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DochEvi [55]
3 years ago
6

11.A heat engine operates between two reservoirs at 800 and 20°C. One-half of the work output of the heat engine is used to driv

e a Carnot heat pump that removes heat from the cold surroundings at 2°C and transfers it to a house maintained at 22°C. If the house is losing heat at a rate of 62,000 kJ/h, determine the minimum rate of the heat supply to the heat engine required to keep the house at 22°C.
Engineering
1 answer:
statuscvo [17]3 years ago
3 0

Answer: Meh account banned

Explanation:

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In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by
Stells [14]

Answer:

\dfrac{\bar{h}}{h}=\dfrac{5}{4}

Explanation:

Given that

Nu_x=0.035Re_x^{0.8} Pr^{1/3}

We know that

Rex=ρvx/μ

So

Nu_x=0.035Re_x^{0.8} Pr^{1/3}

Nu_x=0.035\times\left(\dfrac{\rho vx}{\mu}\right)^{0.8}Pr^{1/3}

All other quantities are constant only x is a variable in the above equation .so lets take all other quantities as a constant C

Nu_x=C.x^{0.8}=C.x^{4/5}

We also know that

Nux=hx/K

C.x^{4/5}=\dfrac{hx}{k}

m is the constant

h=mx^{-1/5}

This is local heat transfer coefficient

The average value of h given as

\bar{h}=\dfrac{\int_{0}^{L}hdx}{L}

\bar{h}=\dfrac{5m}{4}\times\dfrac{L^{4/5}}{L}

\bar{h}=\dfrac{5m}{4}L^{-1/5}             ---------1

The value of local heat transfer coefficient at x=L

h=mx^{-1/5}

h=mL^{-1/5}            -----------2

From 1 and 2 we can say that

\dfrac{\bar{h}}{h}=\dfrac{5}{4}

8 0
3 years ago
A plant has ten machines and currently operates two 8-hr shift per day, 5 days per week, 50 weeks per year. the ten machines pro
Zarrin [17]

Answer:

  83.6%

Explanation:

<h3>(a)</h3>

On its current schedule, the plant can theoretically produce ...

  (30 pc/h/machine)(10 machine)(8 h/shift)(2 shift/day)(5 day/wk)(50 wk/yr)

  = (30)(10)(8)(2)(5)(50) pc/yr = 1,200,000 pc/yr

__

<h3>(b)</h3>

On the proposed schedule, this production becomes ...

  (30 pc/mach)(10 mach)(8 h/sh)(3 sh/da)(6 da/wk)(51 wk/yr)

  = (30)(10)(8)(3)(6)(51) pc/yr = 2,203,200 pc/yr

The increase in capacity is ...

  (2,203,200/1,200,000 -1) × 100% = 83.6% . . . capacity increase

_____

<em>Additional comment</em>

The number of parts per shift did not change. Only the number of shifts per year changed. It went up by a factor of (3/2)(6/5)(51/50) = 1.836. Hence the 83.6% increase in capacity.

We have to assume that maintenance and repair are done as effectvely as before in the reduced down time that each machine has.

3 0
2 years ago
A hollow pipe is submerged in a stream of water so that the length of the pipe is parallel to the velocity of the water. If the
Arlecino [84]

Answer:

increases by a factor of 6.

Explanation:

Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:

Initial flow rate = area * velocity = A * V = AV m³/s

The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:

Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate

Hence, the volume flow rate of the water passing through it increases by a factor of 6.

8 0
3 years ago
A 150-lbm astronaut took his bathroom scale (aspring scale) and a beam scale (compares masses) to themoon where the local gravit
Nonamiya [84]

Answer:

a) W = 25.5 lbf

b) W = 150 lbf

Explanation:

Given data:

Mass of astronaut = 150 lbm

local gravity = 5.48 ft/s^2

a) weight on spring scale

it can be calculated by measuring force against local gravitational force which is equal to weight of body

W = mg

W = (150 \times 5.48)\times \frac{1 lbm}{32.32 lbm. ft/s^2} = 25.5 lbf

b) As we know that beam scale calculated mass only therefore no change in mass due to variation in gravity

thus W= 150 lbf

7 0
3 years ago
3. It's
jeyben [28]

Answer:

right answer is option no d

6 0
3 years ago
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