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Tcecarenko [31]
3 years ago
10

Air is to be heated steadily by an 8-kW electric resistance heater as it flows through an insulated duct. If the air enters at 5

5°C at a rate of 2 kg/s, determine the exit temperature of air. Solve using appropriate software.
Engineering
1 answer:
Furkat [3]3 years ago
6 0

To solve this problem it is necessary to apply the concepts related to the heat exchange of a body.

By definition heat exchange in terms of mass flow can be expressed as

W = \dot{m}c_p \Delta T

Where

C_p = Specific heat

\dot{m}= Mass flow rate

\Delta T = Change in Temperature

Our values are given as

C_p = 1.005kJ/kgK \rightarrow Specific heat of air

T_1 = 50\°C

\dot{m} = 2kg/s

W = 8kW

From our equation we have that

W = \dot{m}c_p \Delta T

W = \dot{m}c_p (T_2-T_1)

Rearrange to find T_2

T_2 = \frac{W}{\dot{m}c_p}+T_1

Replacing

T_2 = \frac{8}{2*1.005}+(50+273)

T_2 = 326.98K \approx 53.98\°C

Therefore the exit temperature of air is 53.98°C

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X cotx expansion using maclaurins theorem.
Lemur [1.5K]

It is to be noted that it is impossible to find the Maclaurin Expansion for F(x) = cotx.

<h3>What is Maclaurin Expansion?</h3>

The Maclaurin Expansion is a Taylor series that has been expanded around the reference point zero and has the formula f(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!

<h3>What is the explanation for the above?</h3>

as indicated above, the Maclaurin infinite series expansion is given as:

F(x)=f(0)+f′. (0) 1! x+f″ (0) 2! x2+⋯+f[n](0)n!

If F(0) = Cot 0

F(0) = ∝ = 1/0

This is not definitive,

Hence, it is impossible to find the Maclaurin infinite series expansion for F(x) = cotx.

Learn more about Maclaurin Expansion at;
brainly.com/question/7846182
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4 0
2 years ago
It has a piece of 1045 steel with the following dimensions, length of 80 cm, width of 30 cm, and a height of 15 cm. In this piec
Serggg [28]

Answer:

material remove in 3 min is 16790.4 mm³/s

Explanation:

given data

length L = 80 cm = 800 mm

width W = 30 cm

height H = 15 cm

make grove length = 80 cm

width = 8 cm

depth = 10 cm

mill toll diameter = 4 mm

axial cutting depth = 20 mm

to find out

How much material removed in 3 minutes

solution

first we find time taken for length of advance that is

time = \frac{length}{advance}

here advance is given as 0.001166 mts / sec

so  time = \frac{800}}{0.001166*1000}

time = 686.106 seconds

now we find material remove rate that is

remove rate = mill toll rate × axial cutting depth × advance

remove rate = 4 × 20×0.001166 ×1000

remove rate = 93.28 mm³/s

so

material remove in 3 minute = 3 × 60 = 180 sec

so material remove in 3 min = 180 × 93.28

material remove in 3 min is 16790.4 mm³/s

7 0
3 years ago
A 10 hp motor is used to raise a 1000 Newton weight at a vertical distance of 5 meters. What is the work the motor performs?
Aleksandr [31]

The work done by a 10 HP motor when it raises a 1000 Newton weight at a vertical distance of 5 meters is <u>5kJ</u>.

Define work. Explain the rate of doing work.

Work is <u>the energy that is moved to or from an item by applying force along a displacement</u> in physics. For a constant force acting in the same direction as the motion, work is <u>easiest expressed as the product of </u><u>force </u><u>magnitude and distance traveled</u>.

Since the <u>force </u><u>transfers one unit of energy for every unit of </u><u>work </u><u>it performs</u>, the rate at which work is done and energy is used are equal.

Solution Explained:

Given,

Weight = 1000N and distance = 5m

A/Q, the work here is done in lifting then

Work = (weight) × (distance moved)

         = 1000 X 5

         = 5000Nm or 5000J = 5kJ

Therefore, the work done in lifting a 1000 Newton weight at a vertical distance of 5 meters is 5kJ.

To learn more about work, use the link given
brainly.com/question/25573309
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<u />

4 0
1 year ago
All holly plants are dioecious—a male plant must be planted within 45 to 55 feet of the
mel-nik [20]

Answer:

The answer is 0.727

Explanation:

lemme know if that's right

8 0
3 years ago
Assign deliveryCost with the cost (in dollars) to deliver a piece of baggage weighing baggageWeight. The baggage delivery servic
Sholpan [36]

Answer: y = x * 1dollars - 30dollars

Explanation:

Giving that the delivery cost in dollar is potent for all x > = 50 pounds of wght

Y = (x - 50)*1 dollar + c ...equ 1

Y = delivery cost equation in dollars

x = weigt of baggage for delivery

c = 20dollars = down payment for the first 50 pound weight of baggage

Equ 1 becomes

Y = (x)dollars - 50 dollars + 20 dollars

Y = (x) dollars -30 dollars

4 0
4 years ago
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