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3 years ago

When do design engineers start on the design improvement step?

Engineering

1 answer:

ArbitrLikvidat [17]3 years ago

8 0

Answer:

as soon as there is a design to improve

Explanation:

As a design engineer, I started on the "design improvement" step as soon as I had an initial conceptual design.

Then, I started that step again when my boss told me, "make it better."

_____

The more interesting question is, "when do you stop the design improvement step?" (Judging by the constant barrage of software updates, that answer is, "never.")

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2 years ago

Steam at 1 MPa, 300 C flows through a 30 cm diameter pipe with an average velocity of 10 m/s. The mass flow rate of this steam i

stealth61 [152]

Answer:

$\dot m = 2.74 kg/s$

Explanation:

given data:

pressure 1 MPa

diameter of pipe = 30 cm

average velocity = 10 m/s

area of pipe $= \frac[\pi}{4}d^2$

$= \frac{\pi}{4} 0.3^2$

A = 0.070 m2

WE KNOW THAT mass flow rate is given as

$\dot m = \rho A v$

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therefore we have

$\dot m = 4.068 * 0.070* 10$

$\dot m = 2.74 kg/s$

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3 years ago

The metal control joints used to relieve stresses caused by expansion and contraction in large ceiling or wall expenses in inter

timurjin [86]

Answer:

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2 years ago

Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th

Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine = $1 - \frac{Tl}{Th} \\$

where Th =temperature at which the heat enters the engine

Tl is the temperature of the environment

since both engines have the same thermal capacities  $n_{th}$  therefore $n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\$

We have now that

$\frac{-1400}{T}+\frac{T}{300}=0\\$

multiplying through by T

$-1400 + \frac{T^{2} }{300}=0\\$

multiplying through by 300

- $420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000} \\T=648.07k$

The temperature T= 648.07k

5 0

3 years ago

Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee

Juli2301 [7.4K]

Answer:

a) 1 m^3/Kg

b) 504 kJ

c) 514 kJ

Explanation:

Given

-The mass of C_o2 = 1 kg

-The volume of the tank V_tank = 1 m^3

-The added energy E = 14 W

-The time of adding energy t = 10 s

-The increase in specific internal energy Δu = +10 kJ/kg

-The change in kinetic energy ΔKE = 0 and The change in potential energy

ΔPE =0

Required

(a)Specific volume at the final state v_2

(b)The energy transferred by the work W in kJ.

(c)The energy transferred by the heat transfer W in kJ and the direction of

the heat transfer.

Assumption

-Quasi-equilibrium process.

Solution

(a) The volume and the mass doesn't change then, the specific volume is constant.

v= V_tank/m ---> 1/1= 1 m^3/Kg

(b) The added work is defined by.

W =E * t ---> 14 x 10 x 3600 x 10^-3 = 504 kJ

(c) From the first law of thermodynamics.

Q - W = m * Δu

Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ

The heat have (+) sign the n it is added to the system.

7 0

3 years ago