Self Studies

Fluid Mechanics Test 7

Result Self Studies

Fluid Mechanics Test 7
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    At a hydraulic jump, the flow depths are 0.4 m and 5 m at the upstream and downstream, respectively. The channel is wide rectangular. The discharge per unit width is nearly
    Solution

    Concept:

    Hydraulic Jump: It is a phenomena associated with sudden rise in water depth when a flow with high velocity and low depth (supercritical flow) strikes another flow with low velocity and high depth (subcritical flow).

    For any hydraulic jump, the following relation always holds good:

    y1y2(y1+y2)=2q2g

    Where, y1 and y2 are the depths before and after the hydraulic jump respectively.

    q is discharge per unit width and g is acceleration due to gravity.

    Calculation:

    Given, upstream depth which is depth before hydraulic jump = y1 = 0.4 m.

    Downstream depth which is depth after hydraulic jump = y2 = 5 m.

    Acceleration due to gravity = g = 9.81 m/s2.

    q2=y1y2(y1+y2)×g2=0.4×5×(0.4+5)×9.812=52.974q=7.278m2s7.3m2s

    ∴ The discharge per unit width is nearly 7.3 m2/s. 

  • Question 2
    1 / -0
    What would be the critical velocity (in m/s, up to two decimal places) of the water flowing through a rectangular channel of width 10 m, when discharge is 20 m3/s? Take g = 10 m/s2.
    Solution

    Concept:

    The flow is said to be critical, if it satisfies the following equation irrespective of the type of channel:

    Q2TgA3=1

    Where,

    Q is the discharge

    T is the top width of channel

    A is the wetted area of channel

    Calculation:

    Let yc is the critical depth of flow

    Wetted area, A = Byc

    Given: Q = 20 m3/s, B = 10 m, g = 10 m/s2

    Top width, T = B = 10 m

    We know that: Q2TgA3=1

    (20)2×1010×(10×yc)3=1yc=0.736m 

    Critical velocity of flow, Vc=QA=(2010×0.736)

    ⇒ Vc = 2.72 m/s

  • Question 3
    1 / -0
    While conducting the flow measurement using a triangular notch, an error of +2% in head over the notch is observed. The percentage error in the computed discharge would be 
    Solution

    Concept:

    For a triangular notch

    Q=815cd2gtanθ2×H3/2

    Where,

    Q = Dischange, Cd = co-eff of discharge, H = Head above the bottom of notch, and θ = enclosed angle

    Calculation:

    Differentiating equation w.r.t. H, we have

    dQdH=815cd2gtanθ2×52H3/2

    dQ=815cd2gtanθ2×22H32×dH

    dQQ=815cd2gtanθ2×32H3/2×dH815cd2gtanθ2×H3/2=52dHH

    dQQ=52dHH=2.5dHH

    Given error in head =dHH=2%

    So, dQQ=2.5×2=5%

    Important Points:

    Discharge over a rectangular notch

    Q=Cd23B2gH3/2

    Where,

    Q = discharge, Cd = co-eff discharge, B = width of notch, and H = Head above the bottom of notch
  • Question 4
    1 / -0

    A wide rectangular channel 4 m wide and normal bed slope 1 in 5000 conveys a constant discharge of 30.0 m3/s. Let the normal depth of flow at a particular section is yn.  Assume Manning’s n = 0.012. Which of the following is correct? 

    Solution

    Concept:

    The Froude’s number is given as:

    Fr=V2gD

    Where,

    D is the hydraulic depth and it is given as D = A/T

    Where A is the wetted area and T is the top width

    For rectangular channel, A = By and T = B ⇒ D = (By)/B = y i.e. depth of flow

    Now,

    If Fr < 1 then flow is sub-critical.

    If F­r > 1 then flow is supercritical.

    If Fr = 1 then flow is critical.

    As per, Manning’s formula, the velocity of flow for normal depth yn is given as:

    V=1nR2/3S1/2

    Where,

    R is the Hydraulic radius and it is given as R = A/P

    (A is wetted area and P is wetted perimeter)

    S is the bed slope or energy line slope.

