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OLEGan [10]
4 years ago
14

Liquid ethanol (C2H5OH) at 298 K, 1 atm enters a steam generator operating at steady state and burns completely with dry air ent

ering at 400 K, 1 atm. The fuel flow rate is 75 kg/s, and the equivalence ratio is 0.7. If the combustion products exit at 1500 K, 1 atm, what is the rate of heat transfer in W?
Engineering
1 answer:
Gennadij [26K]4 years ago
6 0

Answer:

Q= 6491.100 kJ/s

Explanation:

Air-Fuel Ratio:

For a combustion reaction the proportion of air that is present in a gaseous substance responsible for the reaction,this proportion is known as air-fuel ratio.The air fuel ratio is calculated using the combustion reaction for the substance.

Considering reaction for the Ethanol as

C₂H₅OH +XO₂(O₂+3.76N₂)→ aCO₂+bH₂O+cN₂

Balancing the equation we get;

a=2,

2b=6

∴ b=3

xO₂=3

The air-fuel ratio

A/F = XO₂+H₂O+xN₂× mass of N₂/mass(fuel)

3.31×31.9+11.28×28.013/46.069

= 8.943

Equivalent ratio = 0.7,

so, heat transfer

Q= m ×Cp×ΔT

= 75×0.7×112.4(1500-400)

Q= 6491.100 kJ/s

1kJ/s=1000w

∴ Q= 6491100 W

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A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one v
In-s [12.5K]

Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

6 0
3 years ago
Poached eggs are cooked in bath of boiling water at 100°C. Over time they reach thermal equilibrium with the bath. They are then
barxatty [35]

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above and also note that I derived the whole expressions in variable firm that are required to solve this question and giving reasons for each step so that you can employ these steps in solving new problems.

8 0
3 years ago
16. Which polymers are recyclable?
MrRissso [65]

Answer:

Number 2 is HDPE, or High-Density Polyethylene. It is most commonly found in plastic bags, oil and detergent bottles, and toys. It is one of the simplest polymers to recycle. It is broken down into flakes, decontaminated and then reused for pipes, flower pots or reused as (non-food) bottles.

8 0
3 years ago
A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio
natima [27]

Answer:

1) A=282.6 mm

2)a_{max}=60.35\ m/s^2

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm     ------------1

V=2.5 m/s at x=225 mm   ------------2

Where x measured  from mid position.

We know that velocity in simple harmonic given as

V=\omega \sqrt{A^2-x^2}

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

3.5=\omega \sqrt{A^2-0.15^2}    ------3

2.5=\omega \sqrt{A^2-0.225^2}   --------4

Now by dividing equation 3 by 4

\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}

1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}

So    A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

3.5=\omega \sqrt{A^2-0.15^2}

3.5=\omega \sqrt{0.2826^2-0.15^2}

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

a_{max}=\omega ^2A

a_{max}=14.61 ^2\times 0.2826\ m/s^2

a_{max}=60.35\ m/s^2

Time period T

T=\dfrac{2\pi}{\omega}

T=\dfrac{2\pi}{14.609}

T=0.42 sec

8 0
4 years ago
Why do engineers play a variety of roles in the engineering process?
Crazy boy [7]

Answer: The engineering design process emphasizes open-ended problem solving and encourages students to learn from failure. This process nurtures students abilities to create innovative solutions to challenges in any subject! In addition to their involvement in design and development, many engineers work in testing, production, or maintenance. These engineers supervise production in factories, determine the causes of a component's failure, and test manufactured products to maintain quality.

Explanation:

6 0
3 years ago
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