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1 year ago

What is the potiental energy of a 3 kg ball that is on the ground

Engineering

1 answer:

Llana [10]1 year ago

8 0

Answer:

147.15

Explanation:

147.15 is the answer

You might be interested in

A triangular roadside channel is poorly lined with riprap. The channel has side slopes of 2:1 (H:V) and longitudinal slope of 2.

Oliga [24]

Answer:

Q = 14.578 m³/s

Explanation:

Given

We use the Manning Equation as follows

Q = (1/n)*A*(∛R²)*(√S)

where

Q = volumetric water flow rate passing through the stretch of channel (m³/s for S.I.)

A = cross-sectional area of flow perpendicular to the flow direction, (m² for S.I.)

S = bottom slope of channel, m/m (dimensionless) = 2.5% = 0.025

n = Manning roughness coefficient (empirical constant), dimensionless = 0.023

R = hydraulic radius = A/P (m for S.I.) where :

A = cross-sectional area of flow as defined above,

P = wetted perimeter of cross-sectional flow area (m for S.I.)

we get A as follows

A = (B*h)/2

where

B = 5 m (the top width of the flowing channel)

h = (B/2)*(m) = (5 m/2)*(1/2) = 1.25 m (the deep)

A = (5 m*1.25 m/2) = 3.125 m²

then we find P

P = 2*√((B/2)²+h²) ⇒ P = 2*√((2.5 m)²+(1.25 m)²) = 5.59 m

⇒ R = A/P ⇒ R = 3.125 m²/5.59 m = 0.559 m

Substituting values into the Manning equation gives:

Q = (1/0.023)*(3.125 m²)*(∛(0.559 m)²)*(√0.025)

⇒ Q = 14.578 m³/s

8 0

3 years ago

The engine of a 2000kg car has a power rating of 75kW. How long would it take (seconds) to accelerate from rest to 100 km/hr at

Delvig [45]

Answer: 10.29 sec.

Explanation:

Neglecting drag and friction, and at road level , the energy developed during the time the car is accelerating, is equal to the change in kinetic energy.

If the car starts from rest, this means the following:

ΔK = 1/2 m*vf ²

As Power (by definition) is equal to Energy/Time= 75000 W= 75000 N.m/seg, in order to get time in seconds, we need to convert 100 km/h to m/sec first:

100 (Km/h)*( 1000m /1 Km)*(3600 sec/1 h)= 27,78 m/sec

Now, we calculate the change in energy:

ΔK= 1/2*2000 Kg. (27,78)² m²/sec²= 771,728 J

4 0

3 years ago

An ideal reheat Rankine cycle with water as the working fluid operates the boiler at 15,000 kPa, the reheater at 2000 kPa, and t

solniwko [45]

Answer:

See the explanation below.

Explanation:

First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.

Find h₁:

Using Saturated Water Table and given pressure p₁ = 100 kPa

h₁ = 417.5 kJ/kg

Find h₂:

In order to find h₂, add the $w_{p}$ to h₁, where $w_{p}$ is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.

But first we need to compute $w_{p}$ To computer

Pressures:

p₁ = 100 kPa

p₂ = 15,000 kPa

and

Using saturated water pressure table, the volume of water $v_{f}$ = 1.0432

Dividing 1.0432/1000 gives us:

Volume of water = v₁ = 0.001043 m³/kg

Compute the value of h₂:

h₂ = h₁ + v₁ (p₂ - p₁)

= 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)

= 417.5 + 0.001043 (14900)

= 417.5 + 15.5407

= 433.04 kJ/kg

Find h₃

Using steam table:

At pressure p₃ = 15000 kPa

and Temperature = T₃ = 450°C

Then h₃ = 3159 kJ/kg

The entropy s₃ = 6.14 kJ/ kg K

Find h₄

Since entropy s₃ is equal to s₄ So

s₄ = 6.14 kJ/kgK

To compute h₄

s₄ = $s_{f}$ + $x_{4} s_{fg}$

$x_{4} = s_{4} -s_{f} /s_{fg}$

$x_{4}$ = 6.14 - 2.45 / 3.89

$x_{4}$ = 0.9497

The enthalpy h₄:

h₄ = $h_{f} +x_{4} h_{fg}$

= 908.4 + 0.9497(1889.8)

= 908.4 + 1794.7430

= 2703 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa

Find h₅

Using steam table:

At pressure p₅ = 2000 kPa

and Temperature = T₅ = 450°C

Then h₅ = 3358 kJ/kg

Find h₆:

Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to So

s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK

To compute h₆

s₆ = $s_{f}$ + $x_{6} s_{fg}$

$x_{6} = s_{6} -s_{f} /s_{fg}$

$x_{6}$ = 7.29 - 1.3028 / 6.0562

$x_{6}$ = 0.988

The enthalpy h₆:

h₆ = $h_{f} +x_{6} h_{fg}$

= 417.51 + 0.988 (2257.5)

= 417.51 + 2230.41

h₆ = 2648 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa

Compute power used by pump:

