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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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The local atmospheric pressure is measured with a water barometer. If the water column is measured to be 30 ft, what is the atmo
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Answer:

The atmospheric pressure in atm=0.885 atm

Explanation:

Given that

Local pressure (h)= 30 ft of water height     ( 1 ft= 0.3048 m)

We know that pressure in given by

  P=ρgh

We know that ρ of water is 1000\dfrac{kg}{m^3}

So pressure

P=1000(9.81)(9.144)

P=89.7026 \dfrac{N}{m^2}

We know that 1000 Pa=0.00986 atm

So P=0.885 atm

The atmospheric pressure in atm=0.885 atm

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Using the characteristics equation, determine the dynamic behavior of a PI controller with τI = 4 applied to a second order proc
Sladkaya [172]

Answer:

The values of Kc that render this closed-loop process unstable are in the interval

(Kc < 0)

Explanation:

The transfer function of a PI controller is given as

Gc = Kc {1 + (1/sτI)}

τI = 4

Gc = Kc {1 + (1/4s)}

Gc = Kc {(4s+1)/(4s)}

Divide numerator and denominator by 4

Gc = Kc {(s+0.25)/(s)}

For a second order process, the general transfer function is given by

Gp = Kp {1/(τn²s² + 2ζτns + 1)}

Kp = 2, τn = 5 and ζ = 1.5

Gp = 2/(25s² + 15s + 1)

Divide numerator and denominator by 25

Gp = 0.08/(s² + 0.6s + 0.04)

Ga = 1

Gs = 1

We need to find the value(s) of Kc that makes the closed loop transfer function unstable. Gp*Ga*Gc*Gs + 1 = 0

The closed loop transfer function is unstable when the solution(s) of the characteristic equation obtained is positive.

Gp*Ga*Gc*Gs + 1 = 0

Becomes

[0.08/(s² + 0.6s + 0.04)] × [Kc (s+0.25)/(s)] + 1 = 0

[0.08Kc (s + 0.25)/(s³ + 0.6s² + 0.04s)] = - 1

0.08Kc (s + 0.25) = -s³ - 0.6s² - 0.04s

0.08Kc s + 0.02Kc = -s³ - 0.6s² - 0.04s

s³ + 0.6s² + 0.04s + 0.08Kc s + 0.02Kc = 0

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

We will use the direct substitution method to evaluate the values of Kc that matter. The values of Kc at the turning points of the closed loop transfer function.

For the substitution,

We put s = jw into the equation. (frequency analysis)

Note that j = √(-1)

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

(jw)³ + 0.6(jw)² + (0.08Kc + 0.04)(jw) + 0.02Kc = 0

-jw³ - 0.6w² + (0.08Kc + 0.04)(jw) + 0.02Kc = 0

we then collect terms with j and terms without.

(0.08Kcw + 0.04w - w³)j + (0.02Kc - 0.6w²) = 0

Meaning,

0.08Kcw + 0.04w - w³ = 0 (eqn 1)

0.02Kc - 0.6w² = 0 (eqn 2)

0.02 Kc = 0.6 w²

Kc = 15w²

Substituting this into eqn 1

0.08Kcw + 0.04w - w³ = 0

Kc = 15w²

0.08(15w²)w + 0.04w - w³ = 0

1.2w³ + 0.04w - w³ = 0

0.2w³ + 0.04w = 0

w = 0 or 0.2w² + 0.04 = 0

0.2w² = -0.04

w² = -0.2

w = ± √(-0.2)

w = ± 0.4472j or w = 0

Recall, Kc = 15w² = 15(-0.2) = -3 or Kc = 0

The turning points for the curve of the closed loop transfer function occur when

Kc = 0 or Kc = -3

To investigate, we pick values around these turning points to investigate the behaviour of the closed loop transfer function at those points.

Kc < -3, Kc = -3, (-3 < Kc < 0), Kc = 0 and Kc > 0

Note that, one positive characteristic root or pole is enough to make the system unstable.

We pick a value for Kc in that interval and evaluate the closed loop transfer function.

s³ + 0.6s² + (0.08Kc + 0.04)s + 0.02Kc = 0

- First of, let Kc = - 4 (Kc < -3)

s³ + 0.6s² - 0.28s - 0.08 = 0

Solving the polynomial

s = (-0.22002), 0.44223, (-0.82221)

One positive pole means the closed loop transfer function is unstable in this region

Let Kc = -3

s³ + 0.6s² - 0.20s - 0.06 = 0

s = 0.37183, (-0.21251) or (-0.75933)

One positive pole still means that the closed loop transfer function is still unstable.

Then the next interval

Let Kc = -1

s³ + 0.6s² - 0.04s - 0.02 = 0

Solving this polynomial,

s = 0.18686, (-0.1749) or (-0.61196)

The function is unstable in the region being investigated.

Let Kc = 0

s³ + 0.6s² + 0.04s = 0

s = 0, -0.0769, -0.5236

One zero, all negative roots, indicate that the closed loop transfer function is marginally stable at this point.

Let Kc = 1, Kc > 0

s³ + 0.6s² + 0.12s + 0.02 = 0

s = (-0.42894), (-0.08553 + 0.1983j) or (-0.08553 - 0.1983j)

All the real negative parts of the poles are all negative, this indicates stability.

Hence, after examining the turning points of the closed loop transfer function, it is evident that, the region's of Kc where the closed loop transfer function is unstable is (Kc < 0)

Hope this Helps!!!

8 0
3 years ago
F both wires in the twisted pair CANbus need to be repaired, it is best to:
Ne4ueva [31]

Answer:

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3 0
3 years ago
A 2 m3 rigid tank initially contains air at 100 kPa and 22 degrees C. The tank is connected to a supply line through a valve. Ai
Vaselesa [24]

Answer:

a. 9.58kgs b. 340.32KJ

Explanation:

Volume of tank= 2m³

Initial Pressure Pi= 100KPa

Initial Temperature Ti= 22 C= 295K

Line Pressure P₁= 600 KPa

Line Temperature T= 22 C= 295K

Final Pressure P2= 600 KPa

Final Temperature T2= 77 C= 350K

Use Ideal Gas Equation

PV= mRT

P₁V₁= m₁RT₁

m₁= (100 x 2)/(0.287 x 295) = 2.3622kg

P₂V₂= m₂RT₂

m₂= (600 x 2)/(0.287 x 350) = 11.946 kg

Since valve is closed and no mass leave

m₁ + mi = m₂ + me

as per above condition me= 0

mi= m₂ - m₁ = 11.946 - 2.3622 = 9.5838kg

Applying energy equation

m₁u₁ + mihi + Q = m₂u₂ + mehe + W

me and W=0

m₁u₁ + mihi + Q = m₂u₂

m₁CvT₁ + miCpTi + Q =  m₂CvT₂

Q =   m₂CvT₂- m₁CvT₁ - miCpTi

Q = (11.946 x 0.717 350) - (2.3622 x 0.717 x 295) - (9.5868 x 1.004 x 295)

Q = -340.321 KJ (Negative sign doesn't matter as energy is not a vector quantity)

4 0
3 years ago
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