Need Help with these 3 questions 3. Heat flux (20 pts). Mean seawater temperatur

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Need Help with these 3 questions

3. Heat flux (20 pts). Mean seawater temperatures vary greatly throughout the year in the Baltic Sea with mean water temperatures as high as ~20 °C in August and as low as 3 OC in February. For a patch of water with a depth of 10 m, what is the net heat flux lost (in W m2) to the atmosphere from August to February? (Ignore thermal expansion) 4. Volume fluxes and associated velocities (20 pts) a) (10 pts.) Suppose that the volume transport associated with the wind-driven circulation's western boundary current (such as the Gulf Stream) is 50 x 106 m3/sec. Assume that this volume transport is carried in a current of uniform speed which is 50 km wide and 1 km thick. Calculate the average velocity of the current. b) (10 pts.) Suppose that the velocity associated with the Deep Western Boundary Current (DWBC), which flows southward under and inshore of the Gulf Stream, is approximately 0.2 m/sec (the DWBC is associated with the thermohaline circulation; we will talk more about it later). Assume that the width and thickness of the current are 50 km and 0.5 km, respectively. What is the volume transport associated with this current in m*/sec? 5. Pressure Gradients (20 pts) Consider the tank shown in the figure below. There is water in the tank and the surface of the water is sloped as shown. Assume = 1000 kg m-3 (a) What is the pressure (in Pa) on the bottom of the tank just inside of the right edge? (b) What is the pressure (in Pa) on the bottom of the tank just inside of the left edge? (c) What is the pressure gradient (in Pa/m) at the base of the tank? In which direction is this pressure gradient pointing? (d) In which direction would you expect the water to flow (to the right, left, into the page or out of the page)? (In this case we are considering a non-rotating tank; we will soon see in class how rotation has some, perhaps counter-intuitive, effects on flow directions.) 0.75 m

Explanation / Answer

3. a) Net heat flux loss = Flow of energy/ area/ time

Given, Flow of energy = 200 - 30 = 170

area = 10*10 m2

time= august to february = 6 months= 6 * 30 *24 hours

So, Net heat flux = (20-3 )/ 10*10 / 6*30*24

= 734.4 W/m2

4.a) Volume transport = 50*106 m3/sec

Current width=50km, thickness= 1km

Average velocity of current = ((50 *106 )m3/sec) / (50*1* 106 ) m2

=1m/sec

b) Now, the velocity of current = 0.2m/sec

width of current= 50km, thickness of current = 0.5km

So, the volume transport = 0.2 * 50 * 0.5 * 10 6 m3/sec

= 5,000,000 m3/sec

5.a) The pressure on the bottom of the tank just inside of the right edge is

P= p*g *h

where, P= pressure(Pa)

g= gravitational acceleration =9.8 m^2/sec

h=1.5 m

p= density =1000 kg/ m^3

Thus, P= 1000 * 9.8* 1.5

= 14,700 Pa

b)Similarly, the pressure on the bootom of the tank just inside the left edge is

P= 1000 * 9.8 * 0.75

= 7,350 Pa

c) The pressure gradient at the bottom of the tank = ( 14,700- 7,350 )/5 =7,350 Pa/5m

= 1,470 Pa/m

The pressure gradient is pointing towards the left of the tank.

d) As the pressure gradient slopes to the left so the water will also flow to the left.

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