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3 years ago

What is the biggest expectation when engineers test out designs?

2 answers:

6 0

The answer is B because it could be feasible but it’s not a need it and you got a time frame but it’s not a requirement and it doesn’t have to be unique.

Gnom [1K]3 years ago

3 0

Answer: Designs need to be feasible.

Explanation: don’t know, guessed got it right

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A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂ = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}} \right )^{K-1}$

Otto cycle T-S diagram

T₂ = 288.706* $6.25^{0.393}$ = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

$\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}} \right )^{K-1}$

T₄ = 2888.89 / $6.25^{0.393}$ = 1406.5 K

Work done, W = $c_v$ ×(T₃ - T₂) - $c_v$ ×(T₄ - T₁)

0.718×(2888.89 - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60 = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0

4 years ago

Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large

salantis [7]

Answer:

$Q_{cv} = -1007.86kJ$

Explanation:

Our values are,

State 1

$V=3m^3\\P_1=1bar\\T_1 = 295K$

We know moreover for the tables A-15 that

$u_1 = 210.49kJ/kg\\h_i = 295.17kJkg$

State 2

$P_2 =6bar\\T_2 = 296K\\T_f = 320K$

For tables we know at T=320K

$u_2 = 228.42kJ/kg$

We need to use the ideal gas equation to estimate the mass, so

$m_1 = \frac{p_1V}{RT_1}$

$m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}$

$m_1 = 3.54kg$

Using now for the final mass:

$m_2 = \frac{p_2V}{RT_2}$

$m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}$

$m_2 = 19.59kg$

We only need to apply a energy balance equation:

$Q_{cv}+m_ih_i = m_2u_2-m_1u_1$

$Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i$

$Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)$

$Q_{cv} = -1007.86kJ$

The negative value indidicates heat ransfer from the system

7 0

3 years ago

The first step in treating shock is to

san4es73 [151]

Answer:

Lay the person down and elevate thier legs slightly.

Explanation:

5 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

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

$h_g$ =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

What can be the main disadvantage of pulse amplitude modulation?

Feliz [49]

Answer:

transmission bandwidth required is very large.

Explanation:

4 0

2 years ago