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Ber [7]
3 years ago
11

Air enters a control volume operating at steady state at 1.2 bar, 300K, and leaves at 12 bar, 440K, witha volumetric flow rate o

f 1.3 m3/min. The work input to the control volume is 240 kJ per kg of air flowing. Neglecting kinetic and potential energy effects, determine the heat transfer rate, in kW.
Engineering
1 answer:
topjm [15]3 years ago
4 0

Answer:

Heat transfer = 2.617 Kw

Explanation:

Given:

T1 = 300 k

T2 = 440 k

h1 = 300.19 KJ/kg

h2 = 441.61 KJ/kg

Density = 1.225 kg/m²

Find:

Mass flow rate = 1.225 x [1.3/60]

Mass flow rate = 0.02654 kg/s

mh1 + mw = mh2 + Q

0.02654(300.19 + 240) = 0.02654(441.61) + Q

Q = 2.617 Kw

Heat transfer = 2.617 Kw

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BARSIC [14]

Answer:

vertical load = 10 kN

Modulus of elasticity = 200GPa

Yield stress on the cable = 400 MPa

Safety factor = 2.0

Explanation:

Data

let L = \sqrt{(1.5)^{2} + (1.5)^{2} }

        = 3.35 m

substituting 1.5 m for h and 3 m for the tern (a + b)

\theta = tan⁻¹(\frac{1.5}{1.5})

  = 45⁰

substituting 1.5 for h and 3 m for (a+ b) yields:

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   =25.56⁰

checking all the forces, they add up to zero. This means that the system is balanced and there is no resultant force.

6 0
3 years ago
The basic concept of feedback control is that an error must exist before some corrective action can be made?
weqwewe [10]

Answer:

The correct answer is True.

Explanation:

The feedback control system implies that to make a feedback, there must first be an error, otherwise there will be nothing to correct.

This system works so that there is an output that is controlled through a signal.

This signal will be feedback and it will signal an error which will be detected by a controller that will allow entry into the system.

In basic words, this system processes signals, samples them in the form of an output, and re-enters them feedback to detect the error signal.  

7 0
3 years ago
a steel rod of 30mm diameter is enclosed centerally in a hollow copper tube of external diameter 50mm and internal 40mm. the com
masha68 [24]

Answer:

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Explanation:

6 0
3 years ago
An inductor of ????=6.95 H with negligible resistance is placed in series with a ℰ=12.5 V battery, a ????=3.00 Ω resistor, and a
elena-14-01-66 [18.8K]

Answer:

a) I=0 b) 4.17V c) 0.354 A d) 14.5s

Explanation:

a) consider circuit in the attachment

i(t)= E/R (1- e^(-t/RL))

i(0)= 12.5/3×(1-e^(0/RL))

i(0)=0

b) at t⇒∞

i(∞)= 12.5/3× (1- e^(-∞/RL))

    = 4.17V

c) 1/RL= 1/(6.95×3)= 0.0479616

i(1.85) = 12.5/3 × (1- e^(-1.85×0.0479616)

         = 0.354A

d) I/2= I (1- e^(-t/RL))

t= - RL ln0.5

t= - 3×6.95 × (-0.693)

t= 14.5 s

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