Engineering Mathematics and General Sciences

An astronaut weighs 850N on the surface of the earth where g is 9.806 m/sec². What is the mass of the astronaut? What is his mass on the surface of the moon where g is 1/6 of the earth’s gravity?

m_e= w/g = 850N / 9.806 = 86.682kg

m_m= m_e = 86.682kg

What force in Newton will accelerate a mass of 50kg at the rate of 12 m/sec²? F = m xa = 50kg (12 m/sec²)= 600N

What force in Newton will accelerate a mass of 50kg at the rate of 5 m/sec² assuming friction factor f = 0.015? (The working plane is 30° inclined from the horizontal surface)

F – F_f – W sin ө = ma

F = ma + W sin ө + F_f

= ma + W sin ө + µN

= ma + W sin ө + µW cos ө

= 50(5) + 50(9.81) sin30° + 0.015(50) (9.81) cos30°

= 250 + 245.25 + 6.37

= 501.62 N

The mass of a kite is 0.102kg. What is the force due to gravity acting on the kite?

W = m x g = 0.102kg (9.806 m/sec²) = 1 Newton

The weight of 1 Liter of gasoline is 7.0 N. Determine its density.

m = w/g = 7N / 9.806 m/sec² = 0.7138 kg

ρ = m/v = 0.7138 kg / 1 liter = 0.7138 kg/liter * 1000L / m³ = 713.8kg/m³

The weight of 1 liter of gasoline is 7.0N. Determine its relative density.

S.G.g = ρ_g / ρ_water

where:

m = W / g = 7N / 9.81 m/sec² = 0.7136kg

ρ_gas = m / v = 0.7135kg / 1 liter = 0.7136 kg/L

S.G.gas = ρ_g / ρ_w = 0.7136 kg/liter / 1 kg/l = 0.7136

A horizontal pipe (100mm ø and 7m long) transports sea water. Find the weight of the fluid in the pipe. The specific gravity of sea water is 1.03.

V_pipe = п d² * L =_ п _ (0.1m)² (7m) = 0.05498 m³

4 4

m = ρ * V = (1.03) (1000 kg/m³) (0.05498 m³)

= 56.6294 kg

W = m * g = 56.6294 kg * 9.806 m/sec² = 555.31N

A rigid vessel having a volume of 1.5m³ initially holds 5 kg of air under high pressure. If 10% of the mass of air leaks out what is the final density and specific volume of the air in the vessel?

final mass = 5kg (0.10)(5) = 4.5kg

final density = m/v = 4.5kg / 1.5m³ = 3 kg/m³

specific volume = 1/ρ = 0.333 m³/kg

An open water tank is filled to a depth of 1.5m. What is the absolute pressure of the fluid at the bottom of the tank?

P = ρgh = 1000 kg/m³ * 9.806 m/sec² * 1.5m = 14709 N/m²

= 14.709 KPa

Pabs = P + Patm = 14.709 + 101.325 = 116.034 KPa abs

Define a new scale, say °N, in which the boiling and freezing points of water are 1000°N and 100°N, respectively. Correlate this scale with the Celsius scales.

N-100 / 900 = c / 100 N – 100 = 900C / 100

N = 9°C + 100

A pump discharges 20kg/min of diesel fuel (S.G = 0.8) to a vertical cylindrical tank 3 meters high and 1 meter in diameter. Calculate the time (hrs) to fill the said fuel tank.

V_tank = п/4 (d)²h= 0.7854 (1)² (3) = 2.356m³

V flow rate = mass flow rate / density

= 20 kg/min / 0.8(1000 kg/m³

= 0.025 m³/min

Time = V_tank / V_rate = 2.356 m³ / 0.025 m3/min = 94.24 min = 1.57 hrs

A nuclear submarine is 200m below the surface of the sea. What is the pressure (bars) exerted on the hull (outer casing by the water at this depth?

P = p * g * h = 1000 kg/m³ * 9.81 m/sec² * 200m = 1,962,000 Pa

= 1962 KPa * (1bar / 100KPa) = 19.62 bars

An open tank 1 meter in diameter and 1.5m in height is filled with diesel oil (S.G = 0.85). Assuming 75% of its capacity is occupied by the diesel, calculate the absolute pressure of the fluid at the bottom of the tank.

P_bottom = ρ * g * h = (0.85) (1000) (9.81) (1.5) (0.75)

= 9380.81 Pa

= 9.3808 KPa

P_abs = 101.325 + 9.3808 = 110.706 KPa

A 600 kg hammer of a pile driver is lifted 2 meters above a piling head. If the hammer is released, what will be its velocity at the instant it strikes the piling?

PE = mgh = (600)(9.81)(2) = 11,772 Nm

KE = ½(m)v² v² = 2KE/m = 2(11,772) / 600 = 39.24

v = 6.26 m/sec

A girl weighing 470N holds suspended on the end of a rope 5 meters long. What will be her gain in potential energy when a friend swings her to one side so that the rope makes an angle of 35° with the vertical?

