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3 years ago

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Moist air at 27 deg C, 1atm, and 50% relative humidity enters an evaporative cooling unit operating at steady state consisiting

of a heating section followed by a soaked pad evaporative cooler operating adiabatically. The air passing through the eating section is heated to 45 deg C. Next, air passing through a soaked pad exiting with 50% relative humidity. Using data from psychrometric chart, determine A) the humidity ratio of the entering moist air mixture, in kg(vapor) per kg(dry air). B) the rate of heat transfer to the moist air passing through the heating section, in kj per kg of mixture. C) The humidity ratio and temperature, in deg C at the exit of the evaporative cooling section.

1 answer:

OleMash [197]3 years ago

6 0

Answer:

Explanation:

solution is attached below

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A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end

hjlf

Answer:

a) $v=19.6 m/s$

b) $H=19.58 m$

c) $v_{f}=29.57 m/s$

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

$W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J$

But we know the work is equal to change of kinetic energy, so:

$W=\Delta K=\frac{1}{2}mv^{2}$

$v=\sqrt{\frac{2W}{m}}=19.6 m/s$

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

$v_{f}^{2}=v_{i}^{2}-2gh$

At the maximum height the velocity is 0, so v(f) = 0.

$0=v_{i}^{2}-2gH$

$H=\frac{19.6^{2}}{2*9.81}=19.58 m$

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

$v_{f}^{2}=v_{i}^{2}-2gh$

$v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s$

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

6 0

3 years ago

For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger sh

Paul [167]

Answer:

the pressure drop is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere $dp$ = 0.003 m

particle density = 1300 kg/m³

the bed void fraction $\in =$ 0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas $\rho$ = 0.15 mol/dm ⁻³

viscosity of methane gas $\mu$ = 1.429 x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³ to kg/m³

SO; we have :

Density = 0.15 mol/dm ⁻³

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density = $0.1 5 *\dfrac{16}{0.1^3}$

Density = 2400

Density $\rho_f$ = 2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas = 4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

$Re = \dfrac{dV \rho}{\mu}$

$Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}$

$Re=2276.317705$

For Re > 1000

$\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}$

$\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}$

$\Delta P=8575.755212*2.5$

$\Delta = 21439.38803 \ Pa$

To atm ; we have

$\Delta P = \dfrac{21439.38803 }{101325}$

$\Delta P =0.2115903087 \ atm$

ΔP ≅ 0.21159 atm

Thus; the pressure drop is 0.21159 atm

4 0

3 years ago

As a human resources professional for a large company, Tommy is involved with the ___________ process. His goal is to find the r

zaharov [31]

Answer:

Recruitment

Explanation:

The process of recruitment involves several activities whose main objective is to get the right people with relevant qualifications at the appropriate time to meet the needs of an organisation whenever there’s need. The process involves screening a pool of candidates through several steps to obtain the best fit for the job.

4 0

3 years ago

what problem was the team presented within this episode? What problem mine they have thought they should solve if they hadn’t li

jok3333 [9.3K]

Explanation:

whats

the question choices

8 0

3 years ago

The radius of a circle is increased from 5.00 to 5.04 m Estimate the resulting change in area, and then express the estimate as

musickatia [10]

Answer:

1.016%

Explanation:

We need to calculate the rate in the area.

We know that,

$A=\pi r^2$

$A_1=\pi (5)^2 =78.5398m^2\\A_2= \pi(5.04)^2 = 79.8014m^2$

The change in area is

$A_2-A_1=1.2616m^2$

In percentage that represent,

$Rate = \frac{A_2}{A_1} = \frac{79.8014}{78.5398} = 1.016\%$

3 0

3 years ago