LESSON - 48 NUMERICAL PROBLEMS

 

Problem 48.1: A two stroke cycle internal combustion engine has a mean effective pressure of 6 bar. The speed of the engine is 1000 r.p.m. If the diameter of piston and stroke are 110 mm and 140 mm respectively, find the indicated power developed.

Solution:

Given: No. of strokes per cycle for the engine,  S =  2;

             Actual mean effective pressure, p_am = 6 bar;

     Engine speed, N = 1000 r.p.m;

     Diameter of the piston, D = 110 mm = 0.11 m;       

     Stroke length, L = 140 mm = 0.14 m;

     No. of cylinders = 1;

     No. of missed cycle, n_mc = 0;

Determine the indicated power developed, I.P.:

Formula:   Indicated power developed,  IP =   × (No. of cylinders), kW

      =     × (No. of cylinders)       [  n_mc = 0]

 Answer:  Indicated power, I.P =  × (No. of cylinders) 
                                                      = 13.3 kW

Problem 48.2: A 4-cylinder four-stroke petrol engine develops 14.7 kW at 1000 r.p.m. The mean effective pressure is 5.5 bar, Calculate the bore and stroke of the engine, if the length of  stroke is 1.5 times the bore.

Solution: 

Given:   Number of cylinder, n_cyl = 4;

        No. of strokes per cycle for the engine, S = 4;       

              Power developed,   I.P. = 14.7 kW;

       Engine speed, N = 1000 r.p.m.;

      Actual mean effective pressure, P_am = 5.5 bar;

      Length of stroke, L = 1.5 D (bore);

      No. of missed cycle, n_mc = 0.

Determine the bore, D and stroke, L of the engine: 

Formula:  Indicated power developed, IP =    × (No. of cylinders), kW

              =     × (No. of cylinders)              [  n_mc = 0]

       = 

 or                  

Answer:  Diameter,  D  

  

                                         = 0.0879   or  87.9 mm        and       

       Length of stroke,  L =  1.5 D = 1.5  x  87.9 = 131.8 mm.

Problem 48.3: A rope brake was used to measure the brake power of a single cylinder, four stroke cycle petrol engine. It was found that the torque due to brake load is 175 N-m and the engine makes 500 r. p.m. Determine the brake power developed by the engine.

Solution:

Given:   Torque due to brake load, T = 175 N-m .

 Engine speed, N = 500 r.p.m.

Determine the brake power, B.P. ;

                    Formula:   Brake power, B.P. = 

Answer:    Brake power, B.P. = =  = 9.16 kW

Problem 48.4: A single cylinder, four stroke cycle oil engine is fitted with a rope brake. The diameter of the brake wheel is 600 mm and the rope diameter is 26 mm. The dead load on the brake is 200 N and the spring balance reads 30 N. If the engine runs at 450 r.p.m. what will be the brake power of the engine?

Solution:

Given:  Diameter of the brake wheel, D_b = 600 mm = 0.6 m

       Rope diameter, d = 26 mm = 0.026 m

       Dead load on the brake, W= 200 N

       Spring balance reading, S = 30 N

               Engine speed, N = 450  r.p.m.

Determine the brake power, B.P.

Formula: Brake power, B.P. =  = 

Finding unknown, R_eff :

                      R_eff =   = 0.313 m

Answer:   Brake power,   B.P. =   

                                                      = 2.5 kW

Problem 48.5: The following data refers to an oil engine working on Otto four stroke cycle

Brake power                                   = 14.7 kW

Suction pressure                            = 0.9 bar

Mechanical efficiency                     = 80%

Ratio of compression                     = 5

Index of compression curve           = 1.35

Index of expansion curve                = l.3

Maximum explosion pressure        = 24 bar

Engine speed                                  = 1000 r.p.m.

Ratio of stroke: bore (L/D)            = 1.5

Determine the diameter and stroke of the piston?