X = 5 cos 35º = 4.095 m h = 5 – 4.095 = 0.905 m

ΔPE = mgh = Wh = 470(0.905) = 425.35 Nm

A motor driven pump transfer 5000 liters of oil (SG = 0.8) per hour through an elevation of 16 meters. If the delivery velocity is 10 m/sec, what is the input power of the pump?

m = ρV = 0.8(1000)(5) = 4,000 kg/hr = 1.11 kg/sec

ΔPE = mgh = [1.11 kg/s](9.81m/s²)(16m) = 174.23 Watts

ΔKE = ½(m)v² = ½(1.11)[10]² = 55.5 Watts

P_T = 174.23 + 55.5 = 229.73 Watts = 0.23 kW

Pinput = PE = mgh = 0.8(1000)(5)(9.81)[16 + 10²/2(9.81)] = 827,840 J/hr

= 0.23 kW

TDH = [P_d – P_s] / γ + [V_d – V_s] / 2g + Z = [0] / γ + [V_d – 0] / 2g + Z

= 0 + [(10)² /2(9.81)] + 16 = 5.097 + 16 = 21.097 m

Wp = Q x γ x TDH x S.G. = (5/3600)m³/s)[9.81 kN/m³] 21.907m (0.8)

= 0.23 kW

A tank contains 25 liters of lubricating oil having a relative density of 0.9 and a temperature of 10°C. The oil is heated for 15 minutes by a 2.16 kw electric heater. Determine the final temperature of the oil assuming heat loss through the tank as 160w. (specific heat of oil is 2 KJ/kg-K)

Given:

Heating Power = 2.16kw V_oil = 25 liters = 0.025 m³

SG_oil = 0.9 ti = 10°C + 273 = 283K

Time = 15 min = 900 sec Q_L = 160W = 0.16 kw

C_oil = 2 KJ/kg-K

m_oil = ρ_oil(v) = 0.9(1000)(0.025) = 22.5 kg

Recall: Q = m c ΔT

ΔT = (Heating power – Heat loss)time / m(c)

= (2.16 – 0.16)900 / 22.5(2) = 40ºK = 40ºC

t_f = 10ºC + 40ºC = 50ºC

A power station uses oil-fired boilers which are supplied with fuel from four cylindrical oil storage tanks, each 15m in diameter and 10m high. The calorific value of the oil is 38000 KJ/kg and its relative density is 0.80. The output from the power station is 60MW. Assuming that only 15% of the heat input is converted into electrical energy output, how long will the boilers run on the oil in the tanks?

V_ST = 0.7854 (15)²(10) = 1767.15m³ V_T = 4 V_ST = 7068.6 m³

HV_f= 38,000 kJ/kg_f S.G._f = 0.80

Output Power = 60 MW

m_f = P_f x V_T = 0.80(1000)(7068.6) = 5,654,880kg

Q_f = m_fxHV_f= 5,654,880kg (38,000kJ/kg) = 2.149 x 10¹¹ kJ

Power = Work / time = Q_f / time

time = Q_f / Power = 0.15(2.149 x 10¹¹ kJ) / [60,000(3600)] = 149.24 hrs

An elevator and its contents have a mass of 1500kg. What is the electrical power input to the lifting motor if the elevator is raised through a height of 36 meters in 30 seconds? Assume that only 85% of the electrical power input to the motor is available to lift the elevator.

m = 1500kg h = 36m t = 30 seconds

PE = mgh = 1500 kg(9.81 m/sec²)(36m) = 529,740 N-m

P_Lift = 529,740 J / 30sec = 17,658W = 17.658 kW

P_input = P_lift/ 0.85 = 17.658 kW / 0.85 = 20.77 kW

There are 400 kg/min of water being handled by pump. The lift is from a 20-m deep well and the delivery velocity is 15 m/sec. Calculate the required power of the pumping unit in Kw.

ΔPE = mgh = (400/60)kg/s [9.81m/s²] 20m = 1,308 Watts

ΔKE = ½(m)v² = ½(400/60)kg/sec [15m/s]² = 750 Watts

Power = ΔPE + ΔKE = 1,308 + 750 = 2,058 Watts = 2.058 kW

TDH = [P_d – P_s] / γ + [V_d – V_s] / 2g + Z = [0] / γ + [V_d – 0] / 2g + Z

= 0 + [(15)² /2(9.81)] + 20 = 11.47 + 20 = 31.47 m

Wp = Q x γ x TDH x S.G. = (0.4/60)m³/s [9.81 kN/m³] 31.47 m

= 2.058 kW

A car having a total mass of 1500 kg and traveling at 100 kph is made to stop by applying brakes. All of the kinetic energy of the car is converted into internal energy of the brakes. Assuming each of the car’s four wheels has a steel disc brake with a mass of 10kg, what is the temperature change of the brakes? Specific heat of steel = 0.45 KJ/ kg-K.

