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1 year ago

Please help me with this question

Engineering

1 answer:

Nat2105 [25]1 year ago

5 0

Answer:

The answer is c and the teacher helped me

Explanation:

i had help from the tescger and the assignment is done

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A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125

Gekata [30.6K]

Answer:

a) $\dot m_{3} = 135\,\frac{lbm}{s}$ , b) $h_{3}=168.965\,\frac{BTU}{lbm}$ , c) $T = 200.829\,^{\textdegree}F$

Explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

$\dot m_{1} + \dot m_{2} - \dot m_{3} = 0$

The mass flow rate exiting the tank is:

$\dot m_{3} = \dot m_{1} + \dot m_{2}$

$\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}$

$\dot m_{3} = 135\,\frac{lbm}{s}$

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

$\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0$

$h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}$

Properties of water are obtained from tables:

$h_{1}=180.16\,\frac{BTU}{lbm}$

$h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)$

$h_{2}=29.032\,\frac{BTU}{lbm}$

The specific enthalpy at outlet is:

$h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }$

$h_{3}=168.965\,\frac{BTU}{lbm}$

c) After a quick interpolation from data availables on water tables, the final temperature is:

$T = 200.829\,^{\textdegree}F$

8 0

2 years ago

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La iluminación de la superficie de un patio amplio es 1600 lx cuando el ángulo de elevación del sol 53°. Calcular la iluminación

gregori [183]

Answer:

I = 1205.69 Lx

Explanation:

The irradiation or intensity of the solar radiation on the earth is maximum for the vertical fire, with a value I₀

I = I₀ sin θ

in this case with the initial data we can calculate the initial irradiance

I₀ = $\frac{I}{sin \ \theta }$

I₀ = 1600 /sin 53

I₀ = 2003.42 lx

for when the angle is θ = 37º

I = 2003.42 sin 37

I = 1205.69 Lx

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2 years ago

Air enters a tank through an area of 0.2 ft2 with a velocity of 15 ft/s and a density of 0.03 slug/ft3. Air leaves with a veloci

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Answer:

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NeTakaya

Answer:

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2 years ago

Which type of engineering design uses an already-existing design?

Arte-miy333 [17]

The answer is D I’m 90% sure

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