Air is heated from 50 F to 200 F in a rigid container with a heat transfer of 500 Btu. Assume that the air behaves as an ideal gas. Determine the volume of air [ft3] if the initial pressure is 2 atm. Also show the process on a P-v state diagram. Use the following temperature conversion: T[R] = T[F] + 460.

Answer:

a)[tex]T_2=868.24 K[/tex] ,[tex]P_2=2.32 bar[/tex]

b) [tex]s_2-s_1=0.0206[/tex]KW/K

Explanation:

P=4200  KW ,mass flow rate=20 kg/s.

Inlet of turbine

 [tex]T_1[/tex]=807°C,[tex]P_1=5 bar,V_1=100 m/s[/tex]

Exits of turbine

 [tex]V_2=125 m/s[/tex]

Inlet of diffuser

[tex]P_3=1 bar,V_3=15 m/s[/tex]

Given that ,use air as ideal gas

R=0.287 KJ/kg-k,[tex]C_p[/tex]=1.005 KJ/kg-k

Now from first law of thermodynamics for open system at steady state

[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w[/tex]

Here given that turbine is adiabatic so Q=0

Air treat ideal gas   PV=mRT, Δh=[tex]C_p(T_2-T_1)[/tex]

[tex]w=\dfrac{P}{mass \ flow\ rate}[/tex]

[tex]w=\dfrac{4200}{20}[/tex]

w=210 KJ/kg

Now putting the values

[tex]1.005\times (273+807)+\dfrac{100^2}{2000}=1.005T_2+\dfrac{125^2}{2000}+210[/tex]

[tex]T_2=868.24 K[/tex]

Now to find pressure

We know that for adiabatic [tex]PV^\gamma =C[/tex] and for ideal gas Pv=mRT

⇒[tex]\left (\dfrac{T_2}{T_1}\right )^\gamma=\left (\dfrac{P_2}{P_1}\right )^{\gamma -1}[/tex]

[tex]\left(\dfrac{868.24}{1080}\right )^{1.4}=\left (\dfrac{P_2}{5}\right )^{1.4-1}[/tex]

[tex]P_2=2.32 bar[/tex]

For entropy generation

[tex]s_2-s_1=1.005\ln\dfrac{868.24}{1080}-0.287\ln\dfrac{2.32}{5}[/tex]

[tex]s_2-s_1=0.00103[/tex]KJ/kg_k

[tex]s_2-s_1=0.00103\times 20[/tex]KW/K

[tex]s_2-s_1=0.0206[/tex] KW/K

Answer:

w =  -28.8 kJ/kg

q = 723.13 kJ/kg

Explanation:

Given :

Initial properties of piston  cylinder assemblies

Temperature, [tex]T_{1}[/tex] = -20°C

Quality, x = 70%

           = 0.7

Final properties of piston  cylinder assemblies

Temperature, [tex]T_{2}[/tex] = 180°C

Pressure, [tex]P_{2}[/tex] = 6 bar

From saturated ammonia tables at [tex]T_{1}[/tex] = -20°C  we get

[tex]P_{1}[/tex] = [tex]P_{sat}[/tex] = 1.9019 bar

[tex]v_{f}[/tex] = 0.001504 [tex]m^{3}[/tex] / kg

[tex]v_{g}[/tex] = 0.62334 [tex]m^{3}[/tex] / kg

[tex]u_{f}[/tex] = 88.76 kJ/kg

[tex]u_{g}[/tex] = 1299.5 kJ/kg

Therefore, for initial state 1 we can find

[tex]v_{1}[/tex] = [tex]v_{f}[/tex]+x ([tex]v_{g}[/tex]-[tex]v_{f}[/tex]

                       = 0.001504+0.7(0.62334-0.001504)

                       = 0.43678 [tex]m^{3}[/tex] / kg

[tex]u_{1}[/tex] = [tex]u_{f}[/tex]+x ([tex]u_{g}[/tex]-[tex]u_{f}[/tex]

                       = 88.76+0.7(1299.5-88.76)

                       =936.27 kJ/kg

Now, from super heated ammonia at 180°C, we get,

[tex]v_{2}[/tex] = 0.3639 [tex]m^{3}[/tex] / kg

[tex]u_{2}[/tex] = 1688.22 kJ/kg

Therefore, work done, W = area under the curve

           [tex]w = \left (\frac{P_{1}+P_{2}}{2}  \right )\left ( v_{2}-v_{1} \right )[/tex]

