Mechanical Engineering

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Sample Problems and Solutions

Question 1:

Since a refrigeration system operates more efficiently when the condensing temperature is low, evaluate the possibility of cooling the condenser cooling water of the refrigeration system with another refrigeration system. Will the combined performance of the two systems be better, the same, or worse than one individual system? Explain why.

The answer is the combined performance will be worse than one individual system. The following is proof.

Consider the two systems as described above:

where Q_L and Q_H refer to the heat removed from cooled space and rejected to warm space, respectively.  Note that in the combined system, Q_L2 = Q_H1 because the heat rejected from REF1 (Q_H1) is received as the cooling load for REF2 (Q_L2).

The efficiency of refrigerator is normally defined through the coefficient of performance (COP):

So, we can write the COP for the single system as:

 (1)

The COP for the combined system is:

 (2)

Note that we have the same desired output as single system Q_L1.  But Q_L2 = Q_H1, substitute into 2:

 (3)

We see that because:

Q_H2 = Q_H1 + Q_W2

And so:

Q_H2> Q_H1

By comparing equations 1 and 3, we derrive that COP₂ < COP₁

REFERENCES 

Rajput, R.K., 2000.  Thermal Engineering. S.Chand Publishers.

 

Question 2:

A two-stage ammonia system using flash-gas removal and intercooling operates on the cycle shown below.  The condensing temperature is 35 degrees C.  The saturation temperature of the intermediate-temperature evaporator is 0 degrees C, and its capacity is 150KW.  The saturation temperature of the low-temperature evaporator is -40 degrees C, and its capacity is 250KW.  What is the rate of refrigerant compressed by the high-stage compressor?

 

Add an adiabatic/isentropic efficiency of the compressors of 85% and THEN solve the problem. Should your answer be higher or lower than the one given in the book?

 

Given:

T5 = 35oC

T3 = 0oC

T1 = -40oC

Q1 = 150 kW

Q2 = 250 kW

 = 85% = 0.85

Solution:

Find specific enthalpies at all points shown on the above diagram, using saturated and superheated ammonia vapor tables.

 

Point 5. Here the ammonia is saturated liquid. From saturated vapor table at 35^oC we get the enthalpy, and pressure, respectively:

h5 = hf = 346.8 kJ/kg

P5 = 1350.4 kPa


 

Point 6. The throttle valve is isenthalpic, so the enthalpy is


h6 = h5= 346.8 kJ/kg

 

Point 3. Here the ammonia is saturated gas. From saturated vapor table at 0^oC we get the enthalpy, entropy, and pressure respectively:

h3 = hg = 1442.2 kJ/kg

s3 = sg = 5.3309 kJ/(kg*K)

P3 = 429.6 kPa

 

Point 7. Here the ammonia is saturated liquid. From saturated vapor table at 0^oC we get the enthalpy

h7 = hf = 180.36 kJ/kg

 

Point 8. The throttle valve is isenthalpic, so the enthalpy is

h8 = h7 = 180.36 kJ/kg

 

Point 1. Here the ammonia is saturated gas. From saturated vapor table at -40^oC we get the enthalpy and entropy, respectively:

h1 = hg = 1388.8 kJ/kg

s1 = sg = 5.9569 kJ/(kg*K)

 

Point 2. The high-stage compressor is isentropic so the entropy at point 2 is

s2 = s1 = 5.9569 kJ/(kg*K)

The pressure at point 2 is

P2 = P3 = 429.6 kPa

 

Now, from the superheated vapor table and using linear interpolation between pressures 400 kPa and 500 kPa we get for pressure 429.6 kPa near entropy 5.9569 kJ/(kg*K):

h = 1636.7+(1633.1-1636.7)*(429.6-400)/(500-400) = 1635.6 kJ/kg

at

s = 5.9907+(5.8744-5.9907)*(429.6-400)/(500-400) = 5.9563 kJ/(kg*K)

and

h = 1682.8+(1679.8-1682.8)*(429.6-400)/(500-400) = 1681.9 kJ/kg

at

s = 6.1179+(6.0031-6.1179)*(429.6-400)/(500-400) = 6.0839 kJ/(kg*K)

 

Now using linear interpolation, we find enthalpy at point 2 for an ideal low-stage compressor

h2 = 1635.6+(1681.9-1635.6)*(5.9569-5.9563)/(6.0839-5.9563) = 1635.8 kJ/kg

 

Let h2rfeal be the specific enthalpy at point 2 for real low-stage compressor.  Then its isentropic efficiency is given by

= (h2real - h1)/(h2 - h1)

 

From which we get

h2real  = h1 + *(h2 - h1)

 