Vel = 100 km/hr = 27.78 m/sec

KE = ½(m)v = ½(1500 kg)(27.78 m/s)² = 578,796.3 Nm

KE = Q = mcΔT

ΔT = Q / mc = 578.7963 kJ / 40 kg (0.45 kJ/kg-K) = 32.16ºK

A dam holds 200,000 m³ of water at a height of 150 m above a valley floor. If the water is used to drive a hydraulic turbine generator situated on the valley floor, what is maximum energy that can be generated? If the water is transported to the turbine through a frictionless pipe, what is the maximum velocity of the jet of water leaving the pipe?

PE = mgh = 2*10⁸ kg (9.81 m/sec²)(150m) = 2.943 * 10¹¹N•m

where:

m = PV = 1000 kg/m³ (200,000 m³) = 2 * 10⁸ kg

PE = KE : KE = ½ mV²

V = √2KE/m = √2(2.943*10¹¹)/2*10⁸ = 54.25 m/sec

An airplane has a mass of 196,000 kg and carries 400 passengers having an average mass of 60 kg each. What is the total energy of the loaded airplane when traveling at 820 km/hr at a height of 9,500 m above the sea level?

Total Energy = PE + KE = 2.62 x 10¹⁰ joules = 26,200 MJ

where:

PE = mgh

mgh = (196,000 + 24,000) kg * 9.81 m/sec² * 9,500m = 2.0503 * 10¹⁰ N•m

KE = ½ mv²

½ mv² = ½ (220,000) (820 km/hr * hr/3600sec * 1000/1km)² = 5707098765 N•m

A motor driven pump transfers 5,000 liters of oil per hour through an elevation of 16 m. If the specific gravity of the oil is 0.8, what is the input power to the pump?

P = Fh = mgh = 4,000 kg/hr * hr/3600sec * 16 = 17.78 N•m/sec = 17.78 Watts

where:

Poil = S.G * Pw = 0.8(1000kg/m³) = 800 kg/m³

Moil = Poil Vol = 800 kg/m³(5000 L * 1 m³/1000L) = 4,000 kg

The compressor in a commercial refrigerator requires a power input of 1,200 W when rotating at 1,420 rpm. What is the driving torque exerted on the compressor shaft?

P = torque * ω; Torque = P/ ω = (1,200 kJ/sec) / (148.7 rad/sec) = 8.07 kJ

where:

ω = 1420 rev/min * min/60 sec * 2п rad/ rev = 148.7 rad/sec

In an industrial process, a heater transfers 12 kW of power into a tank containing 250 liters of a liquid which has a specific heat capacity of 2.45 kJ/kgK and a density of 784 kg/m³. Determine the temperature increase after 5 minutes assuming there is no heat loss from the tank.

Q = mc Δ T

where:

Q = 12 kw

m = ρ * V = 784 kg/m³ * 250 L * 1m³/1000L = 196 kg

Δ T = Q/mc = (12kJ * 5min * 6sec/min) / (196* 2.45 kJ/kg. k) = 7.5 ^oK

A 50 kw electric furnace measures 1.2m x 1.0m x 0.8m. When the temperature inside the furnace is 1250^oC, a block of aluminum with a mass of 300kg and a temperature of 16.5^oC is placed inside. Assuming the heat loss from the furnace walls is 500 w/m², how long will it take to heat the block to the furnace temperature? (specific heat of aluminum is 0.9 kJ/kg. K)

0.8m t_furnace = 1250^oC = 1523K

t_al= 16.5^oC = 289.5K

1.0m m_al = 300 kg

1.2m q_L = 500 W/m²

Q_L= 500 W/m² * 5.92m²

= 2, 960 W

= 2.96 kW

A₁ = 2 (1.2 x 1) = 2.4 m²

A₂= 2 (1.2 x 0.8) = 1.92m²

A₃= 2 (1 x 0.8) = 1.6 m²

A_T = 5.92 m²

Q = mc Δ T

(50 – 2.96)t = (300)(0.9)(1233.5)

t = 333,045/47.04 = 7,080 sec

= 1.9666 hr

= 1 hr and 58 mins

A 12-ft³ tank contained hydrogen sulfide gas at 150 psia and 60’F after 5 lbs of the gas had been drawn out. Before any gas left the tank, the temperature was 70’F. What mass of the gas was in the tank originally and what was its pressure?

a. 222 b) 422 c) 242 d) 202

m_L= PV/RT = 150(12) 144/45.33(520 = 10.996 lbs.

m_o = m_L + 5 = 10.996 + 5 = 15.996 lbs.

