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Alja [10]
3 years ago
10

If the old radiator is replaced with a new one that has longer tubes made of the same material and same thickness as those in th

e old unit, what should the total surface area available for heat exchange be in the new radiator to achieve the desired cooling temperature gradient
Engineering
1 answer:
Nookie1986 [14]3 years ago
3 0

Answer: hello some parts of your question is missing attached below is the missing information

The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d  is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube

answer : Total surface area = 3/2 * area of old radiator

Explanation:

we will use this relation

K = \frac{Qd }{A* change in T }

change in T =  ΔT  

therefore New Area  ( A ) = 3/2 * area of old radiator

Given that the thermal conductivity is the same in the new and old radiators

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A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor ack
Elza [17]

Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met  the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

3 0
3 years ago
Out
olchik [2.2K]

ave you ever seen a Rube Goldberg machine in action? You probably have, even if you didn’t know what it was. A Rube Goldberg machine is a contraption that uses a chain reaction to carry out a simple task. It performs a very basic job in a complicated way.

7 0
3 years ago
Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 5
Shkiper50 [21]

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

3 0
3 years ago
How does a 2.5 MW wind turbine costing $ 4 million compare to a 5-kw wind turbine $3 /W? a) Same $/w b) Smaller $/w c) Larger $/
My name is Ann [436]
MW means megawatt, and one megawatt is a million Watts.
The 2.5 MW turbine is 4/2.5=1.6 $/w
Answer B
4 0
3 years ago
Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at 10 kPa with a quality of 90 percent. Neglecting the changes
Anna35 [415]

Answer:

The mass flow rate of steam m=5.4 Kg/s

Explanation:

Given:

  At the inlet of turbine P=10 MPa  ,T=500 C

 AT the exit of turbine  P=10 KPa   ,x=0.9

 Required power=5 MW

From steam table

<u> At 10 MPa and 500 C:</u>

  h=3374 KJ/Kg  ,s=6.59 KJ/Kg-K  (Super heated steam table)

<u>At 10 KPa:</u>

h_g=2675.1 KJ/Kg, h_f=417.51  KJ/Kg

s_g= 7.3  KJ/Kg-K                ,s_f=1.3   KJ/Kg-K

So enthalpy of steam at the exit of turbine

h= h_f+x(h_g- h_f)

Now by putting the values

h= 417.51+0.9(2675.1- 417.51) KJ/Kg

h=2449.34  KJ/Kg

Lets take m is the mass flow rate of steam

So 5\times 10^3=m\times (3374-2449.34)

m=5.4 Kg/s

So the mass flow rate of steam m=5.4 Kg/s

8 0
3 years ago
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