Which is

h2real  = 1388.8+0.85*(1635.8-1388.8) = 1598.8 kJ/kg

 

Point 4. The low-stage compressor is isentropic so the entropy at point 4 is

s4 = s3 = 5.3309 kJ/(kg*K)

 

The pressure at point 4 is

P4 = P5 = 1350.4 kPa = 1.3504 MPa

 

Now from the superheated vapor table using a linear interpolation between pressures 1.2 MPa and 1.4 MPa we get for pressure 1.3504 MPa near entropy 5.3309 kJ/(kg*K):

h = 1606.8+(1598.8-1606.8)*(1.3504-1.2)/(1.4-1.2) = 1600.8 kJ/kg

at

s = 5.3916+(5.2994-5.3916)*(1.3504-1.2)/(1.4-1.2) = 5.3223 kJ/(kg*K)

and

h = 1658.0+(1651.4-1658.0)*(1.3504-1.2)/(1.4-1.2) = 1653.0 kJ/kg

at

s = 5.5325+(5.4443-5.5325)*(1.3504-1.2)/(1.4-1.2) = 5.4662 kJ/(kg*K)

 

Now, using linear interpolation, we find enthalpy at point 4 for an ideal high-stage compressor

h4 = 1600.8+(1653.0-1600.8)*(5.3309-5.3223)/(5.4662-5.3223) = 1603.9 kJ/kg

 

Similarly to point 2, we get the enthalpy at point 4 for real high-stage compressor

h4real  = h3 + *(h4 - h3)

 

Which is

h4real  = 1442.2+0.85*(1603.9-1442.2) = 1579.64 kJ/kg

 

To find the mass flow rates of refrigerant through intermediate-temperature and low-temperature evaporators, respectively:

m1 = Q1/(h3 - h6) = 150/(1442.2-346.8) = 0.13693 kg/s

m2 = Q2/(h1 - h8) = 250/(1388.8-180.36) = 0.20688 kg/s

 

Letting m_h be mass flow rate of refrigerant through the high-stage compressors.  Then the steady flow energy equation for the intercooler and flash tank is

m2*h2real + (mh - m1)*h6 = m2*h7 + (mh - m1)*h3

 

Solving for m_h gives the answer

mh = m1 + m2*(h2real - h7)/(h3 - h6)

 

Which is

mh = 0.13693+0.20688*(1598.8-180.36)/(1442.2-346.8) = 0.4048 kg/s

This is lower than the former value 0.4118 kg/s for ideal compressors

 

REFERENCES

Rajput, R.K., 2000.  Thermal Engineering. S.Chand Publishers.

 

Question 3:  

A R22 flash-gas removal system has a capacity of 180KW at an evaporating temperature of -30 degrees C when the condensing pressure is 1500KPa. Compute the power requirement for a system with a single compressor. Compute the total power required by the two compressors in the system where there is no intercooling but there is flash-gas removal at 600KPa.

Add an adiabatic/isentropic efficiency of the compressors of 85% and THEN solve the problem. Should your answer be higher or lower than the one given in the book?

 

(b) Given:

Qin = 180 kW

t1 = -30oC

P2 = P4 = P5 = 1500 kPa

P3 = P6 = P7 = 600 kPa

 = 85% = 0.85

 

Solution:

 

Find the specific enthalpies at all points on the Figure 16-17.

From saturated table we have for saturated liquid near 1500 kPa

    P              h_f

 

Now using linear interpolation find the specific enthalpy at point 5

h5 = hf = 248.361+(249.686-248.361)*(1500-1496.5)/(1533.5-1496.5) =  248.486 kJ/kg

 

The throttle valve at point 6 is isenthalpic, so the specific enthalpy at point 6 is

h6 = h5 = 248.486 kJ/kg

 

From saturated table at -30^oC we get the specific enthalpy and entropy at point 1 for saturated vapor

h1 = hg = 393.138 kJ/kg

s1 = sg = 1.80329 kJ/(kgK)

 

The saturated table gives near 600 kPa

 

Using linear interpolation we get for 600 kPa:

t = 5+(6-5)*(600-583.78)/(602.28-583.78) = 5.87675oC

hg = 407.143+(407.489-407.143)*(600-583.78)/(602.28-583.78) = 407.446 kJ/kg

hf = 205.899+(207.089-205.899)*(600-583.78)/(602.28-583.78) = 206.942 kJ/kg

sg = 1.74463+(1.74324-1.74463)*(600-583.78)/(602.28-583.78) = 1.74341 J(kgK)

 

