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arlik [135]
3 years ago
7

A direct contact heat exchanger (where the fluid mixes completely) has three inlets and one outlet. The mass flow rates of the i

nlets are 1kg/s, 1.5kg/s and 2 kg/s. The enthalpy of those inlets are the 100kJ/kg, 120kJ/kg, and 500kJ/kg, respectively. What is the enthalpy at the outlet?
Engineering
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s

h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

m_1+m_2+m_3=m

   m=1+1.5+2 Kg/s

  m=4.5 Kg/s

Now from energy conservation

m_1h_1+m_2h_2+m_3h_3=mh

By putting the values

1\times 100+1.5\times 120+2\times 500=4.5\times h

So h=284.44 KJ

You might be interested in
Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r
My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0
4 years ago
1. A cylindrical casting is 0.3 m in diameter and 0.5 m in length. Another casting has the same metal is rectangular in cross-se
Lorico [155]

Based on the Chvorinov's rule, the diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the prism casting.

<h3>How to apply the Chvorinov's rule for casting processes</h3>

The Chvorinov's rule is an empirical method to estimate the cooling time of a casting in terms of a <em>reference</em> time. This rule states that cooling time (<em>t</em>) is directly proportional to the square of the volume (<em>V</em>), in cubic meters, divided to the surface area (<em>A</em>), in square meters. Now we proceed to model each casting:

<h3>Cylindrical casting</h3>

t = C · [0.25π · D² · L/(0.5π · D² + π · D · L)]²

t = C · [0.25 · D · L/(0.5 · D + L)]²    (1)

<h3>Prism casting</h3>

t' = C · [3 · T² · L/(6 · T · L + 2 · T · L + 6 · T²)]²

t' = C · [3 · T · L/(8 · L + 6 · T)]²     (2)

<h3>Relationship between the cross sections of both castings</h3>

3 · T² = 0.25π · D²     (3)

Where:

  • <em>t</em> - Cooling time of the cylindrical casting, in time unit.
  • <em>t'</em> - Cooling time of the prism casting, in time unit.
  • <em>C</em> - Cooling factor, in time unit per square meter.
  • <em>D</em> - Diameter of the cylinder, in meters.
  • <em>L</em> - Length of the casting, in meters.
  • <em>T</em> - Width of the cross section of the prism casting, in meters.

If we know that <em>D =</em> <em>0.3 m</em>, then the thickness of the prism casting is:

T = \sqrt{\frac{\pi}{12} }\cdot D

<em>T ≈ 0.153 m</em>

<em />

And (1) and (2) simplified into these forms:

<h3>Cylindrical casting</h3>

t = C · {0.25π · (0.3 m) · (0.5 m)/[0.5 · (0.3 m) + 0.5 m]}²

t = 0.0329 · C     (1b)

<h3>Prism casting</h3>

t' = C · {3 · (0.153 m) · (0.5 m)/[8 · (0.5 m) + 6 · (0.153 m)]}²

t' = 0.00218 · C     (2b)

Lastly we find the <em>percentual</em> difference in the solidification times of the two castings by using the following expression:

<em>r = (</em>1 <em>- t'/t) ×</em> 100 %

<em>r = (</em>1 <em>-</em> 0.00218<em>/</em>0.0329<em>) ×</em> 100 %

<em>r =</em> 93.374 %

The <em>cooling</em> time of the <em>prism</em> casting is 6.626 % of the <em>solidification</em> time of the <em>cylindrical</em> casting. The diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the <em>prism</em> casting. \blacksquare

To learn more on solidification times, we kindly invite to check this verified question: brainly.com/question/13536247

3 0
3 years ago
Handsaw teeth are very sharp: to avoid being cut by the teeth, keep hands and fingers well away from the
siniylev [52]
Handsaw teeth are very sharp: to avoid being cut by the teeth, keep hands and fingers well away from the
path of the blade
6 0
3 years ago
Read 2 more answers
A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is replaced by a high-efficiency 75-hp motor
Naya [18.7K]

Answer:

4.536hp

Explanation:

The decrease in the heat gain of the room is determined from difference in electrical inputs:

Q = W_{shaft} (\frac{1}{n_{1} } - \frac{1}{n_{2} })\\Q = (75hp)*(\frac{1}{0.91 } - \frac{1}{0.963 })\\\\Q = 4.536 hp

8 0
3 years ago
Comparison of density values determines whether an item will float or sink in water. For each of the values below, determine the
geniusboy [140]

Answer:

a) the object floats

b) the object floats

c) the object sinks

Explanation:

when an object is less dense than in the fluid in which it is immersed, it will float due to its weight and volume characteristics, so to solve this problem we must find the mass and volume of each object in order to calculate the density and compare it with that of water

a)

volumen for a cube

V=L^3

L=1.53in=0.0388m

V=0.0388 ^3=5.8691x10^-5m^3=58.69ml

density=m/v

density=13.5g/58.69ml=0.23 g/ml

The wooden block floats  because it is less dense than water

b)

m=111mg=0.111g

density=m/v

density=0.111g/0.296ml=0.375g/ml

the metal paperclip floats   because it is less dense than water

c)

V=0.93cups=220.0271ml

m=0.88lb=399.1613g

Density=m/v

density=399.1613/220.027ml=1.8141g/ml

the apple sinks   because it is  denser than water

4 0
3 years ago
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