P_o = m_oRT/V = 15.996(45.33)(530)/12(144) = 222.4 psia

A closed vessel A contains 3 ft³ of air at 500 psia and a temperature of 120’F. This vessel connects with vessel B, which contains an unknown volume of air at 15 psia and 50’F. After the valve separating the two vessels is opened, the pressure and temperature of the mixture are 200 psia and 70’F. Calculate the volume of vessel B.

b. 6.14 b) 2.42 c) 4.16 d) 5.5

m_m= m_A +m_B

P_m(V_A + V_B)/RT_M = P_AV_A/RT_A+ P_BV_B/RT_B V_B = 4.16FT³

Air undergoes an isentropic change of state from 300K and 110Kpaa to a final pressure of 550 Kpaa. Find the change in temp.

c. 175.15 b. 395.71 c. 751.39 d. 917.31

T₂/T₁ = (P₂/P₁)^K-1/K

T₂= 300 x (550/110)^0.4/1.4 = 475.14;

∆T = T₂– T₁= 475.14 – 300 = 175.15 ⁰K

An air-standard Diesel Cycle operates with a compression ratio of 15. The maximum cycle temperature is 1,700ºC. At the beginning of compression the air is at 100 Kpaa and 40ºC. What is the cut-off ratio?

d. 2.13 b. 1.32 c. 3.21 d. 2.87

T₂ = T₁r_k^K-I = (313 x15) ^0.4= 924.657 ºK

r_c = V₃/V₂ = T₃/T₂ =197.3/924.657 = 2.13

Steam enters a nozzle steadily with a velocity of 100 m/s and enthalpy of 1,300 KJ/kg and leaves with an enthalpy of 1,100 KJ/kg. Neglecting heat transfer, what is the velocity of steam at the nozzle outlet?

e. 720 b. 640 c. 580 d. 490

PE₁ + KE₁ + H₁ = PE₂ + KE₂ +H₂

∆KE = H₁ – H₂ = – ∆H

∆KE = – (1,100 – 1,300) = 200 kJ/kg

∆KE = 200,000 J/kg

∆KE = ½ (V₂² –V₁²)

(V₂² – V₁²) = 2 (∆KE) = 400,000 J/kg

V₂²= 400,000 + (100)² = 410,000

V₂= (410,000)^1/2 = 640.3 m/sec

An air compressor handles 8.5 m³/min of air with a density of 1.26 kg/m³ and a pressure of 1 atm, and it discharges at 445 Kpa (Gage) with a density of 4.86 kg/m³. The change of internal energy is 82 KJ/kg and the heat loss by cooling is 24 KJ/kg. Neglecting changes of kinetic and potential energies find the power in kw.

f. –16.4 b) –24.6 c) –32.4 d) –41.2

P = Q – ∆PE – ∆KE – ∆WK_f– ∆U

Wherein: m_r = 1.26 (8.5)(1/60) = 0.1785 kg/sec

Q_R = – 24(0.01785) = – 4.284 kw

∆U = 82(0.1785) = 14.637 kw

WK_f1 = P₁ x (1/ρ₁) x m_r= (101.325) x (1/1.26) x (0.1785) = 14.35 kw

WK_f2 =P₂ x (1/ρ₂) x m_r = (546.325) x (1/4.86) x (0.1785) = 20.066kw

∆WK_f = 5.716 kw

P = – 4.284 – 5.716 – 14.637 = –24.64 kw

A spark-ignition engine produces 224 kW while using 0.0169 kg/s of fuel. The fuel has a higher heating value of 44,186 KJ/kg, and the engine has a compression ratio of 8. The frictional work lost is found to be 22.4 kW. Determine the indicated thermal efficiency.

g. 11 b. 22 c. 33 d. 44

e_i₌WK_i/ Q_A= WK_i/ m_f x H_V = (224 + 22.4) / (0.0169 x 44,186) = 33%

The volume in the clearance space of a 152.40 mm by 254.0 mm Otto gas engine is 1.70 liters. Find the ideal thermal efficiency of the engine on the standard air basis, if the exponent of the expansion and compression lines is 1.35. Express in percent.

a) 38.55 b) 36.89 c) 26.98 d) 35.95

V_D =π /4 (0.1524)²(0.254) = 0.004633m³ ;

C = V₂/ V_D = 0.0017/0.004633 = 36.69%

r_k = (1+ C)/C = (1 +0.3669)/0.3669 = 3.726

e = 1 –1/r_k ^k-1 = 1 – [1/3.726] ^0..35 = 36.89%

Calculate from the following data the kg of air per kg of fuel used by an automobile engine: air temperature 21.50ºC, barometer 767.0 mm Hg, air entering 1.75 cu.m/min., measured gasoline 14.25 liters per hr, specific gravity of gasoline 0.735.

h. 12.125 b) 15.45 c) 14.45 d) 21.45

m_a =PV/RT = 767 x 101.325/760) x 1.75/ 0.287)(294.5) = 2.11 Kg/min

m_f = ρ x v =0.735 (1000) x (14.25)/1000(60) = 0.1746 kg/min

r_a/f = 2.11 / 0.1746 = 12.125

If 2 kg metal with 90 ºC temperatures is immersed into 4 kg of water at 11 ºC, what will be the final temperature of the water after immersion? Cp (metal) = 0.913 KJ/kg-K

a) 18.77 C b) 19.88 c) 20.55 d) 21.22

(2)(0.913)(90 – T_f) = 4(4.4187)(T_f – 11)

