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sdas [7]
2 years ago
8

Can someone help me with this maze shown below.

Engineering
1 answer:
Gnoma [55]2 years ago
3 0
We can’t see the maze
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Air at 27°C and a velocity of 5 m/s passes over the small region As (20 mm × 20 mm) on a large surface, which is maintained at T
JulsSmile [24]

Answer:

a) The maximum possible heat removal rate = 2.20w

b) Fin length = 37.4 mm

c) Fin effectiveness = 89.6

d) Percentage increase = 435%

Explanation:

See the attached file for the explanation.

5 0
3 years ago
An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 6 kW. De
Paraphin [41]

Answer:

a. 2.08, b. 1110 kJ/min

Explanation:

The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. <u>Assume that the air conditioner operates steadily.</u>

a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is

COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat

COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08

The COP of this air conditioner is 2.08.

b. The rate of heat discharged to the outside air is determined from the energy balance.

Q(H) = Q(L) + W(net in)

Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min

The rate of heat transfer to the outside air is 1110 kJ for every minute.

5 0
3 years ago
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

6 0
2 years ago
Seawater containing 3.50 wt% salt passes through a series of 11 evaporators. Roughly equal quantities of water are vaporized in
statuscvo [17]

Answer: the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr

Explanation:

F, W and B are the fresh feed, brine and total water obtained

w = 2 x 10^4 L/h

we know that

F = W + B

we substitute

F = 2 x 10^4 + B

F = 20000 + B .................EQUA 1

solute

0.035F = 0.05B

B = 0.035F/0.05

B = 0.7F

now we substitute value of B in equation 1

F = 20000 + 0.7F

0.3F = 20000

F = 20000/0.3

F = 66666.67 kg/hr

B = 0.7F

B = 0.7 * F

B = 0.7 * 66666.67

B = 46,666.669 kg/hr

the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr

8 0
3 years ago
Determine how much concrete you will need for a slab which is 50 feet by 30 feet wide and 1 foot thick
Levart [38]

Answer:

Amount of concrete need to make slab = 1,500 feet³

Explanation:

Given:

Length of slab = 50 feet

Width of slab = 30 feet

Height of slab = 1 feet

Find:

Amount of concrete need to make slab

Computation;

Amount of concrete need to make slab = Volume of cuboid

Volume of cuboid = (l)(b)(h)

Amount of concrete need to make slab = (50)(30)(1)

Amount of concrete need to make slab = 1,500 feet³

6 0
2 years ago
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