Assessment Task 2: Engineering Problem A Working

Question 1:

First we need to calculate torque keeping the gate closed due to hydrostatic pressure and the weight of the gate.

(τ_↻)=(τ_P)+(τ_W)

(τ_P)=(r_P)*(F_R)

(F_R)=(P_avg)*A

(P_avg)=(P_C)

(P_C)=ρ*g*(h_C)

where

ρ=1000 (density of water)

g=9.81 (acceleration due to gravity)

(h_C)= 1+1/2*(5)=3.5 (1)of water above gate and then half the height of the gate)

Ignoring (P_atm)since it acts on both sides of the gate

Define a as the width of the gate and b as the length of the gate

Thus

A=a*b

Where

a=10 (Given from question)

sin(45)=5/b

b=5/sin(45)

sin(45)=1/√(,2)

Thus

b=5√(,2)*m

Now to find (r_P)

(r_P)=(y_P)-(y_B)

(y_P)=(y_c)+ (I_xx,c)/((y_c)*A) (ignoring ((P_atm))
sin(45)=(h_C)/(y_c)

(y_c)=(h_C)/sin(45)

(y_c) =(h_C)√(,2)

(I_xx,c)=(a*b3)/12

sin(45)=1/(y_B)

(y_B)=1/sin(45)

(y_B)=√(,2)*m

Now to calculate the torque due to gravity

(τ_W)=(r_W)*(F_W)

(F_W)=m*g

Where

m=500

Gravity acts downwards through the centroid, assuming uniform weight distribution

So

cos(45)=(r_W)/(b/2)

(r_W)=(b*cos(45))/2

(r_W)=b/(2√(,2))

Torque opening the gate

(τ_↺)=r*F

To open the gate

(τ_↺)=(τ_↻)

Thus

r*F=(τ_↻)

F=(τ_↻)/r

Force is applied normal to the gate at the centroid

Thus

r=b/2

Hence

F=((τ_P)+(τ_W))/(b/2)

F=(2*((r_P)*(F_R)+(r_W)*(F_W)))/b

F=(2*(((y_P)-(y_B))*((P_avg)*A)+(b/(2√(,2)))*(m*g)))/b

F=(2*(((y_c)+(I_xx,c)/((y_c)*A)-(y_B))*((P_C)*a*b)+(b/(2√(,2)))*(m*g)))/b

F=2*(((h_C)√(,2)+((a*b3)/12)/((h_C)*a*b√(,2))-(y_B))*(ρ*g*(h_C)*a)+(1/(2√(,2)))*(m*g))

F=2*g*(((h_C)√(,2)+(b2)/(12*(h_C)√(,2))-(y_B))*(ρ*(h_C)*a)+(1/(2√(,2)))*(m))

Subbing in values

F=2*(9.81)*(((3.5)√(,2)+((5√(,2)*m)2)/(12*(3.5)√(,2))-√(,2)*m)*(1000)*((3.5))*(10)+(1/(2√(,2)))*(500))

F=3006087.704

F=3.01⨯10

Question 2:

(a):

d(y)/d(x)=v/u=(-2-3*y)/(12+3*x)

d(y)/(-2-3*y)=d(x)/(12+3*x)

(∫_^)(d(y)/(-2-3*y))=(∫_^)(d(x)/(12+3*x))

Let a = -2-3*y, b=12+3*x

d(a)/d(y)=-3, d(b)/d(x)=3

d(y)=-d(a)/3 , d(x)=d(b)/3

-1/3*(∫_^)(d(a)/a)=1/3*(∫_^)(d(b)/b)

-1/3*ln|a|=1/3*ln|b|+C

-ln|−2−3*y|=ln|12+3*x|+C

-ln|−2−3*y|=ln|3*(4+x)|+C

-ln|−2−3*y|=ln|4+x|+ln|3|+C (ln|3| is a constant)

-ln|−2−3*y|=ln|4+x|+C

-ln|−2−3*y|-ln|4+x|=C

Ccan be any real number so

ln|−2−3*y|+ln|4+x| = C

ln(|-2-3*y|*|4+x|)=C

ln|(-2-3*y)*(4+x)|=C

ℇln|(-2-3*y)*(4+x)|=ℇC

|(-2-3*y)*(4+x)|=C where C > 0

(-2-3*y)*(4+x)=±C

(-2-3*y)*(4+x)=C where C ∈ ℝ

-2-3*y=C/(4+x)

y=C/(-3*(4+x))-2/3

Absorbing constants into C:

y=C/(4+x)-2/3

(b):

Question 3:

(a):

(W_electricity)=(η_turbine)*(η_generator)*(W_wind)

Where

(η_turbine) =0.5

(η_generator)=0.95

(W_wind)=(E_kinetic) (no potential energy is transferred)

(E_kinetic)=1/2*m*v2

m=ρ*v*A

A=π*(D/2)2

Where

ρ=1.25

(v_min)=40

(v_min)=40*(1000/3600)

(v_min)=100/9

(v_max)=60

(v_max)=60*(1000/3600)

(v_max)=50/3

D=100

Hence

(W_electricity)=(η_turbine)*(η_generator)*(W_wind)

(W_electricity)=(η_turbine)*(η_generator)*(E_kinetic)

(W_electricity)=(η_turbine)*(η_generator)(1/2*m*v2)

(W_electricity)=1/2*(η_turbine)*(η_generator)((ρ*v*A)*v2)

(W_electricity)=1/2*(η_turbine)*(η_generator)(ρ*v3)*(π*(D/2)2)

Thus

(W_electricity,min)=1/2*(η_turbine)*(η_generator)(ρ*(v_min)3)*(π*(D/2)2)

(W_electricity,max)=1/2*(η_turbine)*(η_generator)(ρ*(v_max)3)*(π*(D/2)2)

Subbing in values

(W_electricity,min)=1/2*(0.5)*(0.95)*(1.25)*(100/9)3*(π*(100/2)2)

(W_electricity,min)=3198423.591

(W_electricity,min)=3.20⨉10

(W_electricity,max)=1/2*(0.5)*(0.95)*(1.25)*(50/3)3*(π*(100/2)2)

(W_electricity,max)=10794679.62

(W_electricity,max)=1.08⨯10

(b):

Horizontal force on mast is equivalent to the drag force

F=1/2*(C_D)*ρ*v2*A

Assuming (C_D)≈1

Then

F=1/2*ρ*v2*A

Where

ρ=1.25

v=60=50/3 (max force when velocity is maximal)

A=D*h (Area horizontally facing wind is a rectangle)

D=2 (Width of the projected rectangle is equal to the diameter of the mast)

h=150 (Height of mast shown in picture, equal to the length of the projected rectangle)

Hence

F=1/2*ρ*v2*D*h

Sub in values

F=1/2*(1.25)*(50/3)2*(2)*(150)

F=52083.33333

F=52.1

Question 4:

(a):

(P_saturation,20)=2339

Pressure is lowest when velocity is and elevation is maximised

Point 2 and 3 should be checked as these are where velocity and elevation are maximised respectively

(P_1)/(ρ*g)+((v_1)2)/(2*g)+(z_1)= (P_2)/(ρ*g)+((v_2)2)/(2*g)+(z_2)=(P_3)/(ρ*g)+((v_3)2)/(2*g)+(z_3)

Where

ρ=1000

g=9.81

(P_1)=(P_atm)

(P_atm)=101300

(v_1)=0

(v_2)=V/(A_2)

(v_3)=V/(A_3)

(z_1)=7

(z_2)=3

(z_3)=8

(A_2)=π*(d/2)2

(A_3)=π*(D/2)2

d=0.05

D=0.12

Thus

(P_atm)/(ρ*g)+((0)2)/(2*g)+7= (P_2)/(ρ*g)+((V/(A_2))2)/(2*g)+3=(P_3)/(ρ*g)+((V/(A_3))2)/(2*g)+8

(P_atm)/(ρ*g)+7= (P_2)/(ρ*g)+(V2)/(2*(A_2)2*g)+3=(P_3)/(ρ*g)+(V2)/(2*(A_3)2*g)+8

Comparing 1 and 2

(V_1,2)=√(,2*(A_2)2*(((P_atm)-(P_2))/ρ+(4)*g))

(V_1,2)=√(,2*(π*(d/2)2)2*(((P_atm)-(P_2))/ρ+(4)*g))

Comparing 1 and 3

(V_1,3)=√(,2*(A_3)2*(((P_atm)-(P_3))/ρ+(-1)*g))

(V_1,3)=√(,2*(π*(D/2)2)2*(((P_atm)-(P_3))/ρ+(-1)*g))

Maximum flow before cavitation is when we set (P_2)=(P_saturation,20) and (P_3)=(P_saturation,20)

Lowest value between the two will be max flow rate

Sub in values

(V_1,2)=√(,2*(π*(0.05/2)2)2*((101300-2339)/1000+(4)*(9.81)))

(V_1,2)=0.03264378288

(V_1,3)=√(,2*(π*(0.12/2)2)2*((101300-2339)/1000+(-1)*(9.81)))

