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AnnyKZ [126]
3 years ago
13

Two streams of air enter a control volume: stream 1 enters at a rate of 0.05 kg / s at 300 kPa and 380 K, while stream 2 enters

at 400 kPa and 300 K. Stream 3 leaves the control volume at 150 kPa and 270 K. The control volume does 3 kW of work on the surroundings while losing 5 kW of heat. Find the mass flow rate of stream 2. Neglect changes in kinetic and potential energy.

Engineering
1 answer:
alex41 [277]3 years ago
5 0

Answer:

0.08kg/s

Explanation:

For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.

The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.

 

finally you use the two previous equations to make a system and find the mass flows

I attached procedure

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A manometer consists of an inclined glass tube which communicates with a metal cylinder standing upright; liquid fills the appar
Vanyuwa [196]

Answer:

48.61

Explanation:

See attached diagram.

The level rise in the tube is l sin α.

The level drop in the cylinder (let's call it y) is:

π/4 D² y = π/4 d² l

D² y = d² l

y = l (d/D)²

The elevation difference is the sum:

l sin α + l (d/D)²

l (sin α + (d/D)²)

From Bernoulli's principle:

P = ρgl (sin α + (d/D)²)

Divide both sides by density of water (ρw) and gravity:

P/(ρw g) = (ρ/ρw) l (sin α + (d/D)²)

h = S l (sin α + (d/D)²)

If we disregard the level change in the cylinder:

h = S l (sin α)

We want the percent error between these two expressions for h to be 0.1% when α = 25°.

[ S l (sin α + (d/D)²) − S l (sin α) ] / [ S l (sin α + (d/D)²) ] = 0.001

[ S l sin α + S l (d/D)² − S l sin α ] / [ S l (sin α + (d/D)²) ] = 0.001

[ S l (d/D)² ] / [ S l (sin α + (d/D)²)]  = 0.001

(d/D)² / (sin α + (d/D)²) = 0.001

(d/D)² = 0.001 (sin α + (d/D)²)

(d/D)² = 0.001 sin α + 0.001 (d/D)²

0.999 (d/D)² = 0.001 sin α

d/D = √(0.001 sin α / 0.999)

When α = 25°:

d/D ≈ 0.02057

D/d ≈ 48.61

7 0
3 years ago
What would be the most likely scale factor to use for an n-gauge model train setup? (An n-gauge layout uses locomotives that are
slega [8]
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3 years ago
Concepto gestión del tiempo?
Arturiano [62]

Answer:

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Explanation:

it says time management concept.

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3 years ago
A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled
Vikentia [17]

Answer:

Force per unit plate area is 0.1344 N/m^{2}

Solution:

As per the question:

The spacing between each wall and the plate, d = 10 mm = 0.01 m

Absolute viscosity of the liquid, \mu =1.92\times 10^{- 3} Pa-s

Speed, v = 35 mm/s = 0.035 m/s

Now,

Suppose the drag force that exist between each wall and plate is F and F' respectively:

Net Drag Force = F' + F''

F = \tau A

where

\tau = shear stress

A = Cross - sectional Area

Therefore,

Net Drag Force, F = (\tau ' +\tau '')A

\frac{F}{A} = \tau ' +\tau ''

Also

F = \frac{\mu v}{d}

where

\mu = dynamic coefficient of viscosity

Pressure, P = \frac{F}{A}

Therefore,

\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}

\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}

8 0
3 years ago
Q1. Basic calculation of the First law (2’) (a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of hea
Bogdan [553]

Answer:

(a) ΔU = 125 kJ

(b) ΔU = -110 kJ

Explanation:

<em>(a) Suppose that 150 kJ of work are used to compress a spring, and that 25 kJ of heat are given off by the spring during this compression. What is the change in internal energy of the spring during the process?</em>

<em />

The work is done to the system so w = 150 kJ.

The heat is released by the system so q = -25 kJ.

The change in internal energy (ΔU) is:

ΔU = q + w

ΔU = -25 kJ + 150 kJ = 125 kJ

<em>(b) Suppose that 100 kJ of work is done by a motor, but it also gives off 10 kJ of heat while carrying out this work. What is the change in internal energy of the motor during the process?</em>

<em />

The work is done by the system so w = -100 kJ.

The heat is released by the system so q = -10 kJ.

The change in internal energy (ΔU) is:

ΔU = q + w

ΔU = -10 kJ - 100 kJ = -110 kJ

4 0
3 years ago
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