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ddd [48]
2 years ago
12

A wall that cannot be moved because it is carrying the weight of the roof is considered a wall

Engineering
1 answer:
ad-work [718]2 years ago
5 0

Answer:

Blank wall

Explanation:

A wall that cannot be moved because it is carrying the weight of the roof is considered a blank wall.

You might be interested in
Dear sir i want to ask something about the solution of my question?
Eva8 [605]
No you may not ask the question
3 0
3 years ago
A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i
dlinn [17]

Answer:

h_{max} = 51.8 cm

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

\theta = water surface angle with horizontal surface

a_x =accelration in x direction

a_z =accelration in z direction

slope in xz plane is

tan\theta = \frac{a_x}{g +a_z}

tan\theta = \frac{4}{9.81+0}

tan\theta =0.4077

the maximum height of water surface at mid of inclination is

\Delta h = \frac{d}{2} tan\theta

            =\frac{0.4}{2}0.4077

\Delta h  0.082 cm

the maximu height of wwater to avoid spilling is

h_{max} = h_{tank} -\Delta h

            = 60 - 8.2

h_{max} = 51.8 cm

the height requird if no spill water is h_{max} = 51.8 cm

3 0
3 years ago
A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an
Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here  thermodynamic equation that

energy equation that is

h1 + \frac{v1}{2}  + q = h2 + \frac{v2}{2}  + w

put here value with

turbine is insulated so q = 0

so here

3015.4 *1000 + \frac{10^2}{2}  =  2431.7 * 1000 + \frac{90^2}{2}  + w

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0
3 years ago
A. A 3-kg plastic tank that has a volume of 0.2 m^3 is lled with liquid water. Assuming the density of water is 1000 kg=m^3, det
Tpy6a [65]

Answer:

The answer is below

Explanation:

a) The weight of the combined system is the sum of the weight of the water and the weight of the tank

m_{water}=V_{tank}.\rho_{wtaer}\\\\m_{water}=0.2m^3*1000kg/m^3\\\\m_{water}=200 \ kg\\\\m_{total} = m_{water}+m_{tank}\\\\But\ m_{tank}=3kg,therefore:\\\\m_{total} =200kg+3kg\\\\m_{total} =203\ kg\\\\weight_{total}=m_{total}g\\\\weight_{total}=203kg*9.81m/s^2\\\\weight_{total}=1991.43\ N

b) Since the weight of a system can be divided into smaller portions, hence weight is an extensive property.

c) When analyzing the acceleration of gases as they flow through a nozzle, the geometry of the nozzle which is an open system can be chosen as our system.

d) Given that:

\rho_{water}=1000kg/m^3\\\\1kg/m^3=0.062428lb/ft^3\\\\1000kg/m^3=1000kg/m^3*\frac{0.062428lb/ft^3}{kg/m^3}=62.43lb/ft^3\\ \\\rho=SG*\rho_{water}=1.03*62.43=64.272lb/ft^3\\\\P=P_{atm}+\rho g H\\\\P=14.7\ psia+64.272\ lb/ft^3*32.2\ ft/s^2*175\ ft*\frac{1\ ft^2}{12^2\ in^2}*\frac{1\ lbf}{32.2\ lbm.ft/s^2}  \\\\P=92.8\ psia

6 0
3 years ago
An ideal Otto cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression proces
Allushta [10]

Answer:

(a) The amount of heat transferred to the air, q_{out} is 215.5077 kJ/kg

(b) The net work output, W_{net}, is 308.07 kJ/kg

(c) The thermal efficiency is 58.8%

(d) The Mean Effective Pressure, MEP, is 393.209 kPa

Explanation:

(a) The assumptions made are;

c_p = 1.005 kJ/(kg·K), c_v = 0.718 kJ/(kg·K), R = 0.287 kJ/(kg·K),

Process 1 to 2 is isentropic compression, therefore;

T_{2}= T_{1}\left (\dfrac{v_{1}}{v_{2}}  \right )^{k-1} = 300.15\times 9.2^{0.4} = 729.21 \, K

From;

\dfrac{p_{1}\times v_{1}}{T_{1}} = \dfrac{p_{2}\times v_{2}}{T_{2} }

We have;

p_{2} = \dfrac{p_{1}\times v_{1}\times T_{2}}{T_{1} \times v_{2}} = \dfrac{98\times 9.2\times 729.21}{300.15 } = 2190.43 \, kPa

Process 2 to 3 is reversible constant volume heating, therefore;

\dfrac{p_3}{T_3} =\dfrac{p_2}{T_2}

p₃ = 2 × p₂ = 2 × 2190.43 = 4380.86 kPa

T_3 = \dfrac{p_3 \times T_2}{p_2} =\dfrac{4380.86  \times 729.21}{2190.43} = 1458.42 \, K

Process 3 to 4 is isentropic expansion, therefore;

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

1458.42= T_{4} \times \left (9.2 \right )^{0.4}

T_4 = \dfrac{1458.42}{(9.2)^{0.4}}  = 600.3 \, K

q_{out} = m \times c_v \times (T_4 - T_1) = 0.718  \times (600.3 - 300.15) = 215.5077 \, kJ/kg

The amount of heat transferred to the air, q_{out} = 215.5077 kJ/kg

(b) The net work output, W_{net}, is found as follows;

W_{net} = q_{in} - q_{out}

q_{in} = m \times c_v \times (T_3 - T_2) = 0.718  \times (1458.42 - 729.21) = 523.574 \, kJ/kg

\therefore W_{net} = 523.574 - 215.5077 = 308.07 \, kJ/kg

(c) The thermal efficiency is given by the relation;

\eta_{th} = \dfrac{W_{net}}{q_{in}} \times 100=  \dfrac{308.07}{523.574} \times 100= 58.8\%

(d) From the general gas equation, we have;

V_{1} = \dfrac{m\times R\times T_{1}}{p_{1}} = \dfrac{1\times 0.287\times 300.15}{98} =0.897\, m^{3}/kg

The Mean Effective Pressure, MEP, is given as follows;

MEP =\dfrac{W_{net}}{V_1 - V_2} = \dfrac{W_{net}}{V_1 \times (1- 1/r)}= \dfrac{308.07}{0.897\times (1- 1/9.2)} = 393.209 \, kPa

The Mean Effective Pressure, MEP = 393.209 kPa.

3 0
3 years ago
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