Now we get he enthalpy and entropy at point 3, and enthalpy at point 7:

h3 = hg = 407.446 kJ/kg

s3 = sg = 1.74341 J(kgK)

h7 = hf = 206.942 kJ/kg

 

The throttle valve at point 8 is isenthalpic, so the specific enthalpy at point 8 is

h8 = h7 = 206.942 kJ/kg

 

The flash-gas compressor is isentropic, so the specific entropy at point 4 is

s4 = s3 = 1.74341 kJ/(kgK)

 

From saturated table at pressure 1500 kPa we find the saturation temperature 39^oC. Now from superheated table using columns with saturation temperatures 38^oC and 40^oC find the enthalpies and entropies at column with saturated temperature 39^oC near the entropy 1.74341 kJ/(kgK):

h = (427.155+425.871)/2 = 426.513 at s = (1.7365+1.7287)/2 = 1.73260

h = (431.568+430.374)/2 = 430.971 at s = (1.7501+1.7426)/2 = 1.74635

 

Now using liner interpolation find the specific enthalpy at point 4 for ideal flash-gas compressor

h4 = 426.513+(430.971-426.513)*(1.74341-1.73260)/(1.74635-1.73260) = 430.018  kJ/kg

 

Let h_4rfeal be specific enthalpy at point 4 for real flash-gas compressor. Then its isentropic efficiency is given by

= (h4real - h3)/(h4 - h3)

 

From which we get

h4real  = h3 + *(h4 - h3), which is

h4real = 407.446+0.85*(430.018-407.446) = 426.63 kJ/kg

 

The evaporator compressor is isentropic, so the specific entropy at point 2 is

s2 = s1 = 1.80329 kJ/(kgK)

 

Similarly to the above from the superheated table we get near the entropy 1.80329 kJ/(kgK):

h = (448.703+447.771)/2 = 448.237 at s = (1.8008+1.7940)/2 = 1.7974

h = (452.901+452.019)/2 = 452.46 at s = (1.8127+1.8061)/2 = 1.8094

 

Now we can get the enthalpy at point 2 for an ideal evaporator compressor

h2 = 448.237+(452.46-448.237)*(1.80329-1.7974)/(1.8094-1.7974) = 450.301 kJ/kg

 

The specific enthalpy at point 2 for a real evaporator compressor is given by

h2real = h1 + *(h2 - h1) which is

h2real = 393.138+ 0.85*(450.301-393.138) = 441.73 kJ/kg

 

Let m₁ and m₂ be mass flow through evaporator and flash-gas compressors, respectively.  The mass flow m₁ is given by

m1 = Qin/(h1 - h8) = 180/(393.138-206.942) = 0.96672 kg/s

 

The steady flow energy equation for points 3-6-7 gives

(m1 + m2)*h6 = m1*h7 + m2*h3

 

From which we get m₂

m2 = m1*(h6-h7)/(h3-h6) which is

m2 = 0.96672*(248.486-206.942)/(407.446-248.486) = 0.25265 kg/s

 

The power required by the evaporator and flash-gas compressors are given by, respectively:

W1 = m1*(h2real - h1)

W2 = m2*(h4real - h3)

 

Now we can find the total power required by two compressors

W = W1 + W2 = m1*(h2real - h1) + m2*(h4real - h3) 

which is

W = 0.96672*(441.73-393.138)+0.25265*(426.63-407.446) = 51.82 kW

 

This is lower than the former value 60.96 kW for ideal compressors.

 

REFERENCES

Rajput, R.K., 2000.  Thermal Engineering. S.Chand Publishers.

 

Question 4:

In a R22 refrigeration system the capacity is 180KW at a temperature of -30 deg C. The vapour from the evaporator is pumped by one compressor to the condensing pressure of 1500KPa. Later the system is revised to two-stage compression operating on a cycle with intercooling but no removal of flash gas at 600KPa. Calculate the power required from a single compressor in the original system?  Calculate the power required from two compressors in the revised system?

 

(a) Given:

Q_in = 180 kW

t₁ = -30^oC

P₂ = 1500 kPa

 

Solution:

 

Find the specific enthalpies at all points in the above original one evaporator - one compressor refrigeration system, using Table A-6 for saturated and Table A-7 for superheated Refrigerant 22

 

From the saturated table we have for saturated liquid near 1500 kPa

    P              h_f

 

Using linear interpolation find the specific enthalpy at point 3:

h3 = hf = 248.361+(249.686-248.361)*(1500-1496.5)/(1533.5-1496.5) = 248.486 kJ/kg

 

The throttle valve is isenthalpic, so the specific enthalpy at point 4 is

h4 = h3 = 248.486 kJ/kg

 

