Answer pleaseeeee quick

Air at 1 atm enters a thin-walled ( 5-mm diameter) long tube ( 2 m) at an inlet temperature of 100°C. A constant heat flux is ap

alexdok [17]

Answer:

heat rate = 7.38 W

Explanation:

Given Data:

Pressure = 1atm

diameter (D) = 5mm = 0.005m

length = 2

mass flow rate (m) = 140*10^-6 kg/s

Exit temperature = 160°C,

At 400K,

Dynamic viscosity (μ) = 22.87 *10^-6

Prandtl number (pr) = 0.688

Thermal conductivity (k) = 33.65 *10^-3 W/m-k

Specific heat (Cp) = 1.013kj/kg.K

Step 1: Calculating Reynolds number using the formula;

Re = 4m/πDμ

= (4*140*10^-6)/(π* 0.005*22.87 *10^-6)

= 5.6*10^-4/3.59*10^-7

= 1559.

Step 2: Calculating the thermal entry length using the formula

Le = 0.05*Re*Pr*D

Substituting, we have

Le = 0.05 * 1559 * 0.688 *0.005

Le = 0.268

Step 3: Calculate the heat transfer coefficient using the formula;

Nu = hD/k

h = Nu*k/D

Since Le is less than given length, Nusselt number (Nu) for fully developed flow and uniform surface heat flux is 4.36.

h = 4.36 * 33.65 *10^-3/0.005

h = 0.1467/0.005

h = 29.34 W/m²-k

Step 4: Calculating the surface area using the formula;

A = πDl

=π * 0.005 * 2

=0.0314 m²

Step 5: Calculating the temperature Tm

For energy balance,

Qc = Qh

Therefore,

H*A(Te-Tm) = MCp(Tm - Ti)

29.34* 0.0314(160-Tm) = 140 × 10-6* 1.013*10^3 (Tm-100)

0.921(160-Tm) = 0.14182(Tm-100)

147.36 -0.921Tm = 0.14182Tm - 14.182

1.06282Tm = 161.542

Tm = 161.542/1.06282

Tm = 151.99 K

Step 6: Calculate the rate of heat transferred using the formula

Q = H*A(Te-Tm)

= 29.34* 0.0314(160-151.99)

= 7.38 W

the Prandtl number using the formula

5 0

3 years ago

Liquid ethanol (C2H5OH) at 298 K, 1 atm enters a steam generator operating at steady state and burns completely with dry air ent

Gennadij [26K]

Answer:

Q= 6491.100 kJ/s

Explanation:

Air-Fuel Ratio:

For a combustion reaction the proportion of air that is present in a gaseous substance responsible for the reaction,this proportion is known as air-fuel ratio.The air fuel ratio is calculated using the combustion reaction for the substance.

Considering reaction for the Ethanol as

C₂H₅OH +XO₂(O₂+3.76N₂)→ aCO₂+bH₂O+cN₂

Balancing the equation we get;

a=2,

2b=6

∴ b=3

xO₂=3

The air-fuel ratio

A/F = XO₂+H₂O+xN₂× mass of N₂/mass(fuel)

3.31×31.9+11.28×28.013/46.069

= 8.943

Equivalent ratio = 0.7,

so, heat transfer

Q= m ×Cp×ΔT

= 75×0.7×112.4(1500-400)

Q= 6491.100 kJ/s

1kJ/s=1000w

∴ Q= 6491100 W

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6 0

4 years ago

Knowing that the central portion of the link BD has a uniform cross-sectional area of 800 2, determine the magnitude of the load

IceJOKER [234]

Answer: 50

Explanation:

7 0

3 years ago

Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35 MPa m . It has bee

GaryK [48]

Answer: Fracture will not occur since Kc (32.2 MPa√m) ∠ KIc (35 MPa√m).

Explanation:

in this question we are asked to determine if an aircraft will fracture for a given fracture toughness.

let us begin,

from the question we have that;

stress = 325 MPa

fracture toughness (KIc) = 35 MPa√m

the max internal crack length = 1.0 m

using the formula;

Y = KIc/σ√(πα) ---------------(1)

solving for Y we have;

Y = 35 (MPa√m) / 250 (MPa) √(π × 2×10⁻3/2m)

Y = 2.50

so to calculate the fracture roughness;

Kc = Y × σ√(πα) = 2.5 × 3.25√(π × 1×10⁻³/2) = 32.2 MPa√m

Kc = 32.2 MPa√m

From our results we can say that fracture will not occur since Kc (32.2 MPa√m) is less than KIc (35 MPa√m) of the material.

cheers i hope this helps!!!!

8 0

3 years ago

The clepsydra, or water clock, was a device that the ancient Egyptians, Greeks, Romans, and Chinese used to measure the passage

Nookie1986 [14]

Suppose a tank is made of glass and has the shape of a right-circular cylinder of radius 1 ft. Assume that h(0) = 2 ft corresponds to water filled to the top of the tank, a hole in the bottom is circular with radius in., g = 32 ft/s2, and c = 0.6. Use the differential equation in Problem 12 to find the height h(t) of the water.

Answer:

Height of the water = √(128)/147456 ft

Explanation:

Given

Radius, r = 1 ft

Height, h = 2 ft

Radius of hole = 1/32in

Acceleration of gravity, g = 32ft/s²

c = 0.6

Area of the hold = πr²

A = π(1/32)² ---- Convert to feet

A = π(1/32 * 1/12)²

A = π/147456 ft²

Area of water = πr²

A = π 1²

A = π

The differential equation is;

dh/dt = -A1/A2 √2gh where A1 = Area of the hole and A2 = Area of water

A1 = π/147456, A2 = π

dh/dt = (π/147456)/π √(2*32*2)

dh/dt = 1/147456 * √128

dh/dt = √128/147456 ft

Height of the water = √(128)/147456 ft

3 0

4 years ago

Answer pleaseeeee quick​

Answer pleaseeeee quick