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4 years ago

11

True or False: In a technical drawing dimensions are optional. The manufacturer can decide on their own how big or small to make

a design.

1 answer:

Law Incorporation [45]4 years ago

4 0

Answer:

False

Explanation:

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I need unseen passage

monitta

Answer:

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8 0

3 years ago

Read 2 more answers

A 450 MWt combined cycle plant has a Brayton cycle efficiency of 24% and a Rankine cycle efficiency of 29% with no heat augmenta

Romashka-Z-Leto [24]

Answer:

Output of plant=207.18 MW

Explanation:

Given input energy Q=450 MW

Brayton cycle efficiency $\eta_1=24$ %

Rankine cycle efficiency $\eta_2=29$ %

Combined efficiency for two cycle given as follow

$\eta _b=\eta _1+\eta _2-\eta _1\eta _2$

$\eta _b=0.24+0.29-0.24\times 0.29$

$\eta_b=0.4604$

We know that efficiency $\eta =\dfrac{out\ put}{in\ put}$

out put of plant= $450\times 0.4604$

So output of plant=207.18 MW

4 0

3 years ago

The pads are 200mm long, 150 mm wide and thickness equal to 12mm. 1- Determine the average shear strain in the rubber if the for

lord [1]

Answer:

a) 0.3

b) 3.6 mm

Explanation:

Given

Length of the pads, l = 200 mm = 0.2 m

Width of the pads, b = 150 mm = 0.15 m

Thickness of the pads, t = 12 mm = 0.012 m

Force on the rubber, P = 15 kN

Shear modulus on the rubber, G = 830 GPa

The average shear strain can be gotten by

τ(average) = (P/2) / bl

τ(average) = (15/2) / (0.15 * 0.2)

τ(average) = 7.5 / 0.03

τ(average) = 250 kPa

γ(average) = τ(average) / G

γ(average) = 250 kPa / 830 kPa

γ(average) = 0.3

horizontal displacement,

δ = γ(average) * t

δ = 0.3 * 12

δ = 3.6 mm

5 0

4 years ago

A tensile test was made on a tensile specimen, with a cylindrical gage section which had a diameter of 10 mm, and a length of 40

tamaranim1 [39]

Answer:

The answers are as follow:

a) 10 mm

b) 12.730 N/ $mm^{2}$

c) 127.307 N/ $mm^{2}$

d) 0.25

Explanation:

d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9

Force = F =1000 N

let us find initial area first, A1 = pi* $r^{2}$ = 78.55 $mm^{2}$

using reduction in area formula : 0.9 = (A1 - A2 ) / A1

solving it will give, A2 = 0.1 A1 = 7.855 $mm^{2}$

a) The specimen elongation is final length - initial length

50 - 40 = 10 mm

b) Engineering stress uses the original area for all stress calculations,

Engineering stress = force / original area = F / A1 = 1000 / 78.55

Engineering stress = 12.730 $N / mm^{2}$

c) True stress uses instantaneous area during stress calculations,

True fracture stress = force / final area = F / A2 = 1000 / 7.855

True Fracture stress = 127.30 $N / mm^{2}$

e) strain = change in length / original length

strain = 10 / 40 = 0.25

8 0

3 years ago

Refrigerant 134a vapor in a piston-cylinder assembly undergoes a process at constant pressure from an initial state at 8 bar and

jonny [76]

Answer:

- Work done is 2.39 kJ

- heat transfer is 20.23 kJ/kg

Explanation:

Given the data in the question;

First we obtain for specific volumes and specific enthalpy from "Table Properties Refrigerant 134a;

Specific Volume v₁ = 0.02547 m³/kg

Specific enthalpy u₁ = 243.78 kJ/kg

Specific Volume V₂ = 0.02846 m³/kg

Specific enthalpy u₂ = 261.62 kJ/kg

p = 8 bar = 800 kPa

Any changes in kinetic and potential energy are negligible.

So we determine the work done by using the equation at constant pressure

]Work done W = p( v₂ - v₁ )

we substitute

W = 800 kPa( 0.02846 m³/kg - 0.02547 m³/kg )

W = 800 kPa( 0.00299 m³/kg )

W = 2.39 kJ

Therefore, Work done is 2.39 kJ

Heat transfer;

using equation at constant pressure

Heat transfer Q = W + ( u₂ - u₁ )

so we substitute

Q = 2.392 kJ + ( 261.62 kJ/kg - 243.78 kJ/kg )

Q = 2.392 kJ + 17.84 kJ/kg )

Q = 20.23 kJ/kg

Therefore, heat transfer is 20.23 kJ/kg

3 0

3 years ago