From saturated table at -30^oC we get the specific enthalpy and entropy at point 1 for saturated vapor

h1 = hg = 393.138 kJ/kg

s1 = sg = 1.80329 kJ/(kgK)

 

The compressor is isentropic, so the specific entropy at point 2 is

s2 = s1 = 1.80329 kJ/(kgK)

 

From the saturated table at pressure 1500 kPa we find the saturation temperature 39^oC. Now from the superheated table using columns with saturation temperatures 38^oC and 40^oC find the enthalpies and entropies at column with saturated temperature 39^oC near the entropy 1.80329 kJ/(kgK):

h = (448.703+447.771)/2 = 448.237 for s = (1.8008+1.7940)/2 = 1.7974

h = (452.901+452.019)/2 = 452.460 for s = (1.8127+1.8061)/2 = 1.8094

 

Now using liner interpolation find the specific enthalpy at point 2

h2 = 448.237+(452.460-448.237)*(1.80329-1.7974)/(1.8094-1.7974) = 450.310 kJ/kg

 

Find the mass flow rate of refrigerant

m = Qin/(h1 - h4) = 180/(393.138-248.486) = 1.24437 kg/s

 

Now we can find the power of the compressor required

W = m*(h2 - h1) = 1.24437*(450.310- 393.138) = 71.1 kW

 

 

(b) Given:

Qin = 180 kW

t1 = -30oC

P4 = 1500 kPa

P2 = P3 = P6 = 600 kPa

 

Now find the specific enthalpies at all points in the above figure.

 

From the saturated table we have for saturated liquid near 1500 kPa

    P              h_f

 

Now using linear interpolation find the specific enthalpy at point 5

h5 = hf = 248.361+(249.686-248.361)*(1500-1496.5)/(1533.5-1496.5) =  248.486 kJ/kg

 

The throttle valve is isenthalpic, so the specific enthalpy at points 6 and 7 is

h6 = h7 = h5 = 248.486 kJ/kg

 

From saturated table at -30^oC we get the specific enthalpy and entropy at point 1 for saturated vapor

h1 = hg = 393.138 kJ/kg

s1 = sg = 1.80329 kJ/(kgK)

 

The low-stage compressor is isentropic, so the specific entropy at point 2 is

s2 = s1 = 1.80329 kJ/(kgK)

 

The saturated table gives near 600 kPa

Using linear interpolation we get for 600 kPa:

t = 5+(6-5)*(600-583.78)/(602.28-583.78) = 5.87675oC

hg = 407.143+(407.489-407.143)*(600-583.78)/(602.28-583.78) = 407.446 kJ/kg

sg = 1.74463+(1.74324-1.74463)*(600-583.78)/(602.28-583.78) = 1.74341 J(kgK)

 

Now we get he enthalpy and entropy at point 3:

h3 = hg = 407.446 kJ/kg

s3 = sg = 1.74341 J(kgK)

 

From the superheated table using columns with saturation temperatures 5^oC and 10^oC find the enthalpies and entropies at column with saturated temperature 5.87675^oC near the entropy 1.80329 kJ/(kgK):

h = 425.562+(423.974-425.562)*(5.87675-5)/(10-5) = 425.284

for s = 1.8080+(1.7894-1.8080)*(5.87675-5)/(10-5) = 1.80474

and

h = 429.229+(427.724-429.229)*(5.87675-5)/(10-5) = 428.965

for s = 1.8200+(1.8017-1.8200)*(5.87675-5)/(10-5) = 1.81679

 

Now using liner interpolation find the specific enthalpy at point 2

h2 = 425.284+(428.965-425.284)*(1.80329-1.80474)/(1.81679-1.80474) =  424.841 kJ/kg

 

The high-stage compressor is isentropic, so the specific entropy at point 4 is

s4 = s3 = 1.74341 J(kgK)

 

From the saturated table at pressure 1500 kPa we find the saturation temperature 39^oC. Now from the superheated table using columns with saturation temperatures 38^oC and 40^oC find the enthalpies and entropies at column with saturated temperature 39^oC near the entropy 1.74341 kJ/(kgK):

h = (427.155+425.871)/2 = 426.513 at s = (1.7365+1.7287)/2 = 1.73260

h = (431.568+430.374)/2 = 430.971 at s = (1.7501+1.7426)/2 = 1.74635

 

Now using liner interpolation find the specific enthalpy at point 4

h4 = 426.513+(430.971-426.513)*(1.74341-1.73260)/(1.74635-1.73260) = 430.018  kJ/kg

 

Let m₁ and m₂ be mass flow through low-stage and high-stage compressors, respectively.