(V_1,3) =0.1510186142

Hence we take the lower flow rate

V=3.26⨯10

(b):

Maximum elevation occurs when diameter of pipe = D

Call this point 5 and compare with point 1 (need to see if I need to add kinetic correction factor alpha = 1.05 in lecture 5)

(P_1)/(ρ*g)+((v_1)2)/(2*g)+(z_1)=(P_5)/(ρ*g)+((v_5)2)/(2*g)+(z_5)

(z_5)=((P_1)-(P_5))/(ρ*g)+((v_1)2-(v_5)2)/(2*g)+(z_1)

Where

(P_5)=(P_saturation,20) (To test lowest pressure before cavitation)

(v_5)=V/(A_5)

(A_5)=π*(D/2)2

(z_5)=((P_atm)-(P_saturation,20))/(ρ*g)-(V2)/(2*(π*(D/2)2)2*g)+(z_1)

Subbing values

(z_5)=(101300-2339)/((1000)*(9.81))-((3.26⨯10)2)/(2*(π*(0.12/2)2)2*(9.81))+7

(z_5)=16.65703454

(z_5)=16.66

(c):

We can maximise elevation through reducing the velocity head, as this will require either static pressure or elevation to increase according to Bernoulli's law. If elevation increases, we have achieved our goal, if static pressure increases, it allows higher elevations before cavitation occurs. We can reduce the velocity head through increasing the pipe diameter or reducing volume flow rate.

Moreover, we can increase atmospheric pressure by moving the piping system to a lower altitude, as the elevation is proportional to the atmospheric pressure. Elevation is inversely proportional to the saturation pressure, which can be reduces through keeping the water at lower temperatures.

Lastly, the maximum elevation is proportional to the reservoir height. Making the reservoir higher while ensuring we minimise altitude to keep the atmospheric pressure high, will increase maximum elevation.

Question 5:

(d):

(W_max)=m*Δ((e_mechanical))

m=ρ*V

Δ((e_mechanical))=(e_mechanical,reservoir)-(e_mechanical,lake)

(e_mechanical,reservoir)=(P_reservoir)/ρ+((v_reservoir)2)/2+(z_reservoir)*g
(e_mechanical,lake)=(P_lake)/ρ+((v_lake)2)/2+(z_lake)*g

Where

ρ=1000

V=3

(P_reservoir)=(P_lake)=(P_atm)

(v_reservoir)=(v_lake)=0

(z_reservoir)=200

(z_lake)=0

g=9.81

Therefore

(W_max)=ρ*V*(z_reservoir)*g

Subbing in values

(W_max)=(1000)*(3)*(200)*(9.81)

(W_max)=5886000

(W_max)=5.89

(e):

First, we must account for head loss

(W_hydraulic)=(W_max)-(E_mechanical*_loss,piping)

(W_hydraulic)=ρ*V*g*(z_reservoir)-ρ*V*g*(H_L)

(W_hydraulic)=ρ*V*g*((z_reservoir)-(H_L))

(W_electricity,generated)=(W_hydraulic)*(η_turbine)*(η_generator)

Where

(H_L)=13

(η_turbine)=0.75

(η_generator)=0.9

Subbing in values

(W_electricity,generated)=(1000)*(3)*(9.81)*(200-13)*(0.75)*(0.9)

(W_electricity,generated)=3714801.75

(W_electricity,generated)=3714801.75

(W_electricity,generated)=3.71

(f):

Revenue is income - cost

(R_day)=(I_day)-(C_day)

(I_day)=(E_generated)*(P_sold)

(E_generated)=(W_electricity,generated)*(t_generated)

(C_day)=(E_used)*(P_bought)

(E_used)=(W_electricity,used)*(t_used)

(W_electricity,used)=((W_max)+ρ*g*(H_L))/((η_pump)*(η_motor))

Where

(P_sold)=0.5/kWh

(P_bought)=0.15/kWh

(t_generated)=(t_used)=12*h (Both run for 12 hours, fine to keep in hours since price is per hour)
(H_L)=13

(η_pump)=0.8

(η_motor)=0.85

Hence for one day

(R_day)=(W_electricity,generated)*(t_generated)*(P_sold)-(((W_max)+ρ*V*(H_L)*g)*(t_used)*(P_bought))/((η_pump)*(η_motor))

For a year

(R_year)=365*((W_electricity,generated)*(t_generated)*(P_sold)-(((W_max)+ρ*V*(H_L)*g)*(t_used)*(P_bought))/((η_pump)*(η_motor)))

