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FinnZ [79.3K]
4 years ago
15

To cool a summer home without using a vapor compression refrigeration cycle, air is routed through a plastic pipe (k=0.15 W/m*K,

D_inner=0.15m, D_outer=0.17m) that is submerged in an adjoining body of water. the water temperature is nominally at T_infinity=17 degrees Celsius and has a convection coefficient of h=1500W/m^2*K is maintained at the outer surface of the pipe. If air from the home enters at T_m,i=29 degrees Celsius and a volumetric flow rate of V_i=0.025 m^3/s, what pipe length, L, is needed to provide a discharge temperature of T_m,o=21 degrees Celsius?

Engineering
1 answer:
Shalnov [3]4 years ago
8 0

Answer:

See attachment for complete solving

Explanation:

Given that: Brainly.com

What is your question?

mkasblog

College Engineering 5+3 pts

To cool a summer home without using a vapor compression refrigeration cycle, air is routed through a plastic pipe (k=0.15 W/m*K, D_inner=0.15m, D_outer=0.17m) that is submerged in an adjoining body of water. the water temperature is nominally at T_infinity=17 degrees Celsius and has a convection coefficient of h=1500W/m^2*K is maintained at the outer surface of the pipe. If air from the home enters at T_m,i=29 degrees Celsius and a volumetric flow rate of V_i=0.025 m^3/s, what pipe length, L, is needed to provide a discharge temperature of T_m,o=21 degrees Celsius

See attachment fir completed solving

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) A flow is divided into two branches, with the pipe diameter and length the same for each branch. A 1/4-open gate valve is inst
Digiron [165]

Answer:

Va / Vb = 0.5934

Explanation:

First step is to determine total head losses at each pipe

at Pipe A

For 1/4 open gate valve head loss = 17 *Va^2 / 2g

elbow loss = 0.75 Va^2 / 2g

at Pipe B

For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g

elbow loss = 0.75 * Vb^2 / 2g

Given that both pipes are parallel

17 *Va^2/2g +  0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g  + 0.75 * Vb^2 / 2g

∴ Va / Vb = 0.5934

8 0
3 years ago
1. A copper block of volume 1 L is heat treated at 500ºC and now cooled in a 200-L oil bath initially at 20◦C. Assuming no heat
MaRussiya [10]

Answer:

final temperature T = 24.84ºC

Explanation:

given data

copper volume = 1 L

temperature t1 = 500ºC

oil volume = 200 L

temperature t2 = 20ºC

solution

Density of copper \rho cu = 8940 Kg/m³

Density of light oil  \rho oil = 889 Kg/m³

Specific heat capacity of copper Cv = 0.384  KJ/Kg.K

Specific heat capacity of light oil Cv = 1.880 KJ/kg.K

so fist we get here mass of oil and copper that is

mass = density × volume   ................1

mass of copper = 8940 × 1 ×  10^{-3}  = 8.94 kg  

mass of oil = 889 × 200 × 10^{-3}  =  177.8 kg  

so we apply here now energy balance equation that is

M(cu)\times Cv \times (T-T1)_{cu} + M(oil) \times Cv\times (T-T2)_{oil}  = 0

put here value and we get T2

8.94\times 0.384 \times (T-500) + 177.8 \times 1.890\times (T-20)  = 0

solve it we get

T = 24.84ºC

7 0
3 years ago
Two solid yellow center lines on a two-lane highway indicate:
Vladimir79 [104]
That you can pass the person in front of you while using the other lane
4 0
2 years ago
Read 2 more answers
What are practical considerations you might encounter when you increase the moment of inertia (I) while keeping the cross-sectio
Mrrafil [7]

Answer:

The answer is below

Explanation:

The practical considerations you might encounter when you increase the moment of inertia (I) while keeping the cross-sectional area fixed are:

1. Shapes of moment of inertia: Engineers should consider or know the different shapes of moment of inertia for different shape

2. Understanding the orientation of the beam: this will allow engineers to either increase or decrease the moment of inertia of a beam without increasing its cross sectional area.

4 0
3 years ago
A battery with an f.e.m. of 12 V and negligible internal resistance is connected to a resistor of 545 How much energy is dissipa
Alenkasestr [34]

Answer:

When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then  

R

1

 in Figure 1(a) could be the resistance of the screwdriver’s shaft,  

R

2

 the resistance of its handle,  

R

3

 the person’s body resistance, and  

R

4

 the resistance of her shoes.

Figure 2 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)

Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.

Figure 2. Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).

To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2.

According to Ohm’s law, the voltage drop,  

V

, across a resistor when a current flows through it is calculated using the equation  

V

=

I

R

, where  

I

 equals the current in amps (A) and  

R

 is the resistance in ohms  

(

Ω

)

. Another way to think of this is that  

V

 is the voltage necessary to make a current  

I

 flow through a resistance  

R

.

So the voltage drop across  

R

1

 is  

V

1

=

I

R

1

, that across  

R

2

 is  

V

2

=

I

R

2

, and that across  

R

3

 is  

V

3

=

I

R

3

. The sum of these voltages equals the voltage output of the source; that is,

V

=

V

1

+

V

2

+

V

3

.

 

This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation  

P

E

=

q

V

, where  

q

 is the electric charge and  

V

 is the voltage. Thus the energy supplied by the source is  

q

V

, while that dissipated by the resistors is

q

V

1

+

q

V

2

+

q

V

3

.

Explanation:

8 0
3 years ago
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