# Tutor profile: Destiny H.

## Questions

### Subject: Mechanical Engineering

Refrigerant (R134a) is flowing through a 0.5 inch diameter pipe at a rate of 3m/s. The pressure at this initial point is 15KPa. What would be the pressure if the tube were to widen to a 0.75 inch diameter? (density of R134a = 4.25kg/m^3)

We'll start off with Bernoulli's equation: (P1) + 0.5*ρ*(v1)^2 + ρ*g*(h1) = (P2) + 0.5*ρ*(v2)^2 + ρ*g*(h2) Because we have no change in height during this equation (h1 - h2 = 0), we can cancel the two terms 'ρ*g*(h1)' and 'ρ*g*(h2)'. Thus our new equation becomes: (P1) + 0.5*ρ*(v1)^2 = (P2) + 0.5*ρ*(v2)^2 We have 2 unknowns (P2 and v2) and 1 equation, so we'll need to find (v2) now using: Q = (v1)*(A1) = (v2)*(A2) The cross-sectional area of the 0.5 inch diameter tube can be found using the area of a circle: π*d^2 = π*0.5^2 (v1)*(A1) = (v2)*(A2) 3m/s * π*0.5in^2 = v2 * π*0.75in^2 1.3m/s = v2 Now we have all the information needed to solve the equation: (P1) + 0.5*ρ*(v1)^2 = (P2) + 0.5*ρ*(v2)^2 15KPa + 0.5*4.25kg/m^3*(3m/s)^2 = P2 +0.5*4.25kg/m^3*(1.3m/s)^2 9.5KPa = P2

### Subject: Physics

In a kitchen there are 5 appliances. The dishwasher uses 75W, the toaster uses 8W, the garbage disposal uses 20W, and the blender uses 10W. The dishwasher, toaster, garbage disposal, and blender run for 1 hour per day. The refrigerator uses 120W and runs 24 hours per day. How much electrical energy is consumed (in KW*hr) in this kitchen per year (365 days)?

Total the wattage of the dishwasher, toaster, garbage disposal, and blender to operate: 75W + 8W + 20W + 10W = 113W Multiply the the total wattage by the time in hours they will run in a day: 113W * 1hr = 113W*hr/day Multiply the W*hr used in a day by 365 days for the year: 113W*hr/day * 365days/year = 41,245W*hr/year Convert W to KW by dividing by 1000: 41,245W*hr/year / 1000 = 41.2KW*hr Now for the refrigerator multiply the wattage by the time in hours it will run in a day: 120W * 24hr = 2880W*hr/day Multiply the W*hr used in a day by 365 days for the year: 2880W*hr/day * 365days/year = 1,051,200W*hr/year Convert W to KW by dividing by 1000: 1051200W*hr / 1000 = 1,050KW*hr Add the two KW*hr values to get the total electrical energy consumption for the year: 41.2KW*hr + 1,050KW*hr = 1,090KW*hr

### Subject: Algebra

An electrician charges a one time service fee to every home she visits of $59. She then charges $145/hr for each hour she works on the project. Write an algebraic expression that represents the electrician's rate for 'x' hours.

The electrician charges a $59 one-time service fee. This number is not dependent on 'x' thus it will be added to the rate: '+59' For each hour the electrician works, her rate increases by $145. Thus 1 hour = $145 * 1hr, and 2 hours = $145 * 2hr and so on. Now adding the equation all together gives '59 + 145x = y' where 'y' = the electrician's total rate.

## Contact tutor

needs and Destiny will reply soon.