Solution:

Given:   Brake power, b.p. = 14.7 kW;

Suction pressure, p₁ = 0.9 bar;

Mechanical efficiency, η_mech = 80%;

Compression ratio, r_v = = 5;

Index of compression curve, nc = 1.35; Index of expansion curve, ne = l.3; Maximum explosion pressure, p3 = 24 bar; Engine speed, N = 1000 r.p.m.; Ratio of stroke: bore, L/D = 1.5   we have L = 1.5D;                No. of missed cycle, nmc = 0;                Number of cylinder, ncyl = 1;                Number of strokes per cycle for the engine, S = 4;

 Determine the diameter (D) and stroke length (L) of the piston:

Formula:    Indicated power of engine,   

 

 or                           

Finding unknown, p_tm and I.P.;

Finding unknown, p_tm ;

 Theoretical Mean effective pressure, 

 

Finding unknown, W ;

                      Theoretical work done/cycle, W = Area 1-2-3-4

                                                = (Area under the curve 3-4) – (Area under the curve 1-2)

                                                

                                                                  , therefore we have

                                                                 

                                                           

Finding unknown P₂,

Considering compression process 1-2, we have

      = 8.78

 P₂ = P₁ × 8.78 = 0.9 × 8.78 = 7.9 bar

Finding unknown P₄,

Considering compression process 3-4, we have     

   = 8.1

 P₄ =  = 2.96 bar

By substituting p₁, p₂, p₃ and p₄, we have theoretical work done/cycle,

                   W = 

                       = 10⁵ [20.98 V₃]        N-m   

 Theoretical Mean effective pressure,

p_tm =  

= 

= 10⁵ x 5.245   N/m²      or    5.245 bar

Finding unknown, I.P. ;

        

I.P.  =  = 18.37 kW

Answer:   Diameter of piston, D = 

                                                         =  

                                                        = 0.142 m              

                  and length of the stroke, L = 1.5D = 1.5 x 0.142 = 0.213 m         

Problem 48.6: The following readings were taken during the test of a single cylinder four stroke oil engine:

Cylinder diameter = 250 mm,                         Stroke length = 400 mm,           Gross m.e.p. = 7 bar

Pumping m.e.p. = 0.5 bar,                               Engine speed = 250 r.p.m.,      Net load on the brake = 1080 N, 

Effective diameter of the brake = 1.5 m;       Fuel used per hour = 10 kg

Calorific value of fuel = 44300 kJ/kg

Calculate: (i) Indicated power     (ii) Brake power      (iii) Mechanical efficiency     (iv) Indicated thermal efficiency.

Solution:

Given: Cylinder diameter, D =250 mm = 0.25 m;

Stroke length, L= 400 mm = 0.4 m; 

Gross m.e.p., (p_am)_gross= 7 bar;    

Pumping m.e.p., (p_am)_pumping = 0.5 bar;

Net  P_am = (p_am)_gross- (p_am)_pumping  = 7 – 0.5 = 6.5 bar,                  

Engine speed, N = 250 r.p.m.  ,    

Net load on the brake, (W-S) = 1080 N, 

Effective diameter of the brake D_b= 1.5 m  i.e. R_eff =     

Fuel used per hour,    = 0.00277 kg/s      

Calorific value of fuel, C.V. = 44300 kJ/kg;

Number of cylinders, n_cyl = 1;

Number of stroke, S = 4.

Number of missed cycle, n_mc = 0

(i) Determine indicated power, I.P. ;   

Formula: Indicated power,  I.P. = 

Answer:    Indicated power,  I.P. = 

 =     = 26.59 kW

(ii) Determine brake power,   B.P. ;

Formula:    Brake power, B.P. = 

Answer:    Brake power, B.P. = 

       =  = 21.2 kW             

(iii) Determine  mechanical efficiency,  η_mech ;

Formula: Mechanical efficiency, η_{mech =}     

Answer:    Mechanical efficiency, η_mech =  =  = 0.797  or  79.7%

(iv) Determine indicated thermal efficiency, η_i,_th ;

Formula: Indicated thermal efficiency, η_i,_th =  

Answer:    Indicated thermal efficiency, η_i,_th =  = 0.216   or   21.6%

Problem 48.7: The following particulars were obtained during a trial on a 4-stroke gas engine:

Duration of trial = 1 hour;                   Revolutions = 14000;                                          Number of missed cycle = 500

Net brake load = 1470 N;                 Effective brake circumference = 4 m;                Mean effective pressure = 7.5 bar;    

Cylinder diameter = 250 mm;          Stroke = 400 mm;                                                 Compression ratio = 6.5:1

Gas consumption = 20,000 litres     L.C.V of gas at supply condition = 21 kJ/litre                  

Calculate: (i) Indicated power    (ii) Brake power    (iii) Mechanical efficiency   (iv) Indicated thermal efficiency   (v) Relative efficiency, 

Solution: 

Given:  Number of strokes, S = 4;

              Duration of trial = 1 hour; 

              Revolutions, N =   = 66.67   r.p.m.;

