Answer:
T(water)=50.32℃
T(air)=3052.6℃
Explanation:
Hello!
To solve this problem we must use the equation that defines the transfer of heat by convection, which consists of the transport of heat through fluids in this case water and air.
The equation is as follows!

Q = heat
h = heat transfer coefficient
Ts = surface temperature
T = fluid temperature
a = heat transfer area
The surface area of a cylinder is calculated as follows

Where
D=diameter=20mm=0.02m
L=leght=200mm)0.2m
solving

For water
Q=2Kw=2000W
h=5000W/m2K
a=0.01319m^2
Tα=20C

solving for ts


for air
Q=2Kw=2000W
h=50W/m2K
a=0.01319m^2
Tα=20C

Answer:
hi-he = 0
pi-pe = positive
ui-ue = negative
ti-te = negative
Explanation:
we know that fir the sub cool liquid water is
dQ = Tds = du + pdv ............1
and Tds = dh - v dP .............2
so now for process of throhling is irreversible when v is constant
then heat transfer is = 0 in irreversible process
so ds > 0
so here by equation 1 we can say
ds > 0
dv = 0 as v is constant
so that Tds = du .................3
and du > 0
ue - ui > 0
and
now by the equation 2 throttling process
here enthalpy is constant
so dh = 0
and Tds = -vdP
so ds > 0
so that -vdP > 0
as here v is constant
so -dP =P1- P2
so P1-P2 > 0
so pressure is decrease here
Answer:
This question is comprising many parts (a to r). That is impossible to answer in one sheet. Following are attached images having answers to most of the parts.
I hope it will help you a lot.
Explanation:
The drawbar or other connections must be strong enough to pull all the weight of the vehicle being towed. The drawbar or other connection may not exceed 15 feet from one vehicle to the other.