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2 years ago

Pretty ako hjshjs ok.

Engineering

1 answer:

Vadim26 [7]2 years ago

5 0

Answer:

wala naman pong pic po wala kahit walang pic magan da parin

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Who plays blox burg???

xeze [42]

Answer:

I don't have robux

Explanation:

but i love adopt me

7 0

3 years ago

Read 2 more answers

11–17 A long, thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used t

lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

$v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2} }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2} }{4}) } =0.729m/s$

It is necessary to get the Reynold's number:

$Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343$

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

$Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517$

The overall heat transfer coefficient:

$Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} } }$

Here

$h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C$

Substituting values:

$Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278} } =1855.8923W/m^{2} C$

5 0

3 years ago

What unit of measurment would be used to measure current?

Alex_Xolod [135]

Answer:

The S. I unit of current is Amphere

5 0

3 years ago

Energy transfer in mechanical systems: During steady-state operation, a mechanical gearbox receives 70 KW of input power through

Degger [83]

Answer:

Heat transfer rate(Q)= 1.197kW

Power output(W)=68.803kW

3 0

3 years ago

For a turning operation, you have selected a high-speed steel (HSS) tool and turning a hot rolled free machining steel. Your dep

Alisiya [41]

Answer:

MRR = 1.984

Explanation:

Given that

Depth of cut ,d=0.105 in

Diameter D= 1 in

Speed V= 105 sfpm

feed f= 0.015 ipr

Now the metal removal rate given as

MRR= 12 f V d

d= depth of cut

V= Speed

f=Feed

MRR= Metal removal rate

By putting the values

MRR= 12 f V d

MRR = 12 x 0.015 x 105 x 0.105

MRR = 1.984

Therefore answer is -

1.944

8 0

2 years ago