1.826 (90 – T_f) = 16.748 (T_f – 11)

164.34 – 1.826 T_f = 16.748T_f – 184.228

T_f= 348.568 /18.574 = 18.77 ⁰C

A steam engine which takes steam at 689.30 KPa dry and saturated and exhausts to atmospheric pressure is tested for steam rate while developing 25 hp at the brake. It is found that 99.75 kg of steam were used during a 15-minute test. Enthalpy of throttle steam is 2,761.30 kj/kg ; 2,435.60 kj/kg at exhaust pressure after isentropic expansion; and 418.75 kj/kg for liquid at exhaust pressure. Determine the engine efficiency in percent.

i. 12.52 b) 15.45 c) 61.5 d) 51.64

M s = 99.75 /15 = 6.65 kg/min

W_I = m_s(∆h) = 6.65/60 ( 2761.3 – 2435.6 ) = 36.098 kw = 48.389 hp

ή = W_B/W_I= 25 / 48.389 = 51.66%

In an oil refinery, cylindrical tank of 20 m in diameter and 15 m height is used to store crude oil. A process technician notes down the following data.

Time: 8:00 AM

Oil level in the tank 7.00 m

Flowmeter reading at inlet 7,430 litre/minute

Flowmeter reading at outlet 305 m³/hour

Assuming flow rates remain steady, estimate the level in the tank at 10:00 AM.

j. 9.7 b) 7.9 c) 8.5 d) 10.0

Inlet : 7,430 liters/min = 7.43 m³/min = 445.8 m³ /hr

Outlet : 305,000 liters/hr = 305 m³/hr

Volume in after 2 hrs. = 891.6 m³

Volume out = 610.0 m³

Volume added = 281.6 m³

H = V/A = 281.6 m³/ [π/4](20)² = 0.8964 m

@ 10:00 am level in the tank = 7 + 0.8964 = 7.8964 m

A piston cylinder assembly contains 0.25 kg of air with an internal energy of 30kJ/kg. It is compressed to a higher pressure and the final internal energy is 70kJ/kg. During compression, there is a heat loss of 2,000 Joules. Determine the work input needed for compression.

k. – 32kJ/kg b) – 55 c) – 102 d) – 72

Wk + ∆U = Q

WK = Q – ∆U = [2kJ /0.25kg] – (70 – 30)kJ/kg = 8 – 40 = – 32 kJ/kg

What retarding force is required to stop a 0.45 caliber bullet of mass 20 grams and speed of 300m/sec as it penetrates a wooden block to a depth of 2 inches?

A. 17,716 N B. 19,645 C. 15,500 D. 12,500

x = 2 in = 0.0508 m

Energy -- Work relation:

Work = KE

F(x) = ½ mV²

F(0.0508) = ½(0.02)(300)² ;; F = 17,716.53 N

A 50,000 N car traveling with a speed of 150 km/hr rounds a curve whose radius is 150 m. Find the centripetal force.

A. 70 kN B. 25 C. 65 D. 59

Vel = 150 kph = 41.67 m/sec

An = V² / r = (41.67)² / 150 = 11.576 m / sec²

Fc = m An = (W/g)An = (50,000 / 9.81) x 11.576 = 59,000 N = 59 kN

V = 150 kph = 41.67 m/sec

Fc = WV²

Fc = (50,000)(41.67)² = 59,000 N = 59 kN

9.81(150)

Assuming an isothermal condition, how deep at a point where an air bubble,

upon reaching the surface, has 8 times the volume than it had at the bottom?

A. 72.3 m B. 73.3 m C. 71.3 m D. 75.5 m

P₁V₁ = P₂V₂ 101.325 x V₁= [101.325 + δh] V₂

101.325 x 8V₂ = [101.325 + 9.81h]V₂

h = [8(101.325) – 101.325] / 9.81 = 72.3 m

Electrical resistance of 7 ohms and 11 ohms are connected in parallel and the combination of which is then connected in series with a resistance of 15 ohms and a source of 110 volts. Calculate the total current flowing in the circuit.

a) 0.57amp b) 5.7 c) 57 d) 0.057

R_T = 15 + 1 / (1/7 + 1/11) = 19.28 ohms I = E / R = 110 / 19.28 = 5.7 amp

Two wires of the same length and materials have resistances of 10 ohms and 8 ohms, respectively. If the diameter of wire A is 5mm, what is the diameter of wire B?

a) 5.59mm b) 5.95 c) 9.95 d) 9.66

Resistivity constant, K = RA / L

Condition: same length and material

Therefore; R₁A₁ = R₂A₂ A₂ = A₁R₁ / R₂ = 24.54 mm² d₂ = √4A/π = 5.59mm

A battery having an emf of 1.5 volts and internal resistance of 0.2 ohm is connected to a

resistance of 300 ohms. Find the current flowing through the circuit?

