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3 years ago

7

Which option best describes a way engineers can provide maintenance, diagnoses, upgrades, or duplicates even for products they d

idn't design?

1 answer:

guajiro [1.7K]3 years ago

4 0

Answer:

Reverse Engineering

Explanation:

Just took the test

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For a fluid flowing through a pipe assuming that pressure drop per unit length of pipe (P/L) depends on the diameter of the pipe

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A single-cylinder pump feeds a boiler through a delivery

Studentka2010 [4]

Answer:

Net discharge per hour will be 3.5325 $m^3/hr$

Explanation:

We have given internal diameter d = 25 mm

Time = 1 hour = 3600 sec

So radius $r=\frac{d}{2}=\frac{25}{2}=12.5mm=12.5\times 10^{-3}m$

We know that area is given by

$A=\pi r^2=3.14\times (12.5\times 10^{-3})^2=490.625\times 10^{-6}m^2$

We know that discharge is given by $Q=AV$ , here A is area and V is velocity

So $Q=AV=490.625\times 10^{-6}\times 2=981.25\times 10^{-6}m^3/sec$

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8 0

3 years ago

(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5

bezimeni [28]

Answer:

a.) 1.453MW/m2, b.) 2,477,933.33 BTU/hr c.) 22,733.33 BTU/hr d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, $Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\$

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, $Q = \frac{109*0.5*100}{0.0075} = 726,666.667W$

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, $Q=\frac{1*0.5*100}{0.0075} = 6,666.67W$

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, $Q=\frac{109*0.5*100}{0.015} =363,333.33W$

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0

3 years ago