 

The mass flow m₁ is given by

m1 = Qin/(h1 - h7) = 180/(393.138-248.486) = 1.24437 kg/s

 

The steady flow energy equation for intercooler gives

(m2 - m1)*h6 + m1*h2 = m2*h3

 

From which we get m₂

m2 = m1*(h2-h5)/(h3-h5)

 

Which is

m2 = 1.24437*(424.841-248.486)/(407.446-248.486) = 1.38054 kg/s

 

The powers required by low-stage and high-stage compressors are given by, respectively:

W1 = m1*(h2 - h1)

W2 = m2*(h4 - h3)

 

Now we can find the total power required by two compressors

W = W1 + W2 = m1*(h2 - h1) + m2*(h4 - h3)

 

Which is

W = 1.24437*(424.841-393.138) + 1.38054*(430.018-407.446) = 70.6 kW

 

Question 5:

A refrigeration system using R22 is to have a refrigerating capacity of 80KW. The cycle is standard vapour-compression cycle in which the evaporating temperature is -8 deg C and the condensing temperature is 42 deg C.

Determine the volume flow of refrigerant measured in cubic meters per second at the inlet to the compressor.

Calculate the power required by the compressor.

At the entrance to the evaporator what is the fraction of vapour in the mixture expressed both on a mass basis and a volume basis?

Given:

Qin = 40 kW

T1 = -8oC

T3 = 42oC

 

Solution:

Refer to above diagram of the refrigeration system, and find the specific enthalpies at all points.

From the saturated vapor table at 42^oC find the specific enthalpy at point 3 for saturated liquid

h3 = hf = 252.352 kJ/kg

 

The throttle valve is isenthalpic, so the same specific enthalpy is at point 4

h4 = h3 = 252.352 kJ/kg

 

From the saturated vapor table at -8^oC we get the specific enthalpy, specific entropy, specific volume, and pressure at point 1 for saturated vapor, respectively:

h1 = hg = 402.341 kJ/kg

s1 = sg = 1.76394 kJ/(kgK)

v1 = vg = 61.0958 L/kg = 0.0610958 m3/kg

P1 = 380.06 kPa

 

The compressor is isentropic, so the specific entropy at point 2 is

s2 = s1 = 1.76394 kJ/(kgK)

 

From the superheated vapor table at saturation temperature 42^oC we have the enthalpy and entropy near the entropy 1.76394 kJ/(kgK):

         h                     s

 

Using linear interpolation we get the specific enthalpy at point 2

h2 = 438.062+(442.449-438.062)*(1.76394-1.7618)/(1.7747-1.7618) = 438.790 kJ/kg

 

Now we get:

(a) The specific heat received by evaporator is given by

qin = h1 - h4 = 402.341-252.352 = 149.989 kJ/kg

 

So the mass flow of refrigerant is

M = Qin/qin = 40/149.989 = 0.266686 kg/s

 

Now we can get the volume flow rate at the inlet of compressor

V = m*v1 = 0.266686*0.0610958 = 0.0162934 or 0.0163 m3/s

 

(b) Find specific power of the compressor

w = w = h2 - h1 = 438.790-402.341 = 36.4490 kJ/kg

 

Now we can find power of the compressor

W = w*M = 36.4490*0.266686 = 9.72044  9.72 kW

 

(c) The pressure at point 4 is

P4 = P1 = 380.06 kPa

 

From the saturated vapor table at 380.06 kPa we get the specific enthalpies and specific volumes for saturated liquid and vapor, respectively:

hf = 190.718 kJ/kg

hg = 402.341 kJ/kg

vf = 0.76253 L/kg

vg = 61.0958 L/kg

 

The fraction of vapor in the mixture at the entrance of the evaporator, at point 4, expressed on a mass basis is given by:

x4m = (h4 - hf)/(hg - hf) = (252.352-190.718)/(402.341-190.718) = 0.29124 or  0.291

  

(d) Find specific volume at point 4

v4 = (1 - x4m)*vf + x4m*vg = (1-0.29124)*0.76253+0.29124*61.0958 = 18.3340 L/kg

 

The fraction of vapor in the mixture at the entrance of the evaporator expressed on a volume basis is given by:

x4v = (1/v4-1/vf)/(1/g-1/vf) = (1/8.3340-1/0.76253)/(1/61.0958-1/0.76253) = 0.971

 

Question 6:

If in a standard vapor compression cycle using R22 refrigerant the evaporating temperature is -5 deg C and the condensing temperature is 30 deg C, sketch the cycle on pressure-enthalpy coordinates and calculate the work of compression, the refrigerating effect, and the heat rejected in the condenser in KJ/Kg and the corresponding COP as well.

