An actual vapour compression system comprises following process represents

An actual vapour compression system comprises following process represents a. 1-2 Compression process b. 2-3 Condens 1 (or heat rejection from the condenser) c. 3-4 Irreversible expansion d. 4-1 Evaporation (or) heat addition to the evaporator Sketch the processes on T-S diagram.

2 months ago

Solution 1

Guest Guest #9791772
2 months ago

Answer:

Explanation:

The deatailed diagram of VCRS is given below such

1-2=Isentropic compression in which temperature increases at constant entropy

2-3=Isobaric heat rejection i.e. heat rejected at constant pressure(condensation)

3-4=Irreversible expansion or throttling in which enthalpy remains constant

4-1=Isobaric heat addition(Evaporation)

📚 Related Questions

Question
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.
Solution 1

Answer:

[tex]V=68.86ft^3[/tex]

Explanation:

[tex]T_1[/tex] =10°C,[tex]T_2[/tex] =93.33°C

Q=500 btu=527.58 KJ

[tex]P_1= 2atm[/tex]

If we assume that air is ideal gas   PV=mRT, ΔU=[tex]mC_v(T_2-T_1)[/tex]

Actually this is closed system so work will be zero.

Now fro first law

Q=ΔU=[tex]mC_v(T_2-T_1)[/tex]+W

⇒Q=[tex]mC_v(T_2-T_1)[/tex]

527.58 =[tex]m\times 0.71(200-50)[/tex]

m=4.9kg

 PV=mRT

[tex]200V=4.9\times 0.287\times (10+273)[/tex]

[tex]V=1.95m^3[/tex]                ([tex]V=1m^3=35.31ft^3[/tex])              

[tex]V=68.86ft^3[/tex]

Question
Air is entering a 4200-kW turbine that is operating at its steady state. The mass flow rate is 20 kg/s at 807 C, 5 bar and a velocity of 100 m/s. This air then expands adiabatically, through the turbine and exits at a velocity of 125 m/s. Afterwards the air then enters a diffuser where it decelerates isentropically to a velocity of 15 m/s and a pressure of 1 bar. Using the ideal gas model, determine, (a) pressure and temperature of the air at the turbine exit, in units of bar and Kelvin. (b) Entropy production rate in the turbine in units of kW/k, and (c) draw the process on a T-s Diagram.
Solution 1

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

Question
A piston-cylinder assembly contains ammonia, initially at a temperature of-20°C and a quality of 70%. The ammonia is slowly heated to a final state where the pressure is 6 bar and the temperature is 180°C. While the ammonia is heated, its pressure varies linearly with specific volume. For the ammonia, determine the work and heat transfer, each in kJ/kg.
Solution 1

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

Question
A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the piston until the air reaches to 0.1 m3 and 1,000 °C, in which the air undergoes a polytropic process (PV" const). Assume that heat loss from the cylinder, friction of piston, kinetic and potential effects are negligible. 1) Determine the polytropic constant n. 2) Determine the work transfer in ki for this process, and diseuss its direction. 3) sketch the process in T-V (temperature-volume) diagram.
Solution 1

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.

Question
A flat rectangular door in a mine is submerged froa one side in vater. The door dimensions are 2 n high, 1 n vide and the vater level is 1,5 m higher than the top of the door. The door has two hinges on the vertical edge, 160 mm from each corner and a sliding bolt on the other side in the niddle. Calculate the forces on the hinges and sliding bolt. Hint: Consider the door from a side view and from a plaa vies respectively and take moments about a point each time.)
Solution 1

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.

Question
A roller support acts like a contact boundary condition as it can produce a reaction force as a push response to a body but will not produce a pull force to hold a body from moving away. a)True b)- False
Solution 1

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.

Question
A pendulum has an oscillation frequency (T) which is assumed to depend upon its length (L), load mass (m) and the acceleration of gravity (g). Determine the relationship between oscillation frequency, length, load mass and acceleration of gravity. Differentiate as well which variable does not affect the oscillation frequency.
Solution 1

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.

Question
A closed-loop system has a forward path having two series elements with transfer functions 5 and 1/(s + 1). If the feedback path has a transfer function 2/s, what is the overall transfer function of the system?
Solution 1

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]

Question
A refrigeration cycle rejects Qn 500 Btu/s to a hot reservoir at 540 R, while receiving c200 Btu/s at 240°R. This refrigeration cycle a)- is internally reversible b)- is irreversible c)- is impossible d)- cannot be determined
Solution 1

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.

Question
Describe the operational principle of a unitary type air conditioning equipment with a suitable sketch.
Solution 1

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