Answers to Problem set # 2

Grismer Ch. 3 even

Ch. 4 2,4,5,6

    Grismer Ch. 3

    2)Using LaPlace's equation from lab, we have

    h = 2 s cos a / r r g

    If we look at a fluid, this equation gives the height of rise.

    Fluid 2 has a r of 0.67 that of water, and a s pf 0.5 that of water.

    Plugging into equation, we have:

    h = 2 (0.5)s cos a / r (0.67)r g

    or,

    h = (0.5/0.67) *2s cos a / r r g,

    which is 0.75 times the height of water.

    4)

    This problem is to determine the height of rise in water that is in equilibrium with 6 cm of mercury. It does not matter if there is a tank of water balancing Hg, or even just a plain manometer. The solution is:

    X height of water * ¡ water = 6 cm * 13.6 ¡ water

    Rearranging, we have X = 81.6 cm

    6)Hydrostatics problem

    traveling through the manometer, we have

    (downward is positive, negative is upward)

    P4 = 0 gage

    P3 = P4 + (¡ water * 1 m)

    P2 = P3 + (13.6 *¡ water * 0.1 m)
    P4 = P3

    Pa = P4 - (¡ water * 0.3 m), or

    Pa = 0 + (¡ water * 1 m) + (13.6 *¡ water * 0.1 m) - (¡ water * 0.3 m)

    Pa = (9810 N/m3*(1+ 1.36 - 0.3 m)

    Pa = 20.2 KPa

     

     

     

     

     

     

    8)Hydrostatics problem

    Traveling through the tube (downward is positive, upward is negative)

    (next p is at interface starting from left to right)

    Px = ?

    P2=Px - (¡ water * 1.625 m)

    P3=P2

    P4=P3 + (0.9*¡ water * 0.250 m)

    Py=P4 + (¡ water * 0.875 m)

    Rearranging, we have:

    Px-Py = ¡ water * (1.625 - 0.875 - (0.9*0.25m))

    =(9810 N/m3) * (0.525 m)

    = 5.15 KPa

    10)

    Hydrostatic problem - we do not care that the cone is on the right, we only care about pressure head.

    0 height in cone = 13.6 *¡ water * 15 cm

    300 cm = 13.6 *¡ water * (x - 15 cm)

    x = 37 cm

     

    12)

    Area of square plate = (0.85 m)2 = 0.7225 m2

    Pa = ¡ water *0.9 m

    Pb= ¡ water *1.75 m

    Average pressure = (0.9 + 1.75)/2 = 1.325 m and acts at centroid of square plate

    F= PA = (1.325 m) * (9810 N/m3) * (0.7224 m2) = 9391 N

     

     

     

     

     

     

     

     

     

    Grismer Ch. 4
  1. The greatest precipitation intensities would occur on the western side of the mountains due to upward movement and cooling of the moist air mass from the Pacific Ocean.

4) A B C D

Storm 8.4 7.0 * 9.6

(cm)

mean 77.0 88.2 72.6 92.4

ratio 0.109 0.079 0.104

Average = (0.2924/3) * 72.6 cm = 7.1 cm

5)

Arithmetic Average = 32.3 in

Theissen polygon = 30.6 in

Isohyetal method = 35.3 in

Isohyetal is most accurate given to the mountainous topography of watershed.

6)The average precipitation depth is:

2.31 inches