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aliina [53]
3 years ago
8

A closed, rigid tank fitted with a paddle wheel contains 2.0 kg of air, initially at 200oC, 1 bar. During an interval of 10 minu

tes, the paddle wheel transfers energy to the air at a rate of 1 kW. During this time interval, the air also receives energy by heat transfer at a rate of 0.5 kW. These are the only energy transfers. Assume the ideal gas model for the air, and no overall changes in kinetic or potential energy. Do not assume specific heats are constant. Determine the change in specific internal energy for the air, in kJ/kg, and the final temperature of the air, in oC.
Engineering
1 answer:
RUDIKE [14]3 years ago
8 0

Answer:

T=833.8 °C

Explanation:

Given that

m= 2 kg

T₁=200 °C

time ,t= 10 min = 600 s

Work input = 1 KW

Work input = 1 x 600 KJ=600 KJ

Heat input = 0.5 KW

Q= 05 x 600 = 300 KJ

Gas is ideal gas.

We know that for ideal gas internal energy change given as

ΔU= m Cv ΔT

For air Cv= 0.71 KJ/kgK

From first law of thermodynamics

Q  = ΔU +W

Heat input taken as positive and work in put taken as negative.

300 KJ = - 600 KJ + ΔU

ΔU = 900 KJ

ΔU= m Cv ΔT

900 KJ = 2 x 0.71 x (T- 200 )

T=833.8 °C

So the final temperature is T=833.8 °C

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Answer:

Java is called portable because you can compile a java code which will spew out a byte-code, and then you run that code with Java Virtual Machine. Java Virtual Machine is like an interpreter, which reads the compiled byte-code and runs it. So first of all, you need to install the JVM on the system you want.

Explanation:

5 0
3 years ago
A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the
Kay [80]

Answer:

a) V = 0.354

b)  G = 25.34 GPA

Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

Elongation of rod ΔL = 1.4 mm

Normal stress, δ = P/A

Where P = Force acting on the cross-section

A = Area of the cross-section

Using Area, A = π/4 · d²

= π/4 · (0.0020)²  = 3.14 × 10⁻⁴m²

δ = 50/3.14 × 10⁻⁴    = 159.155 MPA

E(long) = Δl/l  = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

= 159.155 × 10⁶/2.33 × 10⁻³    = 68.306 GPA

Also final diameter d(f) = 19.9837 mm

Initial diameter d(i) = 20 mm

Poisson said that V = Е(elasticity)/Е(long)

= -  <u>( 19.9837 - 20 /20)</u>

        2.33 × 10⁻³                  

= 0.354,

∴ v = 0.354

Also G = Е/2. (1+V)

=  68.306 × 10⁹/ 2.(1+ 0.354)

= 25.34 GPA

⇒ G = 25.34 GPA

5 0
3 years ago
What is the output of a system with the transfer function s/(s + 3)^2 and subject to a unit step input at time t = 0?
Dominik [7]

Answer:

0

Explanation:

output =transfer function H(s) ×input U(s)

here H(s)=\frac{s}{(s+3)^2}

U(s)=\frac{1}{s} for unit step function

output =H(s)×U(s)

=\frac{s}{(s+3)^2}×\frac{1}{s}

=\frac{1}{(s+3)^2}

taking inverse laplace of output

output=t×e^{-3t}

at t=0 putting the value of t=0 in output

output =0

3 0
3 years ago
What is the governing ratio for thin walled cylinders?
Ann [662]

Answer:

The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:

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or using the diameter:

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3 0
3 years ago
What are the benefits of using the engineering design process? combining scientific knowledge and creativity in a rigid and stru
Trava [24]

Answer:

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Explanation:

7 0
3 years ago
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