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Ray Of Light [21]
3 years ago
7

What's the best way to plan an organized​

Engineering
1 answer:
Alekssandra [29.7K]3 years ago
5 0

Answer:

Get ready and comfortable.

List all of the tasks you need to accomplish over the next week. .

Next schedule everything.

Get a planner/calender.

Cut those tasks that do not fit into your

You might be interested in
Air at 38°C and 97% relative humidity is to be cooled to 14°C and fed into a plant area at a rate of 510m3/min. (a) Calculate th
Katarina [22]

To develop the problem it is necessary to apply the concepts related to the ideal gas law, mass flow rate and total enthalpy.

The gas ideal law is given as,

PV=mRT

Where,

P = Pressure

V = Volume

m = mass

R = Gas Constant

T = Temperature

Our data are given by

T_1 = 38\°C

T_2 = 14\°C

\eta = 97\%

\dot{v} = 510m^3/kg

Note that the pressure to 38°C is 0.06626 bar

PART A) Using the ideal gas equation to calculate the mass flow,

PV = mRT

\dot{m} = \frac{PV}{RT}

\dot{m} = \frac{0.6626*10^{5}*510}{287*311}

\dot{m} = 37.85kg/min

Therfore the mass flow rate at which water condenses, then

\eta = \frac{\dot{m_v}}{\dot{m}}

Re-arrange to find \dot{m_v}

\dot{m_v} = \eta*\dot{m}

\dot{m_v} = 0.97*37.85

\dot{m_v} = 36.72 kg/min

PART B) Enthalpy is given by definition as,

H= H_a +H_v

Where,

H_a= Enthalpy of dry air

H_v= Enthalpy of water vapor

Replacing with our values we have that

H=m*0.0291(38-25)+2500m_v

H = 37.85*0.0291(38-25)-2500*36.72

H = 91814.318kJ/min

In the conversion system 1 ton is equal to 210kJ / min

H = 91814.318kJ/min(\frac{1ton}{210kJ/min})

H = 437.2tons

The cooling requeriment in tons of cooling is 437.2.

3 0
3 years ago
g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,
irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where   r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where   r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0
3 years ago
What type of engineering do you think would help solve this SDG???
OleMash [197]

Answer:

Explanation:

Planning engineering

4 0
2 years ago
Air enters an adiabatic turbine at 900 K and 1000 kPa. The air exits at 400 K and 100 kPa with a velocity of 30 m/s. Kinetic and
Ivanshal [37]

hooooooooooooooooooooooooooooooooooooooooooooooooooooooe    

3 0
3 years ago
Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a
Ann [662]

Answer:

(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K

Explanation:

Solution

Given that:

The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

Diameter of the tube, D = 1cm =0.01 m

Electrical heat rate, q =76 W/m

Wall Temperature, Ts = 370 K

Now,

From the properties table of engine oil we can deduce as follows:

thermal conductivity, k =0.139 W/m .K

Density, ρ = 854 kg/m³

Specific heat, cp = 2120 J/kg.K

(a) Thus

The wall heat flux is given as follows:

qs = q/πD

=76/π *0.01

= 2419.16 W/m²

Now

The oil mean temperature is given as follows:

Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)

Tb =370 - 11/24 * (2419.16 * 0.005/0.139)

Tb = 330.12 K

(b) The center line temperature is given below:

Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)

Tc =304.73 K

(c) The flow velocity is given as follows:

V = m/ρ (πR²)

Now,

The The axial gradient of the mean temperature is given below:

dTb/dx = 2 *qs/ρ *V*cp * R

=2 *qs/ρ*[m/ρ (πR²) *cp * R

=2 *qs/[m/(πR)*cp

dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120

dTb/dx = 19.81 K/m

(d) The heat transfer coefficient is given below:

h =48/11 (k/D)

=48/11 (0.139/0.01)

h =60.65 W/m². K

8 0
3 years ago
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