              Number of missed cycle, n_mc =  ;

              Net brake load, (W-S) = 1470 N;

              Mean effective pressure,  P_am = 7.5 bar;  

              Gas consumption, V_g = 20,000 litres =  = 5.55 litres/s;

              L.C.V of gas at supply condition, C.V. = 21 kJ/litre

              Cylinder diameter, d = 250 mm = 0.25 m;  

              Stroke length, L = 400 mm = 0.4 m;     

              Effective brake circumference    

            

             No. of cylinder = 1;

             Compression ratio = 6.5:1     or     r_v = 6.5

 (i) Determine indicated power   

Formula: Indicated power,  I.P. =  

Answer:    Indicated power,  I.P. = 

                                                               =  = 24.53 kW.

(ii) Determine brake power,   B.P. ;

Formula:    Brake power, B.P. = 

 Answer:    Brake power, B.P. = 

                                                         =  =  22.86 kW                             

 (iii) Determine  mechanical efficiency,  η_mech ;

Formula: Mechanical efficiency, η_mech  =   

Answer:    Mechanical efficiency, η_mech =    = 0.932   or 93.2%

(iv) Determine indicated thermal efficiency, η_i,_th;

Formula: Indicated thermal efficiency, η_i,_th = 

Answer:    Indicated thermal efficiency, η_i,_th =  =  =  0.212   or   21.2%

(v) Determine relative thermal efficiency, η_relative  ;

Formula:  Relative efficiency,    

Finding unknown, η_{air-standard    ;}

 η_air-standard =   = 0.527    or   52.7%

Answer:   Relative efficiency,     η_relative  =   

  = 0.4023    or 40.23%

Problem 48.8: In a test of a 4-cylinder, 4 stroke engine 75mm bore and 100 mm stroke, the following results were obtained at full throttle at a particular constant speed and with fixed setting of fuel supply of 6.0 kg/h.

B.P. with all cylinder working = 15.6 kW

B.P. with cylinder no. 1 cut out = 11.1 kW

B.P. wirh cylinder no. 2 cut out = 11.03 kW

B.P. with cylinder no. 3cut out = 10.88 kW

B.P. with cylinder no. 4 cut out = 10.66 kW

If  the calorific value of the fuel is 83600 kJ/kg and clearance volume is 0.0001 m³,

Calculate:  (i) Mechanical efficiency, (ii) Indicated thermal efficiency, and  (iii) Air standard efficiency.

Solution:  

Given: Number of cylinders, n_cyl = 4;

     Number of stroke, S = 4.

            Cylinder diameter, D =75  mm = 0.075 m;

            Stroke length, L= 100 mm = 0.1 m; 

            Fuel used per hour,  m_f = 6.0 kg/h

    Calorific value of the fuel, C.V. = 83600 kJ/kg;

    Clearance volume, V_c = 0.0001 m³

    B.P. with all cylinder working, BP_(1+2+3+4) = 15.6 kW

    B.P. with cylinder no. 1 cut out, BP_(2+3+4)   = 11.1 kW

    B.P. wirh cylinder no. 2 cut out BP_(1+3+4)   = 11.03 kW

    B.P. with cylinder no. 3cut out, BP_(1+2+4)   = 10.88 kW

    B.P. with cylinder no. 4 cut out, BP_(1+2+3)   = 10.66 kW

(1)   Determine the mechanical efficiency, η_mech :   

Formula:    

Finding unknown, I.P. _(1+2+3+4) ;

Assuming that the engine is running at constant speed with all cylinder firing/working. As the engine is running at constant speed, the frictional and pumping losses remains constant

                   = FP_(1+2+3+4).

The indicated power when all cylinders firing is given by

                   IP_(1+2+3+4) = BP_(1+2+3+4) + FP_(1+2+3+4)

Now, assuming that the engine is running with cylinder 1 cut-off but at same constant speed as when all cylinders are working. As the engine is running at constant speed, the frictional and pumping losses remains constant

                    = FP_(1+2+3+4).

