Engineering Answers

Question 1

A horizontal "hot air blower" consists of a fan, an electrical heating element and a nozzle

The power input to the fan is 50 J/s and the power input to the heating element is 1.44 kJ/s. The air enters the fan with negligible velocity at a pressure of 1 bar and a temperature of 20 C, the volumetric flow rate being 0.5 m3 /min. After passing over the heating element, the air leaves the nozzle with a velocity of 25 m/s and a pressure of 1.1 bar. Treating the complete arrangement as a steady flow system and ignoring any heat loss from the system, determine:

(a) The mass flow rate of air entering the fan in kg per second (b) The kinetic energy of the air leaving the nozzle per second

(c) The change in enthalpy of the air occurring per second between entering the fan and leaving the nozzle

(d) The temperature of the air as it leaves the nozzle. (e) The density of the air as it leaves the nozzle

(f) The flow area at the nozzle exit.

Take R = 0.287 kJ/kgK and cp = 1.005 kJ/kgK

5.) A student will receive P3,000 at the beginning of each 3 months for 4 years. What is the sum of

this annuity at the end of the 4th year if the interest rate is 6% compounded quarterly?

Ans. 54,604.07

5.) A student will receive P3,000 at the beginning of each 3 months for 4 years. What is the sum of

this annuity at the end of the 4th year if the interest rate is 6% compounded quarterly?

Ans. 54,604.07

4.) An engineer is entitled to receive P25,000 at the beginning of each year for 18 years. If the rate of

interest is 4% compounded annually.

1. What is the present value of this annuity at the time he is supposed to receive the first

payment?

2. What is the sum of this annuity at the end of the 18th year?

3. Find the difference between the sum of this annuity which is paid at the beginning of

each year and an annuity paid at the end of each year.

Ans. 1. P329,141.72 2. P666,780.73 3. P25,645.41

2.) A businessman is faced with the prospect of fluctuating future budget for the maintenance of the

generator. During the first 5 years, P1,000 per year will be budgeted. During the second five

years, the annual budget will be P1,500 per year. In addition, P3,500 will be budgeted for an

overhaul of the machine at the end of the fourth year and another P3,500 for an overhaul at the

end of 8th year. Assuming compound interest at 6% per annum, what is the equivalent annual cost

of maintenance?

Ans. P1,888.87

3.) For some interest rate “i” and some number of interest periods “n” the uniform series capital

recovery factor is 0.3091 and the sinking fund factor is 0.1941.

1. Compute the interest rate.

2. If the interest rate i = 12% compounded annually, compute the capital recovery factor

for a period n = 5 years.

3. If the rate of interest i = 16% compounded quarterly what would be the sinking fund

factor for a period of 6 years.

Ans. 1. 11.5% 2. 0.27741 3. 0.02559

2.) In an ordinary annuity (uniform series of payments) if the nominal rate of interest is 8%

compounded quarterly for 6 years.

1. Compute the value of the capital recovery factor.

2. Compute the value of the sinking fund factor.

3. Compute the value of the present worth factor.

Ans. 1. 0.052871 2. 0.032871 3. 18.914

1.) For a uniform series of payments what is the value of

1. (F/A, 8%, 6) if it is compounded quarterly

2. (P/A, 6%, 8) if it is compounded semi-annually.

3. (F/A, 8%, 4) if it is compounded continuously.

Ans. 1. 30.42186 2. 12.5611 3. 4.528047

4.) An engineer is entitled to receive P25,000 at the beginning of each year for 18 years. If the rate of

interest is 4% compounded annually.

1. What is the present value of this annuity at the time he is supposed to receive the first

payment?

2. What is the sum of this annuity at the end of the 18th year?

3. Find the difference between the sum of this annuity which is paid at the beginning of

each year and an annuity paid at the end of each year.

Ans. 1. P329,141.72 2. P666,780.73 3. P25,645.41