           [tex]w = \left (\frac{1.9019+6\times 10^{5}}{2} \right )\left ( 0.3639-0.43678\right )[/tex]

           [tex]w = -28794.52[/tex] J/kg

                       = -28.8 kJ/kg

Now for heat transfer

[tex]q = (u_{2}-u_{1})+w[/tex]

[tex]q = (1688.2-936.27)-28.8[/tex]

          = 723.13 kJ/kg

Answer:

n=2.32

w= -213.9 KW

Explanation:

[tex]V_1=0.3m^3,T_1=298 K[/tex]

[tex]V_2=0.1m^3,T_1=1273 K[/tex]

Mass of air=1 kg

For polytropic process  [tex]pv^n=C[/tex] ,n is the polytropic constant.

  [tex]Tv^{n-1}=C[/tex]

  [tex]T_1v^{n-1}_1=T_2v^{n-1}_2[/tex]

[tex]298\times .3^{n-1}_1=1273\times .1^{n-1}_2[/tex]

n=2.32

Work in polytropic process given as

       w=[tex]\dfrac{P_1V_1-P_2V_2}{n-1}[/tex]

      w=[tex]mR\dfrac{T_1-T_2}{n-1}[/tex]

Now by putting the values

w=[tex]1\times 0.287\dfrac{289-1273}{2.32-1}[/tex]

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

  We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

Answer:

Force on the bolt = 24.525 kN

Force on the 1st hinge = 8.35 kN

Force on the 2nd hinge = 16.17 kN

Explanation:

Given:

height = 2 m

width =1 m

depth of the door from the water surface = 1.5 m

Therefore,

[tex]\bar{y}[/tex] =1.5+1 = 2.5 m

Hydrostatic force acting on the door is

[tex]F= \rho \times g\times \bar{y}\times A[/tex]

[tex]F= 1000 \times 9.81\times 2.5\times 2\times 1[/tex]

         = 49050 N

         = 49.05 kN

Now finding the Moment of inertia of the door about x axis

[tex]I_{xx}=\frac{1}{12}\times b\times h^{3}[/tex]

[tex]I_{xx}=\frac{1}{12}\times1\times 2^{3}[/tex]

               = 0.67

Now location of force, [tex]y^{*}[/tex]

[tex]y^{*}=\bar{y}+\frac{I_{xx}}{A\times \bar{y}}[/tex]

[tex]y^{*}=2.5+\frac{0.67}{2\times 1\times 2.5}[/tex]

             = 2.634

Therefore, calculating the unknown forces

[tex]F=F_{A}+R_{B}+R_{C} = 49.05[/tex]  ------------------(1)

Now since [tex]\sum M_{R_{A}}=0[/tex]

∴ [tex]R_{B}\times L+R_{C}\times L-F\times \frac{1}{2}=0[/tex]

  [tex]R_{B}+R_{C}-F\times \frac{1}{2}=0[/tex]

  [tex]R_{B}+R_{C}=\frac{F}{2}[/tex]

  [tex]R_{B}+R_{C}=24.525[/tex]        -----------------------(2)

From (1) and (2), we get

[tex]R_{A} = 49.05-24.525[/tex]

                = 24.525 kN

This is the force on the Sliding bolt

Taking [tex]\sum M_{R_{C}}=0[/tex]

[tex]F\times 0.706-R_{A}\times 0.84-R_{B}\times 1.68 = 0[/tex]

[tex]49.05\times 0.706-24.525\times 0.84-R_{B}\times 1.68 = 0[/tex]

[tex]R_{B}[/tex] =8.35 kN

This is the reaction force on the 1st hinge.

Now from (1), we get

[tex]R_{C}[/tex] =16.17 kN

This is the force on the 2nd hinge.

Answer:

a) True

Explanation:

Roller can provide reaction for push support but it can not provide reaction for pull support.  

From the free body diagram of roller and hinge support we can easily find that ,Roller providing vertical reaction and can not provide horizontal reaction.

On the other hand hinge support can provide reaction in both the direction.

So we can say that roller can not proved reaction for pull support.