 

Given:

T1 = -5oC

T3 = 30oC

 

Solution:

 

The diagram of the vapor-compression cycle is shown above, where: P₁=P₄; P₂=P₃

The corresponding cycle on pressure-enthalpy coordinates is drawn below

Find enthalpies at all points using Table A-6 for saturated Refrigerant 22 and Table A-7 for superheated Refrigerant 22

 

From the saturated vapor table at 30^oC we get the specific enthalpy and pressure, respectively, at point 3 for saturated liquid:

h3 = hf = 236.664 kJ/kg

P3 = 1191.9 kPa

 

Since the throttle valve is isenthalpic the specific enthalpy at point 4 is

h4 = h3 = 236.664 kJ/kg

 

From saturated vapor table at -5^oC we get the specific enthalpy and entropy at point 1 for saturated vapor:

h1 = hg = 403.496 kJ/kg

s1 = sg = 1.75928 kJ/(kgK)

 

Since the compressor is isentropic the specific entropy at point 2 is

s2 = s1 = 1.75928 kJ/(kgK)

 

The pressure at point 2 is

P₂ = P₃ = 1191.9 kPa

 

From superheated vapor table at saturation temperature 30^oC we have the enthalpy and entropy near entropy 1.75928 kJ/(kgK):

       h                     s

 

Using linear interpolation we get the specific enthalpy at point 2

h2 = 427.378+(431.549-427.378)*(1.75928-1.7534)/(1.7664-1.7534) = 429.265 kJ/kg

 

Now we get:

(a) The work of compression is given by

w = h2 - h1

which is

w = 429.265-403.496 = 25.769  25.8 kJ/kg

 

(b) The refrigerating effect is given by

qin = h1 - h4

which is

qin = 403.496-236.664 = 166.832  167 kJ/kg

 

(c) The heat rejected in the condenser is given by

qout = h2 - h3

which is

qout = 429.265-236.664 = 192.601  193 kJ/kg

 

 

(d) The coefficient of performance is given by

COP = qin/w

Which is

COP = 166.832/25.769 = 6.47

 

REFERENCES

Rajput, R.K., 2000.  Thermal Engineering. S.Chand Publishers.

 

Sample Problems and Solutions in Advanced Dynamics

Question 1:

Derive the response of viscously damped SDOF system to the force F(t) by means of the convolution integral and plot the response for the system.

Solution

The equation of motion of a single degree of freedom system is 

Natural frequency: 

Damping ratio: 

The impulse response function  when 

Given force ,

The response of the system is

 

Thus,

 

The response of the system with given force is

 

Question 2

Plot the response of a viscously damped force free system.

(a) Given:

x(0) = 2 cm

x'(0) = 0

ω_n = 5 rad/s = frequency of undamped oscillation

ς = 0.1

 

Solution:

We have the differential equation for oscillations

x'' + 2ςω_nx' + ω_n²x = 0                                                                    (1)

 

The general solution is given by

x(t) = X_o exp(-ςω_nt) cos(ω_n√(1- ς²) t + φ)                         (2)

 

Differentiating gives

x'(t) = -X_oω_n exp(ςω_n-t) [ς cos(ω_n√(1- ς²) t + φ) + √(1- ς²) sin(ω_n √(1- ς²) t + φ)]

 

Substituting the initial conditions gives:

x'(0) = -X_oω_n(ς cos φ + √(1- ς² sin φ) = 0

x(0) = X_o cos φ = 2

 

From which we get:

The graph of x(t) for damping factor =0.1  is drawn below

 

(b) Now the formula (2) in part (a) gives the general solution for ς = 1

x(t) = (X_o cos φ) exp(-ω_nt)

 

Substituting the initial conditions gives the system of equations:

x'(0) = -X_o ω_n cos φ = 0

x(0) = X_o cos φ = 2

 

From the first equation we get

cos φ = 0

 

Substituting it into the second equation we get that the X_o is infinitely large. So the system of equations has no physical solution for the given initial conditions

 

(c) Now the differential equation (2) in part (a) gives for ς = 2 and ω_n = 5

x'' + 20x' + 25x = 0

 

Find the solution in form

x = exp(-pt)

 

Put it into the above equation and get the equation for p

p² - 20p + 25 = 0

 

Solving for p gives

p = 10± √(10²-25) = 10 ± √75

 

Thus the general solution is

x(t) = X₁exp[-(10 + √75)t] + X₂exp[-(10 - √75)t]

 

The initial conditions give the system of equations:

x'(0) = -X₁(10 + √75) - X₂(10 - √75) = 0

x(0) = X₁ + X₂ = 2

 

Solving for X₁ and X₂ gives:

X₁ = -0.155

X₂ = 2.15

 

Thus the equation is

x(t) = -0.155exp[-(10 + √75)t] + 2.15exp[-(10 - √75)t]

 

The graph is drawn below

n

 

References 

Meirovitch, L., 2010, Fundamentals of Vibrations, Waveland Pr Inc

 

Sample Problems in Mechanical Engineering

Traditionally mechanical engineering is considered to be made of three mainstream engineering aspects: manufacturing engineering, thermal engineering and design engineering. To excel in the domain a thorough knowledge of all of these is essential. A practicing mechanical engineer must solve many problems every day in all these subdomains.