a) 0.000499amp b) 0.00499 c) 0.0499 d) 0.499

E = IR + Ir = I(R + r) I = E / (R + r) = 1.5 / 300.2 = 0.004996 amp

What current would flow for three hours through a resistance of 50 ohms to produce enough heat to raise the temperature of one kg water from the freezing point to boiling point?

a) 0.881amp b) 8.81 c) 1.88 d) 88.1

Q = mc∆T = 1(4.187)100 = 418.7 kJ in 3 hrs

Heat rate = 418.7 / 3 (3600) = 0.0388 kW = 38.8 W

Therefore; I = √ P/R = √38.8/50 = 0.881 amp

A hoist motor draws 12 amp from the service mains of 240 volts. If it lifts 1,000 kg of load at the rate of 0.25 m/sec, determine the power used by the motor.

a) 2.5W b) 25 c) 2450 d) 2500

Power = Wk / time = (F x S) / t = F x V = 1000 x 9.81 x 0.25 = 2452.5 J / sec

Calculate the approximate mass density of methane gas if the pressure and temperature are one atmosphere and 15.55º C.

a) 675.8kg/m³ b) 6.758 c) 67.58 d) 0.6758

ρ = m / v = p / RT = 101.325 / (0.5196 x 288.55) = 0.6758 kg/m³

One kg of water initially at 15ºC is to be freezed at –3ºC. Calculate the total amount of heat to be removed during the cooling process?

a) 39.831kJ b) 403 c) 3983.1 d) 39831

Q₁ = mc∆T = 1 (4.187)15 = 62.8 Q₂ = 334

Q₃ = 1(0.5)(4.187)3 = 6.28 Q_T = 403Kj

Heat is supplied to 20 pound-mass of ice at 32ºF at the rate of 160 Btu/sec. How long will it take to melt the ice to water at 50ºF?

a) 22.5 sec b) 20.25 c) 30.25 d) 25.2

Q₁ = 20(144) = 2880 Btu Q₂ = 20(1) (50 – 32 ) = 360

T = 3240 / 160 = 20.25sec

A hydraulic turbine receives water from a reservoir at an elevation of 110 m above it. What is the minimum water flow in kg/s to produce a steady turbine output of 30 MW?

l. 27,830 b. 28,700 c. 72,800 d. 87,200

WP = Q x H x γ x S.G

Q = WP / H x γ x S.G.

= 30,000 KN-m /sec / 110m x 1000 kg/m³ x 9.81 m/sec² = 27.8 m³/sec

M_R = 27.8m³/sec x 1000 kg/m³ = 27,800 kg/sec

A stone weighs 105 lbs in air and 83 lb in water. Find the specific gravity of the stone.

A. 2.98 B. 0.35 C. 4.77 D. 2.21

S.G. = Wair / (Wair – Wwater) = 105 / (105 – 83) = 4.77

Volume(stone) = Volume(water displaced) = (105 – 83) / 62.4 = 0.3526 ft³

Specific Weight(stone) = 105 / 0.3526 = 297.82 lb/ft³

Specific Gravity(stone) = 297.82 / 62.4 = 4.77

Determine the friction head loss for fully developed laminar flow of ethylene glycol at 40ºC(ρ = 1,101 kg/m³) through a 5 cm diameter, 50 m long pipe, if friction factor is 0.242 and flowrate of 0.1 kg/s.

A. 6.42 cm B. 2.64 cm C. 3.46 cm D. 5.32 cm

Solution

Q = A x V ; [1 kg/s / 1,101 kg/m³ ]= 0.7854 (0.05)² m² (V); Velocity = 0.04626 m/s

Frictional head loss, hf = fLV² / 2gD

hf = 0.242(50)(0.04626)² / 2(9.81)(0.05) = 2.64 cm

The sun generates 1 kW/m² when used as a source for solar collections. A collector with an area of 1 m² heats water. The flow rate is 3.0 liters/min. What is the temperature rise in the water? The specific heat of water is 4,200 J/kg-°C.

A. 2.76 B. 3.76 C. 4.76 D.5.76

Q = 1 kW / m² x 1m² = 1 kW = 1000 W

m = 3 li / min x 1 kg / li x 1min / 60 sec = 0.05 kg / sec

Q = m Cp Dt 1000 = 0.05 (4,200) Dt Dt = 4.76°C

Steam with an enthalpy of 335 kJ/kg enters a nozzles at a velocity of 80 m/sec. Find the velocity of the steam at the exit of the nozzle if its enthalpy is reduced to 314 kJ/kg, assuming the nozzle is horizontal and disregarding heat losses. Take g = 9.81 m/sec²

A. 220 B. 652 C. 265 D. 625

SOLUTION:

Given: h ₁= 335kJ / kg V₁ = 80 m / sec h₂= 314 kJ / kg

By energy balance (first law of thermodynamics);

h₁ + V₁² / 2 = h₂ + V₂² / 2

[(V₂)²/2] – [(V₁)²/2] = h₁ – h₂

(V₂)² = (V₁)² + 2(h₁ – h₂)

