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2 years ago

Why did you choose agricultural and biosystem as a course? Help me guys

Engineering

1 answer:

Mazyrski [523]2 years ago

7 0

Answer:

To be able to develop as a human and give back to the community. To ensure that people have safe food and water to drink, clean fuel and energy sources, and in general a safe enviornment to live in.

Explanation:

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A system samples a sinusoid of frequency 230 Hz at a rate of 175 Hz and writes the sampled signal to its output without further

Irina-Kira [14]

B I got it right so (;

6 0

2 years ago

How do you calculate the dynamic lift in an aeroplane?

3241004551 [841]

Answer:

the lift equation states that lift L is equal to the lift coefficient CI times the density r times half of the velocity V squared times the wing area A. For given air conditions,shape and inclination of the object, we have to determine a value of CI to determine the lift.

Not really sure but this is all I know

8 0

3 years ago

Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity i

RoseWind [281]

To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.

The stagnation temperature can be defined as

$T_0 = T+\frac{V^2}{2c_p}$

Where

T = Static temperature

V = Velocity of Fluid

$c_p =$ Specific Heat

Re-arrange to find the static temperature we have that

$T = T_0 - \frac{V^2}{2c_p}$

$T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})$

$T = 672.88K$

Now the pressure of helium by using the Adiabatic pressure temperature is

$P = P_0 (\frac{T}{T_0})^{k/(k-1)}$

Where,

$P_0$ = Stagnation pressure of the fluid

k = Specific heat ratio

Replacing we have that

$P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}$

$P = 0.399Mpa$

Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa

<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>

3 0

3 years ago

The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300 lb⋅ft, a

Arturiano [62]

Answer:

Explanation:

Given that:

Torque T = 2300 lb - ft

Bending moment M = 1500 lb - ft

axial thrust P = 2500 lb

yield points for tension σY= 100 ksi

yield points for shear τY = 50 ksi

Using maximum-shear-stress theory

$\sigma_A = \dfrac{P}{A}+\dfrac{Mc}{I}$

where;

$A = \pi c^2$

$I = \dfrac{\pi}{4}c^4$

$\sigma_A = \dfrac{P}{\pi c^2}+\dfrac{Mc}{ \dfrac{\pi}{4}c^4}$

$\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{1500*12c}{ \dfrac{\pi}{4}c^4}$

$\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{72000c}{\pi c^3}}$

$\tau_A = \dfrac{T_c}{\tau}$

where;

$\tau = \dfrac{\pi c^4}{2}$

$\tau_A = \dfrac{T_c}{\dfrac{\pi c^4}{2}}$

$\tau_A = \dfrac{2300*12 c}{\dfrac{\pi c^4}{2}}$

$\tau_A = \dfrac{55200 }{\pi c^3}}$

$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\dfrac{(\sigma_x - \sigma_y)^2}{2}+ \tau_y^2}$

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi c ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi c^3}+ \dfrac{55200}{\pi c^3}} \ \ \ \ \ ------(1)$

Let say :

$|\sigma_1 - \sigma_2| = \sigma_y$

Then :

$2\sqrt{( \dfrac{2500c + 72000}{2 \pi c^3})^2+ ( \dfrac{55200}{\pi c^3})^2 } = 100(10^3)$

$(2500 c + 72000)^2 +(110400)^2 = 10000*10^6 \pi^2 c^6$

$6.25c^2 + 360c+ 17372.16-10,000\ \pi^2 c^6 =0$

According to trial and error;

c = 0.75057 in

Replacing c into equation (1)

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}$

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} + \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}} \ \ \ OR \\ \\ \\ \sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} - \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}$

$\sigma _1 = 22193 \ Psi$

$\sigma_2 = -77807 \ Psi$

The required diameter d = 2c

d = 1.50 in or 0.125 ft

6 0

3 years ago

A refrigerator operating on the Carnot cycle is used to make ice. Water freezing at 32oF is the cold reservoir. Heat is rejected

Mnenie [13.5K]

Answer:

Explanation:

From the given information:

Water freezing temp. $T_L = 32 ^0 \ F$

Heat rejected temp $T_H = 72 ^0 \ F$

Recall that:

The coefficient of performance is:

$COP_{ref} = \dfrac{T_L}{T_H - T_L} \\ \\ = \dfrac{32+460}{(72 +460) -(32+460)} \\ \\ =\dfrac{492}{532 -492} \\ \\ = \dfrac{492}{40} \\ \\ COP_{ref} = 12.3$

Again:

The efficiency given by COP can be represented by:

$COP = \dfrac{Q_L}{W} \\ \\ W = \dfrac{Q_L}{COP} \\ \\ W = \dfrac{2000 \ lbm \times 144 \ Btu/lbm}{12.3} \\ \\ W = 23414.63 \ Btu$

Finally; the power input in an hour can be determined by using the formula:

$Power= \dfrac{W}{t} \\ \\ Power = \dfrac{23414.63 \ Btu}{ 1 \ hr} \\ \\ Power = \dfrac{23414.634 \times 1055.056 \ J}{1 \times 3600} \\ \\ Power = 6.86 \ kW$

In hp; since 1 kW = 1.34102 hp

6.86kW will be = (6.86 × 1.34102) hp

= 9.199 hp

7 0

2 years ago