$P_{p}$ is found by using:

mass flow rate = m = 1.74 kg/s

Volume of water = v₁ = 0.001043 m³/kg

p₁ = 100 kPa

p₂ = 15,000 kPa

$P_{p}$ = ( m ) ( v₁ ) ( p₂ - p₁ )

= (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)

= (1.74 kg/s) (0.001043 m³/kg) (14900)

= 27.04

$P_{p}$ = 27 kW

Compute heat added $q_{a}$ and heat rejected $q_{r}$ from boiler using computed enthalpies:

$q_{a}$ = ( h₃ - h₂ ) + ( h₅ - h₄ )

= ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )

= 2726 + 655

= 3381 kJ/kg

$q_{r}$ = h₆ - h₁

= 2648 kJ/kg - 417.5 kJ/kg

= 2232 kJ/kg

Compute net work

W $_{net}$ = $q_{a}$ - $q_{r}$

= 3381 kJ/kg - 2232 kJ/kg

= 1150 kJ/kg

Compute power produced by the cycle

mass flow rate = m = 1.74 kg/s

W $_{net}$ = 1150 kJ/kg

P = m * W $_{net}$

= 1.74 kg/s * 1150 kJ/kg

= 2001 kW

Compute rate of heat transfer in the reheater

Q = m * ( h₅ - h₄ )

= 1.74 kg/s * 655

= 1140 kW

Compute Thermal efficiency of this system

μ $_{t}$ = 1 - $q_{r}$ / $q_{a}$

= 1 - 2232 kJ/kg / 3381 kJ/kg

= 1 - 0.6601

= 0.34

= 34%

7 0

3 years ago

Please answer fast. With full step by step solution.

lina2011 [118]

Let f(z) = (4z ² + 2z) / (2z ² - 3z + 1).

First, carry out the division:

f(z) = 2 + (8z - 2) / (2z ² - 3z + 1)

Observe that

2z ² - 3z + 1 = (2z - 1) (z - 1)

so you can separate the rational part of f(z) into partial fractions. We have

(8z - 2) / (2z ² - 3z + 1) = a / (2z - 1) + b / (z - 1)

8z - 2 = a (z - 1) + b (2z - 1)

8z - 2 = (a + 2b) z - (a + b)

so that a + 2b = 8 and a + b = 2, yielding a = -4 and b = 6.

So we have

f(z) = 2 - 4 / (2z - 1) + 6 / (z - 1)

f(z) = 2 - (2/z) (1 / (1 - 1/(2z))) + (6/z) (1 / (1 - 1/z))

Recall that for |z| < 1, we have

$\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n$

Replace z with 1/z to get

$\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}$

so that by substitution, we can write

$\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}$

Now condense f(z) into one series:

$\displaystyle f(z) = 2 - \sum_{n=0}^\infty 2^{-n+1} z^{-(n+1)} + 6 \sum_{n=0}^\infty z^{-n-1}$

$\displaystyle f(z) = 2 - \sum_{n=0}^\infty \left(6+2^{-n+1}\right) z^{-(n+1)}$

$\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{-(n-1)+1}\right) z^{-n}$

$\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{2-n}\right) z^{-n}$

So, the inverse Z transform of f(z) is $\boxed{6+2^{2-n}}$ .

4 0

3 years ago

Oil with a density of 800 kg/m3 is pumped from a pressure of 0.6 bar to a pressure of 1.4 bar, and the outlet is 3 m above the i

Naddik [55]

Answer:

23.3808 kW

20.7088 kW

Explanation:

ρ = Density of oil = 800 kg/m³

P₁ = Initial Pressure = 0.6 bar

P₂ = Final Pressure = 1.4 bar

Q = Volumetric flow rate = 0.2 m³/s

A₁ = Area of inlet = 0.06 m²

A₂ = Area of outlet = 0.03 m²

Velocity through inlet = V₁ = Q/A₁ = 0.2/0.06 = 3.33 m/s

Velocity through outlet = V₂ = Q/A₂ = 0.2/0.03 = 6.67 m/s

Height between inlet and outlet = z₂ - z₁ = 3m

Temperature to remains constant and neglecting any heat transfer we use Bernoulli's equation

$\frac {P_1}{\rho g}+\frac{V_1^2}{2g}+z_1+h=\frac {P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\\Rightarrow h=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+z_2-z_1\\\Rightarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+\frac{6.67_2^2-3.33^2}{2\times 9.81}+3\\\Rightarrow h=14.896\ m$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 14.896\\\Rightarrow W_{p}=23380.8\ W$

∴ Power input to the pump 23.3808 kW

Now neglecting kinetic energy

$h=\frac{P_2-P_1}{\rho g}+z_2-z_1\\\Righarrow h=\frac{(1.4-0.6)\times 10^5}{800\times 9.81}+3\\\Righarrow h=13.19\ m\\$

Work done by pump

$W_{p}=\rho gQh\\\Rightarrow W_{p}=800\times 9.81\times 0.2\times 13.193\\\Rightarrow W_{p}=20708.8\ W$

∴ Power input to the pump 20.7088 kW

6 0

3 years ago