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motikmotik
4 years ago
7

A double-pipe heat exchanger is used to cool a hot fluid (cp = 3800 J/kg·K) entering the heat exchanger at 200°C with a flow rat

e of 0.4 kg/s. In the cold side, cooling fluid (cp = 4200 J/kg·K) enters the heat exchanger at 10°C with a mass flow rate of 0.5 kg/s. The double-pipe heat exchanger has a thin-walled inner tube, with convection heat transfer coefficients inside and outside of the inner tube estimated to be 1400 W/m2 ·K and 1100 W/m2 ·K, respectively. The heat exchanger has a heat transfer surface area of 2.5 m2 , and the estimated fouling factor caused by the accumulation of deposit on the surfaces is 0.0002 m2 ·K/W. (a) Determine the effectiveness values for the parallel- and counter-flow configurations. (b) Determine outlet temperatures of the hot fluid for the parallel- and counter-flow configurations. (c) Determine outlet temperatures of the cold fluid for the parallel- and counter-flow configurations.

Engineering
1 answer:
MAXImum [283]4 years ago
3 0

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem

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A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 19.636
luda_lava [24]

Answer:

The original length of the specimen is found to be 76.093 mm.

Explanation:

From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:

Initial Volume = Final Volume

πd1²L1/4 = πd2²L2/4

d1²L1 = d2²L2

L1 = d2²L2/d1²

where,

d1 = initial diameter = 19.636 mm

d2 = final diameter = 19.661 mm

L1 = Initial Length = Original Length = ?

L2 = Final Length = 75.9 mm

Therefore, using values:

L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²

<u>L1 = 76.093 mm</u>

5 0
3 years ago
In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t
gladu [14]

Answer:

F(x) = 0           ;  x < 0

         0.064   ;  0 ≤ x < 1

         0.352   ;  1 ≤ x < 2

         0.784   ;  2 ≤ x < 3

            1        ;    x ≥ 3

Explanation:

Each wafer is classified as pass or fail.

The wafers are independent.

Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.

X ~ Bi(n,p)

Where n = 3 and p = 0.6 is the success probability

The probatility function is given by :

P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}

Where nCx is the combinatorial number

nCx=\frac{n!}{x!(n-x)!}

Let's calculate f(x) :

f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064

f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288

f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432

f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216

For the cumulative distribution function that we are looking for :

P(X\leq x)=F(x)

F(0)=f(0)\\F(1)=f(0)+f(1)\\F(2)=f(0)+f(1)+f(2)\\F(3)=f(0)+f(1)+f(2)+f(3)=1

F(0)=0.064\\F(1)=0.064+0.288=0.352\\F(2)=0.064+0.288+0.432=0.784\\F(3)=0.064+0.288+0.432+0.216=1

The cumulative distribution function for X is :

F(x) = 0           ;  x < 0

         0.064   ;  0 ≤ x < 1

         0.352   ;  1 ≤ x < 2

         0.784   ;  2 ≤ x < 3

            1        ;    x ≥ 3

5 0
3 years ago
What the different methods to turn on thyrister and how can a thyrister turned off​
myrzilka [38]

Answer:

forward voltage triggering

temperature triggering

dv/dt triggering

light triggering

gate triggering

Then turning off;

Turn off is accomplished by a "negative voltage" pulse between the gate and cathode terminals

Explanation:

hope it helps

8 0
3 years ago
A long, circular aluminum rod is attached at one end to a heated wall and transfers heat by convection to a cold fluid.
Trava [24]

Answer:

a. Heat removal rate will increase

b. Heat removal rate will decrease

Explanation:

Given that

One end of rod is connected to the furnace and rod is long.So this rod can be treated as infinite long fin.

We know that heat transfer in fin given as follows

Q_{fin}=\sqrt{hPKA}\ \Delta T

We know that area

A=\dfrac{\pi}{4}d^2

Now when diameter will triples then :

A_f=\dfrac{\pi}{4}{\left (3d \right )}^2

A_f=9A

Q'_{fin}=\sqrt{9hPKA}\ \Delta T

Q'_{fin}=3\sqrt{hPKA}\ \Delta T

Q'_{fin}=3Q

So the new heat transfer will increase by 3 times.

Now when copper rod will replace by aluminium rod :

As we know that thermal conductivity(K) of Aluminium is low as compare to Copper .It means that heat transfer will decreases.

3 0
3 years ago
Which option identifies the type of device the engineer will develop in the following scenario?
Stells [14]
It would be actuator
4 0
3 years ago
Read 2 more answers
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