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Blababa [14]
3 years ago
5

Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450 K, 350 kPa, and velocity of 3

m/s. At the exit, the temperature is 300 K and the velocity is 460 m/s. Using the ideal gas model for air with constant c_p = 1.011 kJ/kg*K, determine
(a) the area at the inlet, in m^2

(b) the heat transfer between the nozzle at its surroundings in kW.

(c) Specify whether the heat transfer is to or from the air.
Engineering
1 answer:
viktelen [127]3 years ago
4 0

Answer:

Given that

Mass flow rate ,m=2.3 kg/s

T₁=450 K

P₁=350 KPa

C₁=3 m/s

T₂=300 K

C₂=460 m/s

Cp=1.011 KJ/kg.k

For ideal gas

P V = m R T

P = ρ RT

\rho_1=\dfrac{P_1}{RT_1}

\rho_1=\dfrac{350}{0.287\times 450}

ρ₁=2.71 kg/m³

mass flow rate

m= ρ₁A₁C₁

2.3 = 2.71 x A₁ x 3

A₁=0.28 m²

Now from first law for open system

h_1+\dfrac{C_1^2}{200}+Q=h_2+\dfrac{C_2^2}{2000}

For ideal gas

Δh = CpΔT

by putting the values

1.011\times 450+\dfrac{3^2}{200}+Q=1.011\times 300+\dfrac{460^2}{2000}

Q=1.011\times 300+\dfrac{460^2}{2000}-\dfrac{3^2}{200}-1.011\times 450

Q= - 45.49 KJ/kg

Q =- m x 45.49 KW

Q= - 104.67 KW

Negative sign indicates that heat transfer from air to surrounding

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Iteru [2.4K]

Answer:

7,217*10^28 atoms/m^3

Explanation:

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The Avogadro's Number, 6,022*10^23, is the number of atoms in one mole of any substance. To calculate the number of atoms in one cubic meter of vanadium we write:

1m^3*(100^3 cm^3/1 m^3)*(6,1 g/1 cm^3)*(1 mol/50,9g)*(6,022*10^23 atoms/1 mol)=7,217*10^28 atoms

Therefore, for vanadium we have 7,217*10^28 atoms/m^3

6 0
3 years ago
A mass weighing 22 lb stretches a spring 4.5 in. The mass is also attached to a damper with Y coefficient . Determine the value
Dominik [7]

Answer:

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Explanation:

Given that

m = 22 lb

g= 32 ft/s²

m = \dfrac{22}{32}=0.6875\ s^2/ft

x= 4.5 in

1 in = 0.083 ft

x= 0.375 ft

Spring constant ,K

K=\dfrac{m}{x}=\dfrac{22}{0.375}

K= 58.66  lb/ft

The damper coefficient for critically damped system

C_c=2\sqrt{mK}

C_c=2\sqrt{0.6875\times 58.66}

Cc= 12.7 lb.sec/ft

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Answer:

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The two long are \rightarrow 122ft1in=122.08ft

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The height and the thickness are 14ft and 0.83ft respectively.

So we only calculate the Quantity of concrete,

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That in cubic yards is equal to 180.15 (1cy=27ft^3)

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8 0
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Answer:

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