/330 + 440t)]
A car passes a dip in the road, going first down, then up. At the very bottom of the dip, when the car’s instantaneous velocity is passing through horizontal, how does the magnitude of the normal force on the car compare to the magnitude of the car’s weight?
A. 
B. 
C. 
D. 
E. We do not have enough information to answer.