Subbing in values

(R_year)=365*((3714801.75)*(12)*(0.5*k*W*h)-((5886000+(1000)*(3)*(13)*(9.81))*(12)*(0.15*k*W*h))/((0.8)*(0.85)))

(R_year)=5695484.029

(R_year)=5.70⨉10

Question 6:

(a):

Forces in x direction, assuming right to be positive:

(F_flange,x)+(F_(P_inlet))=m*((v_out,x)-(v_in,x))

(F_flange,x)=m*((v_out,x)-(v_in,x)) -(F_(P_inlet))

(F_(P_inlet))=(P_inlet)*(A_inlet)

m=ρ*V

(v_out,x)=-(v_out*let)*cos(60)=-1/2*(v_out*let)

(v_out*let)=V/(A_outlet)

(A_outlet)=π*(d/2)2

(v_in,x)=(v_in*let)

(v_inlet)=V/(A_inlet)

(A_inlet)=π*(D/2)2

Where

ρ=1000

V=0.2

d = 0.12

D=0.4

Find (P_inlet) using Bernoulli's theorem by comparing point at inlet with point at outlet.

(P_inlet)/(ρ*g)+((v_in*let)2)/(2*g)+(z_inlet)=(P_outlet)/(ρ*g)+((v_outlet)2)/(2*g)+(z_outlet)

We are finding in terms of gage pressure, assuming (P_outlet) is open to atmosphere

(P_outlet)=0

(z_inlet)=h

(z_outlet)=0

Where

g=9.81

h=0.6

So

(P_inlet)/ρ+((v_in*let)2)/2+h*g=((v_outlet)2)/2

(P_inlet)=ρ*(((v_out*let)2-(v_inlet)2)/2-h*g)

(P_inlet)=ρ*((8*V2)/(π2)*(1/(d4)-1/(D4))-h*g)

Forces in z direction, assuming up to be positive:

(F_flange,z)-(F_W,water)=m*((v_out,z)-(v_in,z))

(F_flange,z)=m*((v_out,z)-(v_in,z))+(F_W,water)

(F_W,water)=ρ*V*g

(v_out,z)=(v_out*let)*sin(60)=-((v_out*let)√(,3))/2

(v_in,z)=0

Where

V=0.04

(F_flange)=√(,(F_flange,x)2+(F_flange,z)2)

(F_flange,x)=ρ*V*(-V/(2*π*(d/2)2)-V/(π*(D/2)2))-ρ*((8*V2)/(π2)*(1/(d4)-1/(D4))-h*g)*(π*(D/2)2)

(F_flange,z)=-(ρ*V2√(,3))/(2*π*(d/2)2)+ρ*V*g

Subbing in values

(F_flange,x)=-20836.64503

(F_flange,z)=-((1000)*(0.2)2√(,3))/(2*π*(0.12/2)2)+(1000)*(0.04)*(9.81)

(F_flange,z)=−2670.538308

(F_flange)=√(,(-20836.64503)2+(-2670.538308)2)

(F_flange)=21007.08335
(F_flange)=21

For angle

θ=tan(-1)((F_flange,z)/(F_flange,x))

Subbing in values

θ=tan(-1)((-2670.538308)/(-20836.64503))

θ=7.3 or 0.13

below the horizontal to the left, since component forces are negative

(b)

Forces on x axis remain the same, however for the z axis

(F_flange,z)-(F_W,water)-(F_W,elbow)=m*((v_out,z)-(v_in,z))

(F_flange,z)=m*((v_out,z)-(v_in,z))+(F_W,water)+(F_W,elbow)

(F_W,elbow)=m*g

Where

m = 7

Hence,

(F_flange,z)=−2670.538308+m*g

Subbing in values

(F_flange,z)=−2670.538308+(7)*(9.81)

(F_flange,z)=−2601.868308

Thus

(F_flange)=√(,(F_flange,x)2+(F_flange,z)2)

Subbing in values

(F_flange)=√(,(-20836.64503)2+(−2601.868308)2)

(F_flange)=20998.46411

(F_flange)=21

For angle

θ=tan(-1)((F_flange,z)/(F_flange,x))

Subbing in values

θ=tan(-1)((-2601.868308)/(-20836.64503))

θ=7.12 or 0.12

below the horizontal to the left, since component forces are negative

Question 7:

(η_pump)=(m*Δ((e_mechanical,oil)))/(W_shaft)

m=ρ*V

Δ((e_mechanical,oil))=(e_mechanical,outlet)-(e_mechanical,inlet)

