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2 years ago

Refrigerant-134a is to be cooled by water in a condenser. The refrigerant enters

Engineering

1 answer:

KonstantinChe [14]2 years ago

4 0

Answer:

Refrigerant R-134a is to be cooled by waterin a condenser.The refrigerant enters thecondenser with a mass flow rate of 6 kg/minat 1 MPa and 70 C and exits at 35 C. The cool-ing water enters at 300 kPa and 15 C andleaves at 25 C. Neglecting pressure drops,determine a) the required mass flow rate ofthe cooling water, and b) the heat transferrate from the refrigerant to the water.SolutionFirst consider the condenser as the control volume. The process is steady,adiabatic and no work is done. Thus over any time intervalΔt,ΔEΔt=0and thusXin˙E=Xout˙Ewhere˙E=˙m h+12V2+gz650:351 Thermodynamics·Prof. Doyle Knight37

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The y component of velocity in a steady, incompressible flow field in the xy plane is v = -Bxy3, where B = 0.4 m-3 · s-1, and x

love history [14]

Answer:

attached below

Explanation:

3 0

2 years ago

The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm

aksik [14]

Answer:

Tmax=14.5MPa

Tmin=10.3MPa

Explanation:

T = 600 * 0.15 = 90N.m

$T_max =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0175}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}$

=14.5MPa

$T_{min} =\frac{T_c}{j} = \frac{x}{y} = \frac{90 \times 0.0125}{\frac{\pi}{2} \times (0.0175^4-0.0125^4)}$

=10.3MPa

7 0

2 years ago

g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,

irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0

2 years ago

A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at

DiKsa [7]

Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

$dQ = m \times c_p \times (T_2 -T_1)$

For ideal gas, $c_p$ = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

3 0

2 years ago

What Happens If A Sonic Boom Is Created?

Volgvan

Answer:

Explanation:

A sonic boom is a loud sound kind of like an explosion. It's caused by shock waves created by any object that travels through the air faster than the speed of sound. Sonic booms create huge amounts of sound energy. When an object moves through the air, it makes pressure waves in front of and behind it.

8 0

2 years ago