The indicated power when cylinder ‘1’ cut off  is given by

                   IP_(2+3+4) = BP_(2+3+4) + FP_(1+2+3+4)    

The Indicated power of cylinder ‘1’ when all cylinder firing can be calculated by subtracting above two equations

                   IP₁ = BP_(1+2+3+4)  -  BP_(2+3+4)  = 15.6 - 11.1 = 4.5 kW

Similarly Indicated power of cylinders ‘2’, ‘3’, and ‘4’ when all cylinder firing can be calculated

           IP₂ = BP_(1+2+3+4)  - BP_(1+3+4) = 15.6 - 11.03 = 4.57 kW

           IP₃ = BP_(1+2+3+4)  -  BP_{(1+ 2+4)} = 15.6 - 10.88 = 4.72 kW

           IP₄ = BP_(1+2+3+4)  - BP_(1+2+3)  = 15.6 – 10.66 = 4.94 kW 

Therefore, total indicated power of all the engine when all cylinder firing/working is

 IP_(1+2+3+4) = IP₁ + IP₂  + IP₃  + IP₄         

                                     = 4.5 + 4.57 + 4.72 + 4.94 = 18.73 kW

       Answer:      Mechanical efficiency, η_mech  

         = 0.833    or        83.3%

(2)   Determine indicated thermal efficiency, η_i,_th ;

Formula: Indicated thermal efficiency, η_i,_th = 

Answer:    Indicated thermal efficiency, η_i,_th=    = 0.1344   or   13.44%

(3)   Air-standard efficiency, η_air-standard  ;  

Formula: Indicated thermal efficiency, η_air-standard = 1 − 

Finding unknown, r_v ;

Compression ratio, r_v = 

Finding unknown, V_S ;

Stroke volume, V_S = 

 = 0.0004417 m³

                                r_v =   = 5.4

     Answer:   Air-standard efficiency, η_air-standard = 1 − =  

                                                                                                              = 0.49     or       49%

Problem 48.9:  During a trial of a single cylinder oil engine working on dual cycle, the following observations were made:

Compression ratio = 15,                Oil consumption 10.2 kg/h,     Calorific value of fuel= 43890 kJ/kg

Air consumption = 3.8 kg/min        Speed = 1900 r.p.m.               Torque on the brake drum  = 186 N-m

Quantity of cooling water used = 15.5 kg/min,                                Temperature rise of cooling water = 36°C,

Exhaust gas temperature = 410°C,         Room temperature = 20°C,           C_P for exhaust gases 1.17 kJ/kg K

Calculate: (i) Brake power, 

                  (ii) Brake specific fuel consumption, and

                 (iii) Brake thermal efficiency.

                 (iv) Also draw heat balance sheet on minute basis.

Solution:

Given:  Compression ratio,   r_v = 15;    

             Oil consumption,  = 10.2 kg/h;

             Calorific value of fuel, C.V.= 43890 kJ/kg;

             Air consumption,   = 3.8 kg/min;

             Speed, n = 1900 rpm;

             Torque on the brake drum, T  = 186 N-m;

             Quantity of cooling water used,  = 15.5 kg/min;

             Temperature rise of cooling water, (T_wo -T_Wi)  = 36°C;

             Exhaust gas temperature, T_g = 410°C;

             Room temperature, T_a = 20°C;

             C_P for exhaust gases = 1.17;

              Number of missed cycle, n_mc  = 0;

              No. of cylinder = 1.

(i) Determine brake power,   B.P. ;

Formula:    Brake power, B.P. = 

Answer:    Brake power, B.P. = 

                                                                   =  = 37 kW

 (ii) Brake specific fuel consumption, b.s.f.c. ;

Formula:    Brake specific fuel consumption, b.s.f.c. =           ,kg/kWh

Answer:    Brake specific fuel consumption, b.s.f.c. = 

       =  = 0.2756 kg/kWh

(iii) Determine brake thermal efficiency, η_i,_th;

Formula:  Brake thermal efficiency, η_i,_th = 

Answer:    Brake thermal efficiency, η_i,_th =   =  = 0.2975   or   29.75%

(iv) Determine heat balancing (minute basis):

Heat balance sheet (minute basis)

Finding unknown of the sheet:

Heat supplied by the fuel per minute

            =  = 7461 kJ/min

     (a) Heat equivalent of B.P.

                   = B.P. x 60 = 37 x 60 = 2220 kJ/kg

     (b) Heat carried away by cooling water

                     = x C_pw (T_wo -T_Wi) = 15.5 x 4.18 x 36   = 2332 kJ/min

     (c) Heat carried away by exhaust gases

                   = x C_pg x (T_g- T_a) = = 1811  kJ/min

      (d) Heat unaccounted for (by difference)

     = 7461 – [2220 + 2332 + 1811] = 1098 kJ/min

Answer:       Heat balance sheet (minute basis)