Δ((e_mechanical,oil))=((P_outlet)/ρ+α((v_outlet)2)/2+(z_outlet)*g)-((P_inlet)/ρ+α((v_in*let)2)/2+(z_inlet)*g)

(W_shaft)=(W_electricity)*(η_motor)

Where

ρ=850

V=0.15

(z_inlet)≈(z_outlet)

(v_inlet)=V/(A_inlet)

(v_out*let)=V/(A_outlet)

(A_inlet)=π*((D_inlet)/2)2

(A_outlet)=π*((D_outlet)/2)2

Δ(P)=(P_outlet)-(P_inlet)=300000

(D_inlet)=0.09

(D_outlet)=0.15

α=1.05

(W_electricity)=20000

(η_motor)=0.9

Finding change in mechanical energy

Δ((e_mechanical,oil))=Δ(P)/ρ+α((v_outlet)2-(v_in*let)2)/2

Δ((e_mechanical,oil))=Δ(P)/ρ+α(8*V2)/(π2)*(1/((D_outlet)4)-1/((D_inlet)4))

Hence

(η_pump)=(ρ*V*(Δ(P)/ρ+α(8*V2)/(π2)*(1/((D_outlet)4)-1/((D_inlet)4))))/((W_electricity)*(η_motor))

Subbing in values

(η_pump)=((850)*(0.15)*(300000/(850)+(1.05)*((8*(0.15)2)/(π2))*(1/((0.15)4)-1/((0.09)4))))/((20000)*(0.9))

(η_pump)=0.7005135407

(η_pump)=70.1

Question 8:

(a):

(W_pump)*(η_pump)=m*Δ((e_mechanical,water))

m=ρ*V

V=((W_pump)*(η_pump))/(ρ*Δ((e_mechanical,water)))

Δ((e_mechanical,water))=(e_mechanical,roof)-(e_mechanical,ground) (We can use surface of roof rather than discharge point as change in mechanical energy will similar, and taking discharge point requires solving a cubic equation)

Δ((e_mechanical,water))=((P_roof)/ρ+((v_roof)2)/2+(z_roof)*g)-((P_ground)/ρ+((v_ground)2)/2+(z_ground)*g)

Where

(W_pump)=5000

(η_pump)=0.8

ρ=1000

(P_roof)≈(P_ground)≈(P_atm)

(v_roof)≈(v_ground)≈0

(z_roof)=30

(z_ground)=0

g=9.81

Therefore

Δ((e_mechanical,water))=(z_roof)*g

V=((W_pump)*(η_pump))/(ρ*(z_roof)*g)

Subbing in values

V=((5000)*(0.8))/((1000)*(30)*(9.81))

V==0.01359157322

V=1.36⨉10

(b):

Compare mechanical energy at inlet and outlet of pump

Δ((e_mechanical,water))=(e_mechanical,outlet)-(e_mechanical,inlet)

Δ((e_mechanical,water))=((P_outlet)/ρ+((v_outlet)2)/2+(z_outlet)*g)-((P_inlet)/ρ+((v_inlet)2)/2+(z_inlet)*g)

Where

Δ(P)=(P_outlet)-(P_inlet)

(v_outlet)=V/(A_outlet)

(v_in*let)=V/(A_inlet)

(A_outlet)=π*((D_outlet)/2)2

(A_inlet)=π*((D_inlet)/2)2

(z_outlet)≈(z_inlet)

(D_outlet)=0.075

(D_inlet)=0.1

So

Δ((e_mechanical,water))=Δ(P)/ρ+((v_outlet)2-(v_in*let)2)/2

Δ(P)=ρ*(Δ((e_mechanical,water))-((v_out*let)2-(v_in*let)2)/2)

Δ(P)=ρ*((z_roof)*g-(8*V2)/(π2)*(1/((D_outlet)4)-1/((D_inlet)4)))

Subbing in values

Δ(P)=(1000)*((30)*(9.81)-(8*(0.01359157322)2)/(π2)*(1/((0.075)4)-1/((0.1)4)))

Δ(P)=291064.9371

Δ(P)=291.1

(c):

Taking mechanical energy at surface of water at roof and ground

(W_pump)*(η_pump)=m(Δ((e_mechanical,water))+(H_L)*g)

Thus using working from (a)

V=((W_pump)*(η_pump))/(ρ*g*((z_roof)+(H_L)))

Where (H_L)=5

Subbing in values

V=((5000)*(0.8))/((1000)*(9.81)*(30+5))

V=0.01164991991

V=698.9951944*LPM

V=699*LPM