_______________

V₂ = √ (V₁)² + 2(h₁ – h₂)

__________________________

= √ (80)² + [2(335,000 – 314,000)]

= 220 m / s

A refrigeration system operates on the reverse Carnot cycle. The minimum and maximum temperatures are minus 25ºC and plus 72ºC respectively. If the heat rejected to the condenser is 6,000 kJ/min., draw the T – S diagram and find: Tons of refrigeration developed

A. 24.04 B. 20.44 C.17.14 D. 5.08

SOLUTION:

T₁ = – 25 + 273 = 248 K T₂ = 72 + 273 = 345 K

(S₂ – S₃) (345) = 6,000 kJ/min ; but S₂ = S₁ and S₃ = S₄; (S₁ – S₄) = 6,000 / 345

Refrigerating Effect = (S₁ – S₄) T₁ = (6,000 / 345) (248) = 4,313 kJ/min

Power Input Required = 6,000 – 4,313 = 1,687 kJ/min = 28.12 kW

Tons of Refrigeration = (4,313 / 60) / 3.516 = 20.44 tons ref.

A certain gas at 101.325 kPa and 16°C whose volume is 2.83 m³are compressed into a storage vessel of 0.31 m³capacity. Before admission, the storage vessel contained the gas at a pressure and temperature of 137.8 kPa and 24°C; after admission the pressure has increased to 1,171.8 KPa. What should be the final temperature of the gas in the vessel in Kelvin?

A. 298.0 B. 319.8 C. 180.0 D. 420.0

Solution:

Solving for the mass of gas which is to be compressed:

PV = mRT ; 101.325 (2.83) = m₁ (R) (16+273) ; m₁ = 0.9922/R kg

Solving for the mass of gas initially contained in the vessel:

PV = mRT 137.8 (0.31) = m₂ R (24 + 273) ; m₂ = 0.1438/R kg

solving for the final temperature

PV = mRT

1,171.8 (0.31) = RT

T = 319.8 °K

An engine has an efficiency of 26%. It uses 2 gallons of gasoline per hour. Gasoline has a heating value of 20,500 Btu/lb and a specific gravity of 0.8. What is the power output of the engine?

A. 20.8Kw B. 29Kw C. 19.8Kw D. 18.9Kw

Solution:

A diesel power station is to supply power demand of 30 KW. If the overall efficiency of the power generating unit is 40%, calculate the amount of diesel oil required, kg/hr

A. 4.37 B. 6.37 C. 5.37 D. 7.37

Note: The calorific value of fuel oil used is 12,000 kcal/kg. 1 kWh = 860 kcal

Efficiency = Output / Input = 30 / 0.4 = 75 kW

Input per hour = 75 kW x 1 hr = 75 kWh

Since 1 kWh = 860 kcal then 75kWh x 860 kcal / kWh = 64,500 kcal

Fuel oil required = 64,500 kcal / 12,000 kcal/kg = 5.37 kg

A supercharged six-cylinder four stroke cycle diesel engine of 10.48 cm bore and 12.7 cm stroke has a compression ratio of 15. When it is tested on a dynamometer with a 53.34 cm arm at 2500 rpm, the scale reads 81.65 kg, 2.86 kg of fuel of 45,822.20 kJ/kg heating value are burned during a 6 min test, and air metered to the cylinders at the rate of 0.182 kg/s. Find the brake thermal efficiency.

A. 0.327 B. 0.367 C. 0.307 D. 0.357

Solution:

T = 81.65 (0.00981) (0.5334) = 0.42725 kN-m

Brake Power = 2pNT = 2p(0.42725 kN-m) (2500/60)

Brake Power = 111.854 kw

mf = 2.86 / 6(60) = 0.00794 kg/sec

Brake thermal efficiency = brake power/ mfQh

= 111.854 kw / [0.00794 kg/sec x 45,822.20 kJ/kg]

brake thermal efficiency = 0.307 = 30.7%

A diesel engine is operating on a 4-stroke cycle, has a heat rate of 11,315.6 kJ/kWh brake. The compression ratio is 13. The cut-off ratio is 2. Using k = 1.32, what is the brake engine efficiency?

A. 63.5 B. 51.2 C. 73.5 D. 45.3

Solution:

e = cycle efficiency = W / Qa = 1 – 1 r_c^k – 1

r_k^{k – 1} k(r_c –1)

= 1 2 ^1.32– 1 = 50.1%

13 ^0.321.32(2 –1)

e_b = brake thermal efficiency = W_b / Q_a= 3600 / HR_b = 3600 / 11,315.6 = 31.81%

hb = brake engine efficiency = Wb / W = e_b / e = 0.3181 / 0.501 = 63.5%

A hydroelectric generating station is supplied from a reservoir of capacity 6,000,000 m³ at a head of 170 m. Assume hydraulic efficiency of 80% and electrical efficiency of 90%. The fall in the reservoir level after a load of 15 MW has been supplied for 3 hours, if the area of the reservoir is 2.5 sq. km is closest to:

A. 5.39 cm B. 5.98 C. 4.32 D. 4.83

Solution:

Output = Q w H h_h h_e

15,000 = Q (9.81) (170) (0.80) (0.90)

Q = 12,492 m³/ s

In 3 hours, volume of water consumed

= 12,492 m³/ s (3hrs) (3600sec/hr)

= 134,914 m³

Volume = area x height

134,914 m³ = (2.5 x 10⁶ m²) H

H = 0.0539m = 5.39cm

23.5 kg of steam per second at 5 MPa and 400°C is produced by a steam generator. The feedwater enters the economizer at 145°C and leaves at 205°C. The steam leaves the boiler drum with a quality of 98%. The unit consumes 2.75 kg of coal per second as received having a heating value of 25,102 kJ/kg. What would be the overall efficiency of the unit in percent?

A. 65 B. 95 C. 88 D. 78

Steam properties:

At 5 Mpa and 400 C; h= 3,195.7 kJ/kg

At 5 Mpa; hf = 1,154.23 hfg = 1,640.1

At 205C; hf = 875.04

At 145C; hf = 610.63

Solution:

Overall efficiency = heat absorbed / heat supplied

= m_s (h_s – h_f) / m_f (Q_h) = 23.5 (3,195.7 – 610.63) / 2.75(25,210) = 88%

A hydro-electric plant having 50 sq. km reservoir area and 100 m head is used to generate power. The energy utilized by the consumers whose load is connected to the power plant during a five-hour period is 13.5 x 10⁶ kwh. The overall generation efficiency is 75%. Find the fall in the height of water in the reservoir after the 5-hour period.

A. 2.13 gpm B. 1.32 C. 3.21 D. 0.53

Solution:

Q = flow in m³/sec

Energy Output = Q(r)H x n_T x Time

13.5 x 10⁵ = Q(9.81)(100)(0.75)(5)

Q = 3,669.725 m³/sec

In 5 hours, the volume of water consumed:

V = 3,669.725(5 x 3600) = 66,055.050 m³

Volume = Area x Height

66,055,050 = (50 x 10⁶)h h = 1.321 m

A gas turbine unit operates at a mass flow of 30 kg/s. Air enters the compressor at a pressure of 1 bar and temperature of 15°C and is discharged from the compressor at a pressure of 10.5 bar. Combustion occurs at constant pressure and results in a temperature rise of 420°K. If the flow leaves the turbine at a pressure of 1.2 bar, determine: the net power output from the unit, KW

A. 4,388.74 B. 5,388.74 C. 6,388.74 D. 7,388.74

Take Cp = 1.005 kJ/kg °K and g = 1.4

T₂ = T₁ (P₂/P₁)⁽^g^-1)/^g = 288 x (10.5)^0.286 = 564.22°K

T₃ = 564.22 + 420 = 984.22°K

T₄ = T₃ / (P₃/P₄)⁽^g^-1)/^g = 984.22 / (10.5/1.2)^0.286 = 529.27 ºK

W_c = mCp(T₂ – T₁) = 30 x 1.005 x (564.22 – 288) = 8328 kW

W_T = mCp(T₃ – T₄) = 30 x 1.005 x (984.22 – 529.29) = 13716.74 kW

W_N = W_T – W_C = 13716.74 – 8328 = 5,388.74 kW

Calculate the enthalpy in KJ of 1.50 kg of fluid that occupy a volume of 0.565 m³ if the internal energy is 555.60 K cal per kg and the pressure is 2 atm abs.

A. 2,603.7 kJ B. 3,603.7 C. 4,603.7 D. 5,603.7

h = u + pv = [555.6 kcal x (1.055/0.252)] + 2(101.325)(0.565/1.5) = 2402.35 kJ/kg

H = mh = 1.5(2402.35) = 3,603.5kJ

A wheel starts from rest and acquires a speed of 500 rad/sec in 15 seconds. Calculate the approximate number of turns made by the wheel.

A. 119.4 rev B. 1194 C. 11,940 D. 119,400

N = 500 rad/sec (15sec)(1 rev/2π) = 1194.26 rev

Heat is supplied to 20 lbs of ice at 32ºF at the rate of 160 Btu/sec. If the heat fusion of ice is 144 Btu/lb, how long will it take to convert the ice to water at 50ºF?

A. 1.525 min B. 5.125 C. 2.515 D. 12.55

Q_T = Q₁ + Q₂ + Q₃ = 300(0.5)(4) + 300(144) + 300(1)(18) = 49,200 Btu

T = Q_T / Q_rate = 49,200 Btu / 160 Btu/sec = 307.5 sec = 5.125 min

A ball dropped from the roof of a building 40 m high will hit the ground with a velocity of:

A. 20 m/s B. 28 C. 38 D. 40

V = √2gh = √2(9.81)(40) = 28

Power and Industrial Plant Engineering

Mechanical Engineering Reviewer