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=1000kg/m^3 (density of water)

g=9.81m/s^2 (acceleration due to gravity)

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

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

Thus

(P_C)=(1000kg/m^3)*(9.81m/s^2)*(3.5m)

(P_C)=34335Pa

(P_avg)=34335Pa

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

Thus

A=a*b

Where

a=10*m (Given from question)

sin(45°)=(5*m)/b

b=(5*m*)/sin(45°)

sin(45°)=1/√(,2)

Thus

b=5√(,2)*m

Hence

A=(10m)*(5√(,2)*m)

A=50√(,2)*m^2

Thus

(F_R)=(34335Pa)*(50√(,2)*m^2)

(F_R)=1716750√(,2)*N

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) =3.5√(,2)*m

(I_xx,c)=(a*b^3)/12

(I_xx,c)=((10m)*(5√(,2)*m)^3)/12

(I_xx,c)=(625√(,2)*m^4)/3

Thus

(y_P)=(3.5√(,2)*m)+((625√(,2)*m^4)/((3)*(3.5√(,2)*m)*(50√(,2)*m^2)))

(y_P)=((86√(,2)*m)/21)

sin(45°)=1m/(y_B)

(y_B)=1m/sin(45°)

(y_B)=√(,2)*m

Thus

(r_P)=((86√(,2)*m)/21)-(√(,2)*m)

(r_P)=(65√(,2)*m)/21

Hence torque on the gate due to water pressure is

(τ_P)=((65√(,2)*m)/21)*(1716750√(,2)*N)

(τ_P)=10627500N*m

Now to calculate the torque due to gravity

(τ_W)=(r_W)*(F_W)

(F_W)=m*g

Where

m=500kg

(F_W)=(500*kg)*(9.81m/s^2)

(F_W)=4905N

Gravity acts through the centroid assuming uniform weight distribution

cos(45°)=(r_W)/(b/2)

(r_W)=(b*cos(45°))/2

(r_W)=(5*m)/2

(r_W)=2.5m

Thus

(τ_W)=(4905N)*(2.5m)

(τ_W)=12262.5N*m

Hence

(τ_↻)=10627500N*m+12262.5N*m

(τ_↻)=10639762.5N*m

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=(r_P) (As pressure also acted on the centroid normal to the surface)

r=(65√(,2)*m)/21

Hence

F=10639762.5N*m/((65√(,2)*m)/21)

F=3437461.731N/√(,2)

F=2430652.5N

F=2.43⨉10^6N

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|*|+x| = C

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

ℇ^ln|(-2-3*y)*(4+x)|=ℇ^C

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

(-2-3*y)*(+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): UNDONE DO THIS DONT SUBMIT BEFORE YOU DO THIS

AHHHHH

AHHHH*AHH

AHHHH

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*v^2

m=ρ*v*A

Where

ρ=1.25kg/m^3

(v_min)=40km/h

(v_min)=40*(1000m/3600s)

(v_min)=100m/s/9

(v_max)=60km/h

(v_max)=60*(1000m/3600s)

(v_max)=50m/s/3

A=π*(D/2)^2

D=100m

A=π*(100m/2)^2

A=2500*π*m^2

Thus

(m_min)=(1.25kg/m^3)*(100m/s/9)*(2500*π*m^2)

(m_min)=(312500*π*k*g/s)/9

(m_max)=(1.25kg/m^3)*(50m/s/3)*(2500*π*m^2)

(m_max)=(156250*π*k*g/s)/3

Hence

(E_kinetic,min)=1/2*((312500*π*k*g/s)/9)*(100m/s/9)^2

(E_kinetic,min) =2143347.051*π*W

(E_kinetic,max)=1/2*((156250*π*k*g/s)/3)*(50m/s/3)^2

(E_kinetic,max) =(195312500*π*W)/27

Thus

(W_wind,min)=2143347.051*π*W

(W_wind,max)==(195312500*π*W)/27

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

(W_electricity,min)=(0.5)*(0.95)*(2143347.051*π*W)

(W_electricity,min)=3198423.591*W

(W_electricity,min)=3.20⨉10^6*W

(W_electricity,min)=3.20⨉10^3*k*W

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

(W_electricity,max)=(0.5)*(0.95)*((195312500*π*W)/27)

(W_electricity,max)=10794679.62

(W_electricity,max)=1.08⨉10^7*W

(W_electricity,max)=1.08⨯10^4*k*W

(b):

F=m*a
m=m*Δ(t)

a=d(v)/d(t)

m=ρ*v*A

Where

ρ=1.25kg/m^3

v=60km/h=50m/s/3 (max force when velocity is maximal as this will create the greatest transfer of momentum)

A=D*h (Area horizontally facing wind is a rectangle assuming force is from one side only)

D=2000⨉10^(-3)m

h=150m

Thus

A=(2000⨉10^(-3)m)*(150m)

A=300m

Maximum force is when momentum from air is transferred and air loses all its momentum, thus losing all its velocity. Thus compute acceleration at this point.

d(v)/d(t)=Δ(v)/Δ(t)
Δ(v)=(0km/h-60km/h)

Δ(v)=-(60*(1000m))/3600s

Δ(v)=-50m/s/3

Hence

F=(m*Δ(t))*(Δ(v)/Δ(t))

F=(ρ*v)*Δ(v)

F=(1.25kg/m^3)*(-50m/s/3)

F=-125N/6

Assuming F is in positive