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White raven [17]
3 years ago
13

An electrical heater is a form of sensible heating process, and heats 0.1m/s of air from 15°C and 80% RH to 50°C? The barometric

pressure is 101.325 kPa. Using the psychrometric chart given, determine the heater wattage.
Engineering
1 answer:
lawyer [7]3 years ago
3 0

Answer:

The heater load =35 KJ/kg

Explanation:

Given that

At initial condition

Temperature= 15°C

RH=80%

At final condition

Temperature= 50°C

We know that in sensible heating process humidity ratio remain constant.

Now from chart

At temperature= 15°C and RH=80%

h_1=38 \frac{KJ}{kg},v=0.8 \frac{m^3}{kg}

At  temperature= 50°C

h_2=73 \frac{KJ}{kg}

So\ the\ heater\ load =h_2-h_1

The heater load = 73 - 38 KJ/kg

The heater load =35 KJ/kg

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An add tape of 101 ft is incorrectly recorded as 100 ft for a 200-ft distance. What is
baherus [9]

Answer:

the correct distance is 202 ft

Explanation:

The computation of the correct distance is shown below:

But before that correction to be applied should be determined

= (101 ft - 100 ft) ÷ (100 ft) × 200 ft

= 2 ft

Now the correct distance is

= 200 ft +  2 ft

= 202 ft

Hence, the correct distance is 202 ft

The same would be relevant and considered too

4 0
3 years ago
Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7 m/s. Assume the width of the plate (into the paper)
mezya [45]

Complete Question

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a co temperature of 100C,find:

The total heat transfer rate from the plate to the air

Answer:

q=1.7845

Explanation:

From the question we are told that:

Air Temperature T_1=40c

Length l=2m

Velocity v=7m/s

Width w=0.5

Constant temperature T_t= 100C

Generally the equation for Total heat Transfer is mathematically given by

 q=hA(T_s-T_\infty)

Where

h=Convective heat transfer coefficient

 h=29.9075w/m^2k

Therefore

 q=h(L*B)(T_s-T_\infty)

 q=29.9075*(2*0.5)(100+273-(40+273))

 q=1794.45w

 q=1.7845

5 0
3 years ago
Incremental software development could be very effectively used for customers who do not have a clear idea about the systems nee
avanturin [10]

<u>Software Development and Client Needs</u>

In Incremental method of software development customers who do not have a basic idea of the development process are being carried along on like other methods that will relegate them to the background until a product is ready.

With this model and structure in place, when softwares/ products are built from several stages e.g prototype, testing, and when new features are added customers are always carried along with their valuable feedback and suggested greatly considered to achieve the customers satisfactions

This model will work well for the customers/clients who does not have a clear idea on the systems needed for their operations.

In summary the incremental model combines features from the waterfall and prototyping model.

For more information on soft ware development process kindly visit

brainly.com/question/20369682

5 0
2 years ago
Assuming the point estimate in Problem 6.36 is the true population parameter, what is the probability that a particular assay, w
sp2606 [1]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability is 0.958

Explanation:

The explanation is shown on the second and third uploaded image

3 0
4 years ago
A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric
diamong [38]

Answer:

Explanation:

Given

q_1=11.5\ nC charge is placed at x=0\ cm

another charge of q_2=-1.2\ nC is at x=3\ cm

We know that Electric field due to positive charge is away  from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to q_2 at some distance r from it

E_2=\frac{kq_2}{r^2}

Now Electric Field due to q_1 is

E_1=\frac{kq_1}{(3+r)^2}

Now E_1+E_2=0

\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0

\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}

\frac{3+r}{r}=3.095

thus r=1.43\ cm

Thus Electric field is zero at some distance r=1.43 cm right of q_2

3 0
3 years ago
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