Answer:

Mass does not affect oscillation frequency.                                                    

Explanation:

Let the bob of the pendulum makes a small angular displacement θ. When the pendulum is displaced from the equilibrium position, a restoring force tries to act upon it and it tries to bring the pendulum back to its equilibrium position. Let this restoring force be F.

Therefore, F = -mgsinθ  

Now for pendulum, for small angle of θ,

sinθ[tex]\simeq[/tex]θ

Therefore, F = -mgθ

Now from Newton's 2nd law of motion,

F = m.a = -mgθ

[tex]\Rightarrow m.\frac{d^{2}x}{dt^{2}} = - mg\Theta[/tex]

Now since, x = θ.L

[tex]\Rightarrow L.\frac{d^{2}\Theta }{dt^{2}}= -g\Theta[/tex]

[tex]\Rightarrow \frac{d^{2}\Theta }{dt^{2}}= -\frac{g}{L}.\Theta[/tex]

[tex]\Rightarrow \frac{d^{2}\Theta }{dt^{2}}+\frac{g}{L}.\Theta =0[/tex]

Therefore, angular frequency

 [tex]\omega ^{2}[/tex] = [tex]\frac{g}{L}[/tex]

ω = [tex]\sqrt{\frac{g}{L}}[/tex]

Also we know angular frequency is , ω = 2.π.f

where f is frequency

Therefore

2πf = [tex]\sqrt{\frac{g}{L}}[/tex]

f = [tex]\frac{1}{2 \pi }\sqrt{\frac{g}{L}}[/tex]

So from here we can see that frequency,f is independent of mass, hence it does not affect frequency.

Answer:

Transfer function for feedback path is given by:

[tex]\frac{C(s)}{R(s)}[/tex]=[tex]\frac{G(s)}{1+G(s)H(s)}[/tex]       (1)

Explanation:

with reference to fig1:

two blocks in series are multiplied:

[tex]\frac{5}{s(s+1)}[/tex]

for feedback function:

1+G(s)H(s)=[tex]1+\frac{5}{s+1}.\frac{2}{s}[/tex]

Now from eqn (1):

[tex]\frac{C(s)}{R(s)} = \frac{5}{s(s+1)+10}[/tex]

Answer:

(b)     Irreversible cycle.

Explanation:

 Given;

 [tex]T_2=540R ,Q_2= 500 Btu/s ,T_1=240 R ,Q_1= 200 Btu/s [/tex]

To find the validity of cycle

      [tex]\oint _R\frac{dQ}{T}\leq0[/tex]

If it is zero then cycle will be reversible cycle and if it is less than zero then cycle will be irreversible cycle.These are possible cycle.

If it is greater than zero ,then cycle will be impossible .

Now find

[tex]\dfrac{Q_1}{T_1}-\dfrac{Q_2}{T_2}=\dfrac{200}{240}-\dfrac{500}{540}[/tex]

[tex]\dfrac{Q_1}{T_1}-\dfrac{Q_2}{T_2}[/tex]= -0.09

It means that this  cycle is a irreversible cycle.

Answer:

Operational Principle of a Unitary Type Air Conditioning Equipment:

A unitary air conditioning system is basically a room type air conditioning system which comprises of an outdoor unit, a compressor for compressing

coolant, a heat exchanger (outdoor) for heat exchange, an expander attached to the heat exchanger for expansion of coolant and a duct.

It continuously removes heat and moisture from inside an occupied space and cools it with the help of heat exchanger and condensor in the condensing unit and discharges back into the same occupied indoor space that is supposed to be cooled.

The cyclic process to draw hot air, cool it down and recalculation of ther cooled air keeps the indoor occupied space at a lower temperature needed for cooling at home, for industrial processes and many other purposes.

refer to fig 1

Answer:

The application of force is the main difference between truss and frame.

Explanation:

Truss:

Truss is a collection of beams,which use to handle the tensile and   compression loads . That collection of beams creates rigid structure.

The load on the truss will be acting always at the  at the hinge. Truss is     widely used in the construction areas.                

Frame:

       Like truss, it is also a combination of beams and used to handle the loads. The main difference between truss and frame is the application of load. In the frame load can apply at the any point of the member of frame along  with hinge.

Truss are connected by pin joint and can not transfer moment ,on the other hand frames are connected by rigid joint like welding so frame can transfer moment.

Truss and frame both forms a rigid structure and is used in the construction areas.