 

Shaper Cutting Time

Consider a shaper machine is working to smooth a surface plate of length 400mm. Assume the feed of the shaper tool is 2mm/stroke and it is working at a speed of 50 strokes per minute.  Determine the time it takes for the shaper to complete machining of the surface of the shaper machine.

The length of the travel needed on the job, L_w is 400mm

Assuming a reasonable approach (A) and overrun (O) as 2 mm each, the length of cut required for shaper machine is Lw+A+O = 400 + 2+ 2 = 404 mm.

The feed of the shaper (S_o) is 2mm/stroke

The speed of the shaper is (N_s) = 50 strokes/minute, hence the time it takes to complete the cutting process = (L_w+A+O)/N_s*S_o = 404/2*50 = 4.04 minutes.

 

Milling Cutting Time Calculation

Determine the milling time required for milling a rectangular plate of length 100mm, width 50mm by a helical fluted plain HSS milling cutter of diameter 60mm and length 75mm and 6 teeth.

 

Assume A = O = 5mm, Vc = 40m/min and So = 0.1mm/teeth.

Cutting time Tc = Length of travel (Lc)/Table Feed (Sm)

Length of travel (Lc) = Length of work piece (Lw) + Approach (A) + Overrun (O) + (1/2 x Diameter of the rod)

Table feed(Sm) = So x Zc x N

So = Feed per teeth

Zc = Number of teeth on the cutter

N = Speed of the cutter

Lc = 100mm + 5mm + 5mm + 30mm = 140mm

Sm = 0.1 x 6 x N

N = (1000 x Vc)/(∏ x Dc) = (1000 x 40)/(∏ x 60) = 212.31rpm or ~213rpm

Sm = 0.1 x 6 x 213 = 127.8mm/min

Hence the Cutting time = Lc/Sm = 140/127.88 = 1.09minutes.

 

Casting and Metal Flow Problems

Consider the case of a metal mold with a sprue length of 20cm and a cross sectional area at the base of the sprue of 2.5 cm². If the sprue feeds a horizontal runner leading into a mold cavity whose volume is 1560 cubic centimeter. Determine

1. a) the velocity of the molten metal at the base of the sprue;
2. b) volume flow rate of the molten metal; and
3. c) the time it takes to fill the mold.

For determining the flow rate of the molten metal at the bottom of the sprue pin, Bernoulli’s principle is required. As per Bernoulli’s principle the pressure at the top of the sprue should be converted to the velocity at the bottom of the sprue, so it will be given by the equation:

V = sqrt (2*g*h), where h is the height of the sprue, V is the velocity of the molten metal at the bottom of the sprue,

 

So in the given problem

9. A) Flow velocity of the molten metal, V = sqrt (2*9.81*100*20) = 198.1 cm/s
10. B) The flow rate of the molten metal at the bottom of the sprue pin is given by Q = A * V

Where A is the area of cross section of the sprue pin and V is the velocity of the molten metal,

dQ/dt = 2.5 * 198.1 = 495.25 cubic centimeters/s.

495. C) Time it takes to fill the mold = Q’/(dQ/dt) = 1560/495.25 = 3.15 seconds

 

Thermal Engineering

Steady flow Energy Equation (first law of thermodynamics) and the principle of energy conservation

0.5 kg/s of fluid flows in a steady state process. The properties of fluid at the entrance are measured as

p₁ = 1.4bar, and the density of the fluid is 2.5kg/m3.  

U_i = 920kJ/kg while at exit the properties are P₂ = 5.6bar, density = 5kg/m3, U₂ = 720KJ/kg.

The velocity at entrance is 200m/s, while at exit it is 180m/s. It rejects 60KW of heat and raises through 60m during the flow process. Determine the change of enthalpy as well as the work done during the process.

 

The key to the solution is the fact that the total energy remains constant during the process.

