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kolezko [41]
3 years ago
14

A preheater involves the use of condensing steam at 100o C on the inside of a bank of tubes to heat air that enters at I atm and

25o C. The air moves at 5 m/s in cross flow over the tubes. Each tube is 1 m long and has an outside diameter of 10 mm. The bank consists of 196 tubes in a square, aligned array for which ST = SL = 15 mm. What is the total rate of heat transfer to the air? What is the pressure drop associated with the airflow?
(b) Repeat the analysis of part (a), but assume that the tubes have a staggered arrangement with ST = 15 mm and SL = 10 mm.
Engineering
1 answer:
Anna35 [415]3 years ago
8 0

Answer:

Please see the attached file for the complete answer.

Explanation:

Download pdf
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The snowmobile has a weight of 250 lb, centered at G1, while the rider has a weight of 150 lb, centered at G2. Ifh=3ft, determin
KATRIN_1 [288]

Answer:

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Explanation:

Please kindly check attachment for the step by step solution of the given problem.

5 0
4 years ago
Q1: The first option should always be to get out safely (RUN)
nekit [7.7K]

Answer:

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6 0
3 years ago
When an object is moving, we use the following coefficient for friction calculations a)-μk b)-μs c)-γk d)- γs
Reika [66]

Answer:\mu_{k}

Explanation:

We use kinetic friction when a body is moving i.e. \mu_{k} for calculations.

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Static friction value increases as we apply more force while kinetic friction occurs when there is relative motion between bodies.

3 0
4 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
A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

5 0
3 years ago
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