    Calculation:

    Given: B = 4 m, S =15000, Q = 30 m3/s, n = 0.012

    V=1n(R)23S12

    R=AP=(BynB+2yn)=yn                                   (For wide rectangular channel)

    Q = A V

    Q=(Byn)1n(yn)23S12

    Q=(BnS12)(yn53)

    30=(40.012)(15000)12(yn53) ⇒ yn = 3.04 m

    Now, velocity of flow, V

    V=1n(yn)23S12

    V=10.012×(3.04)23(15000)V2

    ⇒ V = 2.47 m/s

    D = yn (for rectangular channel)

    Fr=Vgyn=2.4710×3.04 ⇒ Fr = 0.447

    Fr < 1 ⇒ Flow is sub-critical.

  • Question 5
    1 / -0

    What would be the specific force (in kN m3/kN, up to two decimal places) for a triangular channel of apex angle 90°  and top width of 2 m if the water is flowing at the rate of 10 m3/s in the channel?  Take g = 10 m/s2.

    Solution

    Concept

    The specific force for a channel is given as,

    F=AZ¯+Q2Ag

    Where,

    A is the wetted area

    Q is the discharge in the channel

    Z̅ is the centroid of the channel section measured from the top of the channel.

    Calculation

    Given:

    Q = 10 m3/s; B = 2 m

    tan 45° = y/1 ⇒ y = 1 m

    Z¯=13y=13×1=0.33m

    Wetted area, A =12×2×1=1m2

    Fs=AZ¯+Q2Ag  =(1×0.67)+(10)21×10 

    ⇒ Fs = 10.33 kN m3/kN

  • Question 6
    1 / -0
    For a constant specific energy of 3.0 N-m/N, what would be the maximum discharge that may be possible in a rectangular channel of 2 m width? Take g = 10 m/s2.
    Solution

    Concept

    For a constant specific energy, the discharge will be maximum when flow is critical and it satisfies the following condition:

    Q2TgA3=1

    Where,

    Q is the discharge

    T is the top width of channel

    A is the wetted area of channel

    Specific energy under critical conditions for a given critical depth yc is given as:

    Ec=yc+V22g

    Where,

    V is the velocity of flow

    yc is the depth flow at a particular section

     Calculation:

    Given: Ec = 3 m, B = 2 m

    Ec=yc+V22g 

    Ec=yc+(QByc)22g 

    Ec=yc+Q2B2yc2(2g)     …1)

    For critical flow,

    Q2TgA3=1 ⇒ Q2Bg(Byc)3=1 

    Q2yc2=gB2yc      …2)

    Substitute the value of Q2yc2 in equation 1)

    Ec=yc+gB2yc2gB2 

    Ec=yc+yc2=32yc 

    3=32×ycyc=2m 

    Substitute yc = 2 m in equation 2)

    Q2(2)2=10×(2)2×(2)3 

    ⇒ Q = 17.89 m3/s

    ∴ The maximum discharge is 17.89 m3/s

  • Question 7
    1 / -0

    A rectangular channel of width 2 m is provided with hydraulic jump as an energy dissipator device at its downstream side.  It dissipates an energy equivalent to 2 m of water. Assume that energy dissipation is only due to hydraulic jump and there is no other loss in energy. The Froude number before the jump is 10 .  What would be the sequent depth (post jump depth) (in m, up to two decimal places) ?  

    Solution

    Concept

    The energy loss due to hydraulic jump is given as:

    E=(y2y1)34y1y2

    Where,

    y1 is the pre jump depth

    y2 is the post jump depth

    The relation between pre and post jump depth is given as,

    y2y1=12{1+1+8F12}

    Where,

    F1 is the Froude number before jump.

    Calculation:

    Given: F1 = 10, B = 2 m

    Let y2 = Ky1

    y2y1=12(1+1+8F12) 

    y2y1=K=12(1+1+8×102) 

    ⇒ K = 13.65

    Energy loss is,

    E=(y2y1)34y1y2 

    2=(Ky1y1)3(4y1)×(Ky1) 

    2=(K1)3y14K 

    Putting K = 13.65 in above,

    2=((13.651)34×13.65)×y1 

    ⇒ y1 = 0.054 m

    Now, y2 = K y1 = 13.65 × 0.054 = 0.736 m

    y2 = 0.74 m

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now