 

Mass flow rate of fluid in the process = mf = 0.5kg/s

Pressure at the inlet to the process = 1.4bar

Density of the fluid = 2.5kg/m3

Initial internal energy of the fluid = 920KJ/kg

Final pressure of the fluid = 5.6bar

Final density at the exit = 5kg/m3

Final internal energy of the fluid =720kJ/kg

Velocity of the fluid at the entrance = 200m/s

Velocity of the fluid at the exit = 180m/s

The amount of the heat given to the fluid during the process = 60KW

The rise in height of the fluid during the process = 60m

Enthalpy change during the process (h2-h1) = ∆(U2-U1) + ∆PV

Therefore, ∆PV = p2/ρ2 – p1/ρ1.

 

Specific enthalpy change = (720 - 920)  x 10³ + (5.6/5 - 1.4/2.5) x 10⁵ = -144KJ/kg

Hence Enthalpy change = ∆H = mf x (h2-h1) = 0.5 x (-144) = -72 KJ/s

 

The steady flow energy equation (SFEE) can be applied as:

SFEE - Q - WS = mf [(h2 – h1) + ½( V₂² –V₁²) + g(Z2 –Z1)]

60 x 10³ - Ws = 0.5 x (-144 + 0.5 x (180² -200²) + 9.81 x 60)

Hence Ws = 13605.7W = 136.1KW

 

Principle of Energy Conservation

0.25kg/s of water is heated from 30^oC to 60⁰C by hot gasses that enter at 180⁰C and leave at 80⁰C. Calculate the mass flow rate of gases when its specific heat at constant pressure is 1.08 KJ/kg - K. Find the entropy change of water and of hot gases. Take the specific heat of water as 4.186KJ/kg - K

 

The total energy entering the system should be equal to the energy leaving the system. The system is being heated by the hot gases entering and it is being cooled by the water leaving the system.

The temperature of the exhaust gas entering into the system = 180 degree Celsius

The temperature of the exhaust gas leaving the system = 80 degree Celsius.

The specific heat at constant pressure of the exhaust gases = 1.08KJ/kg - K

The specific heat of water = 4.1868KJ/kg - K

Water flow rate = 0.25kg/s

The heat intake from the exhaust gases = mg x Cpg x (Tg1-Tg2) (1)

.mg is the mass of the exhaust gases entering into the system

Cpg is the specific heat of the exhaust gases

Tg1 is the temperature of the exhaust gases entering into the system

Tg2 is the temperature of the exhaust gases leaving the system

 

The heat carried away by water = mw x Cpw x (Tw2 - Tw1) (2)

Mw is the mass of water = 0.25kg/s

CPw is the specific heat of water = 4.1868KJ/kg-K

Tw1 = 30^oC

Tw2 = 60⁰C

 

Equating equations 1 and 2, Mg x 1.08 x (180-80) = 0.25 x 4.1868 x (60 - 30), which yields that the mass flow rate of exhaust gases = 0.291kg/s.

 

Entropy change of hot water is calculable by: mw x Cpw x ln(Tw2/Tw1) = 0.25 x 4.1868 x ln(273 + 60/273 + 30)

S2 – S1 = 0.25 x 4.186 x ln(Tw2/Tw1) = 0.099KJ/kg-K

             

Entropy change of Exhaust gases = 0.291 x 1.08 x ln(80+273)/(180+273) = -0.0783KJ/kg-K.

 

 

Machine Design

You are required to design a pulley, B, given that the diameter of a pulley A is 300mm and the speed of rotation of pulley A is 250rpm. The belt thickness is 6mm. Determine the diameter of the pulley B, if it is rotating at a speed of 750rpm

 

For the belt drive to function the surface velocity of the pulley A should be same as the surface velocity of the pulley B.

 

The surface velocity of the pulley A = ∏ x 0.3 x 250/60 = 3.925m/s

Considering the surface velocity of the pulley as same, its diameter will be given by ∏ x D x 750/60 = 3.925m/s, where D = 100mm.

 

Suppose a shaft is made of a material with shear stress of 50Mpa and if the maximum torque that can act on the shaft is 200N-m, design the shaft diameter considering a factor of safety of 3.

 

Let T be the torque of the shaft = 200N-m,

The shear stress of the shaft material (fs) will be given by, T = ∏/16 x fs x d³

In the given scenario diameter d is 2.73cm = 3 cm.

Considering the factor of safety of 3, diameter of the shaft = 3 x 3 = 9cm.

  

References:

Grote, K.H. and Antonsson, E.K., 2009. Springer handbook of mechanical engineering (Vol. 10). Springer Science & Business Media.

Kreith, F., 1999. Mechanical engineering handbook.

Jain, R.K., 1995. Machine design. Khanna Publishers.

Ballaney, P.L., 1981. Thermal Engineering. Khanna.

 

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