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yawa3891 [41]
3 years ago
9

A 150-lbm astronaut took his bathroom scale (aspring scale) and a beam scale (compares masses) to themoon where the local gravit

y is g 5.48 ft/s2. Determine how much he will weigh (a) on the spring scale and(b) onthe beam scale.?
Engineering
1 answer:
Nonamiya [84]3 years ago
7 0

Answer:

a) W = 25.5 lbf

b) W = 150 lbf

Explanation:

Given data:

Mass of astronaut = 150 lbm

local gravity = 5.48 ft/s^2

a) weight on spring scale

it can be calculated by measuring force against local gravitational force which is equal to weight of body

W = mg

W = (150 \times 5.48)\times \frac{1 lbm}{32.32 lbm. ft/s^2} = 25.5 lbf

b) As we know that beam scale calculated mass only therefore no change in mass due to variation in gravity

thus W= 150 lbf

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sesenic [268]

Answer:

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6 0
3 years ago
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Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m
zvonat [6]

Answer:

T = 858.25 s

Explanation:

Given data:

Reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3,  c 550 J/kg K, k 48 W/m K),

initial uniform temperature ( Ti ) = 200 c

Final temperature = 550 c

convection coefficient  = 250 w/m^2 k

products combustion temp = 800 c

calculate how long the plate should be left in the furnace ( to attain 550 c )

first calculate/determine the Fourier series Number ( Fo )

\frac{T_{0}-T_{x}  }{T_{1}-T_{x}  } = C_{1} e^{(-0.4888^{2}*Fo )}

= 0.4167 = 1.0396e^{-0.4888*Fo}

therefore Fo =  3.8264

Now determine how long the plate should be left in the furnace

Fo = (\frac{k}{pc_{p} } ) ( \frac{t}{(L/2)^2} )

k = 48

p = 7830

L = 0.1

Input the values into the relation and make t subject of the formula

hence t = 858.25 s

8 0
3 years ago
A two-bus power system is interconnected by one transmission line. Bus 1 is a generator bus with specified terminal voltage magn
mixas84 [53]

Answer:

a) | v2 | ,  β2   ( load, bus voltage at bus 2 )

  p1 ,  q1 ( slack, bus power at bus 1 )

b) q2 , β2  

  p1 and q1 ( slack, bus power at bus 1 )

Explanation:

Attached below is a schematic representation of the solution

<u>a) Identify the variables in the solution vector assume Bus 2 is a load bus</u>

The specified parameters are ; P2 and q2

while | v2 | and β2 are not specified

given that bus 2 is a load bus, bus 1 is a slack bus with ; | v1 |  and β1 been specified while p1 and q1 are not specified

<em>Hence the variables in the solution </em>

<em>= | v2 | ,  β2   ( load, bus voltage at bus 2 )</em>

<em>   p1 ,  q1 ( slack, bus power at bus 1 ) </em>

<u>b) Identify the variables in the solution vector ( assume Bus 2 is a PV bus )</u>

specified at Bus 2 are ; | p2 | , | v2 |

unspecified : q2 , β2

Bus 1 ( still a slack bus )

specified parameter : | v1 |  and β1

unspecified : p1 and q1

<em>Hence the variables in the solution </em>

<em>= q2 , β2  </em>

<em>   p1 and q1 ( slack, bus power at bus 1 ) </em>

7 0
3 years ago
Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320
lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

n = \frac{S_{uts}}{\tau_{max}}

Where:

n - Safety factor, dimensionless.

S_{uts} - Ultimate shear strength, measured in pascals.

\tau_{max} - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: (n = 4.2, S_{uts} = 320\times 10^{6}\,Pa)

\tau_{max} = \frac{S_{uts}}{n}

\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}

\tau_{max} = 76.190\times 10^{6}\,Pa

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}

Where:

\tau_{max} - Maximum allowable shear stress, measured in pascals.

V - Shear force, measured in kilonewtons.

A - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

V = \frac{P}{5}

V = \frac{450\,kN}{5}

V = 90\,kN

The minimum allowable cross section area is cleared in the shearing stress equation:

A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}

If V = 90\,kN and \tau_{max} = 76.190\times 10^{3}\,kPa, the minimum allowable cross section area is:

A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}

A = 1.640\times 10^{-3}\,m^{2}

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

A = \frac{\pi}{4}\cdot D^{2}

The diameter is now cleared and computed:

D = \sqrt{\frac{4}{\pi}\cdot A}

D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})

D = 0.0457\,m

D = 45.7\,mm

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0
3 years ago
Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ
shusha [124]

Answer:

A)W'/m = 311 KJ/kg

B)σ'_gen/m = 0.9113 KJ/kg.k

Explanation:

a).The energy rate balance equation in the control volume is given by the formula;

Q' - W' + m(h1 - h2) = 0

Dividing through by m, we have;

(Q'/m) - (W'/m) + (h1 - h2) = 0

Rearranging, we have;

W'/m = (Q'/m) + (h1 - h2)

Normally, this transforms to another equation;

W'/m = (Q'/m) + c_p(T1 - T2)

Where;

W'/m is the rate at which power is developed

Q'/m is the rate at which heat is flowing

c_p is specific heat at constant pressure which from tables at a temperature of 980k = 1.1 KJ/kg.k

T1 is initial temperature

T2 is exit temperature

We are given;

Q'/m = -30 kj/kg (negative because it leaves the turbine)

T1 = 980 k

T2 = 670 k

Plugging in the relevant values;

W'/m = -30 + 1.1(980 - 670)

W'/m = 311 KJ/kg

B) The Entropy produced from the entropy balance equation in a control volume is given by the formula;

(Q'/T_boundary) + m(s1 - s2) + σ'_gen = 0

Dividing through by m gives;

((Q'/m)/T_boundary) + (s1 - s2) + σ'_gen/m = 0

Rearranging, we have;

σ'_gen/m = -((Q'/m)/T_boundary) + (s2 - s1)

Under the conditions given in the question, this transforms normally to;

σ'_gen/m = -((Q'/m)/T_boundary) - c_p•In(T2/T1) - R•In(p2/p1)

σ'_gen/m is the rate of entropy production in kj/kg

We are given;

p2 = 100 kpa

p1 = 400 kpa

T_boundary = 315 K

For an ideal gas, R = 0.287 KJ/kg.K

Plugging in the relevant values including the ones initially written in answer a above, we have;

σ'_gen/m = -(-30/315) - 1.1(In(670/980)) - 0.287(In(100/400))

σ'_gen/m = 0.0952 + 0.4183 + 0.3979

σ'_gen/m = 0.9113 KJ/kg.k

6 0
3 years ago
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